----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



                        1    1    1
                       /    /    /
                      |    |    |         1
                      |    |    |   ------------- dx dy dz
                      |    |    |   x y z - z + 1
                     /    /    /
                       0    0    0

                                  divided by

                              1    1    1
                             /    /    /
                            |    |    |
                            |    |    |   1 dx dy dz
                            |    |    |
                           /    /    /
                             0    0    0



that equals, up to, 200,  digits , 2.4041138063191885707994763230228999815299\
    725846809977635845431106836764115726261803729117472186705162923983155905\
    214388369839919973465664275527936744158003229078835658987201334383831510\
    444849884879231



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                               that is EXACTLY, 3





                       yielding an irrationality measure

                                            1/2
                                ln(17 + 12 2   ) + 3
                            1 + --------------------
                                            1/2
                                ln(17 + 12 2   ) - 3



                          that to 10 decimals, equals

                                  13.41782024

           Comment: Note that this constant appears to be , 2 Zeta(3)

                                   Prove it!

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



            3                       2
     (1 + n)  X(n)   (2 n + 3) (17 n  + 51 n + 39) X(1 + n)
     ------------- - -------------------------------------- + X(2 + n) = 0
              3                            3
       (2 + n)                      (2 + n)



                       Subject to the initial conditions



                              A(0) = 0, A(1) = 12



                               B(0) = 1, B(1) = 5



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



               1    1    1 /x (1 - x) y (1 - y) z (1 - z)\n
              /    /    /  |-----------------------------|
             |    |    |   \        x y z - z + 1        /
             |    |    |   -------------------------------- dx dy dz
             |    |    |            x y z - z + 1
            /    /    /
              0    0    0

                                   divided by

                              1    1    1
                             /    /    /
                            |    |    |
                            |    |    |   1 dx dy dz
                            |    |    |
                           /    /    /
                             0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



                                          3
                                    LCM(n)



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

                                       1

                           whose log divided by n is



                                       0.



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

           log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 3



                          that equals in decimals, 3.



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[3.009066494, 3.005797947, 3.005172539, 3.005235113, 3.001565064, 3.006218447,

    3.003330641, 3.004205579, 2.998050015, 3.000371705, 2.996947934,

    2.998127299, 2.998647268, 2.993766601, 2.993979398, 2.991683140,

    2.992996713, 2.991260036, 2.989631928, 2.997072166, 2.999729869]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - 3
                      where , delta = --------------------
                                                  1/2
                                      ln(17 + 12 2   ) + 3



                      that in decimals is, , 0.08052943118



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------



           For the following we have a proposed INTEGERATING factor

For GBCZ3ck(0,1/2,1/2,0,1/2) ( that equals (conjecturally) -198 + 288 ln(2))\
    :



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



               1    1    1
              /    /    /   1/2        1/2  1/2        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |            x y z - z + 1
            /    /    /
              0    0    0

                                  divided by

               1    1    1
              /    /    /
             |    |    |    1/2        1/2  1/2        1/2
             |    |    |   x    (1 - x)    y    (1 - z)    dx dy dz
             |    |    |
            /    /    /
              0    0    0



that equals, up to, 200,  digits , 1.6263880012642491121628509799548516057440\
    386957535131867558427340973631272720780944886381749685820120964265339243\
    574913014298278983416004254091764709018770360495403985709733348084271236\
    808019596835248



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                               that is EXACTLY, 2





                       yielding an irrationality measure

                                            1/2
                                ln(17 + 12 2   ) + 2
                            1 + --------------------
                                            1/2
                                ln(17 + 12 2   ) - 2



                          that to 10 decimals, equals

                                  4.622100833

       Comment: Note that this constant appears to be , -198 + 288 ln(2)

                                   Prove it!

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



       2               2
(1 + n)  (3 + 2 n) (6 n  + 30 n + 37) X(n)
------------------------------------------
      2                               2
  (6 n  + 18 n + 13) (5 + 2 n) (3 + n)

                      4        3         2
       8 (2 + n) (51 n  + 408 n  + 1189 n  + 1492 n + 678) X(1 + n)
     - ------------------------------------------------------------ + X(2 + n)
                      2                               2
                  (6 n  + 18 n + 13) (5 + 2 n) (3 + n)

     = 0



                       Subject to the initial conditions



                             A(0) = 0, A(1) = 39/2



                              B(0) = 1, B(1) = 12



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1
  /    /
 |    |
 |    |
 |    |
/    /
  0    0

       1  1/2        1/2  1/2        1/2 /x (1 - x) y (1 - y) z (1 - z)\n
      /  x    (1 - x)    y    (1 - z)    |-----------------------------|
     |                                   \        x y z - z + 1        /
     |   ---------------------------------------------------------------- dx dy
     |                            x y z - z + 1
    /
      0

    dz

                                   divided by

               1    1    1
              /    /    /
             |    |    |    1/2        1/2  1/2        1/2
             |    |    |   x    (1 - x)    y    (1 - z)    dx dy dz
             |    |    |
            /    /    /
              0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



                                    LCM(2 n)



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

                                     (3001)

                           whose log divided by n is



                                 0.002668900281



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

           log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 2



                          that equals in decimals, 2.



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[1.998872285, 1.998492030, 1.994209502, 1.996565009, 1.997004164, 1.994673635,

    1.991817146, 1.994787047, 1.992621001, 1.992422203, 1.994095066,

    1.992471560, 1.994270184, 1.993900470, 1.996523476, 1.995066610,

    1.989257184, 1.993464876, 1.987459919, 1.988118448, 1.991516255]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - 2
                      where , delta = --------------------
                                                  1/2
                                      ln(17 + 12 2   ) + 2



                      that in decimals is, , 0.2760828719



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------



     For GBCZ3ck(0,0,0,1/2,1/2) ( conjecturally -(2-4*ln(2))/(3-4*ln(2))):



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



                      1    1    1
                     /    /    /          1/2
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  3/2
                   /    /    /    (x y z - z + 1)
                     0    0    0

                                  divided by

                      1    1    1
                     /    /    /          1/2
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  1/2
                   /    /    /    (x y z - z + 1)
                     0    0    0



that equals, up to, 200,  digits , 3.3973192967782128351916747382613431926455\
    194497161505816281132837465909916639803216810005496908243791527496440794\
    151909530851386066530563437782403023560470414393410212267742310708911815\
    473562940410120



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                               that is EXACTLY, 2





                       yielding an irrationality measure

                                            1/2
                                ln(17 + 12 2   ) + 2
                            1 + --------------------
                                            1/2
                                ln(17 + 12 2   ) - 2



                          that to 10 decimals, equals

                                  4.622100833

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



          3                        2
16 (1 + n)  (1 + 2 n) (2 + n) (12 n  + 48 n + 47) X(n)
------------------------------------------------------
           2                       3          2
      (12 n  + 24 n + 11) (5 + 2 n)  (3 + 2 n)

                       4         3          2
       4 (2 + n) (816 n  + 4896 n  + 10456 n  + 9336 n + 2903) X(1 + n)
     - ----------------------------------------------------------------
                                 3      2
                        (5 + 2 n)  (12 n  + 24 n + 11)

     + X(2 + n) = 0



                       Subject to the initial conditions



                                               176
                              A(0) = 0, A(1) = ---
                                               27



                                               52
                              B(0) = 1, B(1) = --
                                               27



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



            1    1    1  1/2 /x (1 - x) y (1 - y) z (1 - z)\n
           /    /    /  y    |-----------------------------|
          |    |    |        \        x y z - z + 1        /
          |    |    |   ------------------------------------- dx dy dz
          |    |    |                           3/2
          |    |    |            (x y z - z + 1)
         /    /    /
           0    0    0

                                   divided by

                      1    1    1
                     /    /    /          1/2
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  1/2
                   /    /    /    (x y z - z + 1)
                     0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



                                                                   3
            PP(0, 1/2, 0, 1, {0}, 1, n) PP(1/2, 1, 0, 2, {0}, 1, n)
            --------------------------------------------------------
                                      (4 n)
                                     2



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

    22     8     3      3      4      6      3      3            3      3
 (2)    (5)   (7)   (11)   (13)   (17)   (23)   (29)   (37)  (41)   (43)   (47)

               3      3      3      3      3       2
     (53)  (59)   (61)   (67)   (71)   (73)   (353)

                           whose log divided by n is



                                 0.06545031499



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

           log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 2



                          that equals in decimals, 2.



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[1.998872410, 1.999653417, 1.999666014, 1.999018761, 1.996441973, 1.998429986,

    1.998128753, 1.997459923, 1.993207000, 1.994294212, 1.994854333,

    1.993252802, 1.996828220, 1.994685178, 1.998032543, 1.992920214,

    1.994178076, 1.994933133, 1.992465228, 1.989997909, 1.989143832]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - 2
                      where , delta = --------------------
                                                  1/2
                                      ln(17 + 12 2   ) + 2



                      that in decimals is, , 0.2760828719



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------



    For GBCZ3ck(0,1/4,0,3/4,1/2), conjecturally  -(-1+ln(2))/(-2+3*ln(2)) :



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



                      1    1    1
                     /    /    /       1/4  1/2
                    |    |    |       x    y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  7/4
                   /    /    /    (x y z - z + 1)
                     0    0    0

                                  divided by

                      1    1    1
                     /    /    /       1/4  1/2
                    |    |    |       x    y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  3/4
                   /    /    /    (x y z - z + 1)
                     0    0    0



that equals, up to, 200,  digits , 3.8626241756073689289497644204227424381588\
    308654429276591496976609402560095129935633774149052654227792389605961993\
    755616852079521683087863008620793806533450509185262667005215271512956753\
    000709869739368



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                               that is EXACTLY, 2





                       yielding an irrationality measure

                                            1/2
                                ln(17 + 12 2   ) + 2
                            1 + --------------------
                                            1/2
                                ln(17 + 12 2   ) - 2



                          that to 10 decimals, equals

                                  4.622100833

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                        2                               3
       256 (n + 2) (24 n  + 96 n + 95) (1 + 2 n) (1 + n)  X(n)
----------------------------------------------------------------------
               2
(4 n + 5) (24 n  + 48 n + 23) (5 + 2 n) (7 + 4 n) (11 + 4 n) (9 + 4 n)

                         4         3          2
       16 (n + 2) (1632 n  + 9792 n  + 21124 n  + 19308 n + 6269) X(1 + n)
     - -------------------------------------------------------------------
                    2
               (24 n  + 48 n + 23) (5 + 2 n) (11 + 4 n) (9 + 4 n)

     + X(n + 2) = 0



                       Subject to the initial conditions



                                               368
                              A(0) = 0, A(1) = ---
                                               105



                                               32
                              B(0) = 1, B(1) = --
                                               35



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



          1    1    1  1/4  1/2 /x (1 - x) y (1 - y) z (1 - z)\n
         /    /    /  x    y    |-----------------------------|
        |    |    |             \        x y z - z + 1        /
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              7/4
        |    |    |               (x y z - z + 1)
       /    /    /
         0    0    0

                                   divided by

                      1    1    1
                     /    /    /       1/4  1/2
                    |    |    |       x    y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  3/4
                   /    /    /    (x y z - z + 1)
                     0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



                                                                         2
     LCM(4 n) PP(1/4, 3/4, 0, 2, {0}, 1, n) PP(3/4, 1, 0, 4/3, {0}, 1, n)
     ---------------------------------------------------------------------
                                     (6 n)
                                    2



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

    12     4      3      3      2                        2
 (2)    (5)   (11)   (13)   (17)   (19)  (23)  (29)  (41)   (43)  (47)  (59)

     (61)  (67)  (71)  (73)  (1091)

                           whose log divided by n is



                                 0.02924576416



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

           log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 2



                          that equals in decimals, 2.



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[1.996289088, 1.996966583, 1.996792004, 1.999285585, 1.995468666, 1.997497507,

    1.993310671, 1.994571616, 1.993725915, 1.991815973, 1.991235664,

    1.986494096, 1.995360008, 1.996078408, 1.994456149, 1.993743457,

    1.994632543, 1.995376050, 1.991113420, 1.992701368, 1.986589497]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - 2
                      where , delta = --------------------
                                                  1/2
                                      ln(17 + 12 2   ) + 2



                      that in decimals is, , 0.2760828719



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------





For GBCZ3ck(0,0,0,1/3,2/3), not yet identified, possibly a first irrationali\
    ty proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



                      1    1    1
                     /    /    /          2/3
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  4/3
                   /    /    /    (x y z - z + 1)
                     0    0    0

                                  divided by

                      1    1    1
                     /    /    /          2/3
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  1/3
                   /    /    /    (x y z - z + 1)
                     0    0    0



that equals, up to, 200,  digits , 2.6471192443562322644146794393489034730059\
    539503857696418205328929898996405862809289122323099978530616551009746774\
    478653424523212284445255820640901507506226907188877958140412756981543905\
    060093192904408



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                              that is EXACTLY, 9/4



                       that in decimals is , 2.250000000



                       yielding an irrationality measure

                                           1/2
                               ln(17 + 12 2   ) + 9/4
                           1 + ----------------------
                                           1/2
                               ln(17 + 12 2   ) - 9/4



                          that to 10 decimals, equals

                                  5.528043858

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                  2                                 3
  81 (2 + n) (36 n  + 147 n + 148) (2 + 3 n) (1 + n)  X(n)
------------------------------------------------------------
               2                       2
(3 n + 4) (36 n  + 75 n + 37) (8 + 3 n)  (7 + 3 n) (5 + 3 n)

                         4          3          2
       18 (2 + n) (1836 n  + 11475 n  + 25776 n  + 24481 n + 8214) X(1 + n)
     - --------------------------------------------------------------------
                          2                       2
                     (36 n  + 75 n + 37) (8 + 3 n)  (7 + 3 n)

     + X(2 + n) = 0



                       Subject to the initial conditions



                                               333
                              A(0) = 0, A(1) = ---
                                               25



                                               126
                              B(0) = 1, B(1) = ---
                                               25



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



            1    1    1  2/3 /x (1 - x) y (1 - y) z (1 - z)\n
           /    /    /  y    |-----------------------------|
          |    |    |        \        x y z - z + 1        /
          |    |    |   ------------------------------------- dx dy dz
          |    |    |                           4/3
          |    |    |            (x y z - z + 1)
         /    /    /
           0    0    0

                                   divided by

                      1    1    1
                     /    /    /          2/3
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  1/3
                   /    /    /    (x y z - z + 1)
                     0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



LCM(3 n) PP(1/3, 1, 0, 3/2, {0}, 1, n) PP(1/3, 1, 3/2, 3, {2}, 3, n)

                                   (3 n)
    PP(2/3, 1, 0, 3/2, {0}, 1, n)/3



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

    11     3     4     3      2      2      2      2      2                  2
 (2)    (3)   (5)   (7)   (11)   (13)   (17)   (23)   (41)   (43)  (47)  (59)

         2      2
     (61)   (67)   (643)  (9001)

                           whose log divided by n is



                                 0.03348499256



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

          log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 9/4



                      that equals in decimals, 2.250000000



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[2.236648544, 2.235591137, 2.236655726, 2.237337887, 2.234692201, 2.233228596,

    2.235793695, 2.235623674, 2.240832024, 2.240309468, 2.244312771,

    2.240159891, 2.243329459, 2.242670030, 2.235932102, 2.242102929,

    2.241809637, 2.238026833, 2.239823390, 2.240118488, 2.240944768]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                 1/2
                                     ln(17 + 12 2   ) - 9/4
                     where , delta = ----------------------
                                                 1/2
                                     ln(17 + 12 2   ) + 9/4



                      that in decimals is, , 0.2208459174



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------





For GBCZ3ck(0,0,0,2/3,1/3), not yet identified, possibly a first irrationali\
    ty proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



                      1    1    1
                     /    /    /          1/3
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  5/3
                   /    /    /    (x y z - z + 1)
                     0    0    0

                                  divided by

                      1    1    1
                     /    /    /          1/3
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  2/3
                   /    /    /    (x y z - z + 1)
                     0    0    0



that equals, up to, 200,  digits , 5.2085219274937867218094320899398631882603\
    010886573222116423882881312623626645070011258704664563455604476863297503\
    353527066951041765512585829489979171500401747038084510964454091910989356\
    920150603395871



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                              that is EXACTLY, 9/4



                       that in decimals is , 2.250000000



                       yielding an irrationality measure

                                           1/2
                               ln(17 + 12 2   ) + 9/4
                           1 + ----------------------
                                           1/2
                               ln(17 + 12 2   ) - 9/4



                          that to 10 decimals, equals

                                  5.528043858

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                  2                                 3
  81 (2 + n) (36 n  + 141 n + 136) (1 + 3 n) (1 + n)  X(n)
------------------------------------------------------------
     2                       2
(36 n  + 69 n + 31) (7 + 3 n)  (5 + 3 n) (8 + 3 n) (4 + 3 n)

                         4          3          2
       18 (2 + n) (1836 n  + 10557 n  + 21645 n  + 18638 n + 5646) X(1 + n)
     - --------------------------------------------------------------------
                          2                       2
                     (36 n  + 69 n + 31) (7 + 3 n)  (8 + 3 n)

     + X(2 + n) = 0



                       Subject to the initial conditions



                                               279
                              A(0) = 0, A(1) = ---
                                               80



                                               27
                              B(0) = 1, B(1) = --
                                               40



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



            1    1    1  1/3 /x (1 - x) y (1 - y) z (1 - z)\n
           /    /    /  y    |-----------------------------|
          |    |    |        \        x y z - z + 1        /
          |    |    |   ------------------------------------- dx dy dz
          |    |    |                           5/3
          |    |    |            (x y z - z + 1)
         /    /    /
           0    0    0

                                   divided by

                      1    1    1
                     /    /    /          1/3
                    |    |    |          y
                    |    |    |   ------------------ dx dy dz
                    |    |    |                  2/3
                   /    /    /    (x y z - z + 1)
                     0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



                                      2
LCM(3 n) PP(2/3, 1, 0, 3/2, {0}, 1, n)  PP(1/3, 2/3, 0, 3/4, {0}, 1, n)

                                      /  (3 n)
    PP(1/3, 2/3, 3/2, 3, {1}, 3, n)  /  3
                                    /



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

    10     3     4     2      2      3                              2
 (2)    (3)   (5)   (7)   (11)   (13)   (17)  (19)  (23)  (29)  (41)   (43)

               2      2      2                                      2
     (47)  (59)   (61)   (67)   (71)  (73)  (79)  (83)  (89)  (9001)

                           whose log divided by n is



                                 0.04173096913



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

          log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 9/4



                      that equals in decimals, 2.250000000



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[2.258125552, 2.249537029, 2.254708401, 2.254306246, 2.252416556, 2.252311032,

    2.253250019, 2.256194024, 2.252105250, 2.253154852, 2.251738044,

    2.255667252, 2.249334800, 2.250371571, 2.249958175, 2.251180630,

    2.251629115, 2.252974088, 2.252887876, 2.251665039, 2.252010070]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                 1/2
                                     ln(17 + 12 2   ) - 9/4
                     where , delta = ----------------------
                                                 1/2
                                     ln(17 + 12 2   ) + 9/4



                      that in decimals is, , 0.2208459174



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------





For GBCZ3ck(0, 1/3, 2/3, 1/3, 2/3), not yet identified, possibly a first irr\
    ationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



               1    1    1
              /    /    /   1/3        2/3  2/3        2/3
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        4/3
            /    /    /          (x y z - z + 1)
              0    0    0

                                  divided by

               1    1    1
              /    /    /   1/3        2/3  2/3        2/3
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        1/3
            /    /    /          (x y z - z + 1)
              0    0    0



that equals, up to, 200,  digits , 1.7169588321632764037319465007440094092152\
    844931642745279714578452186241038707666117030622554001302564589839452156\
    874327993278109521620466345653629332856591995756432625798241107664162877\
    087654991729732



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

                               that is EXACTLY, 3





                       yielding an irrationality measure

                                            1/2
                                ln(17 + 12 2   ) + 3
                            1 + --------------------
                                            1/2
                                ln(17 + 12 2   ) - 3



                          that to 10 decimals, equals

                                  13.41782024

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                             2        2
9 (2 n + 5) (2 + n) (5 + 3 n)  (1 + n)  X(n)
--------------------------------------------
                                3
   (2 n + 3) (4 + 3 n) (7 + 3 n)  (3 + n)

                2                       3
       12 (153 n  + 612 n + 574) (2 + n)  X(1 + n)
     - ------------------------------------------- + X(2 + n) = 0
                                 3
              (2 n + 3) (7 + 3 n)  (3 + n)



                       Subject to the initial conditions



                             A(0) = 0, A(1) = 45/4



                                               105
                              B(0) = 1, B(1) = ---
                                               16



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1
  /    /
 |    |
 |    |
 |    |
 |    |
/    /
  0    0

       1  1/3        2/3  2/3        2/3 /x (1 - x) y (1 - y) z (1 - z)\n
      /  x    (1 - x)    y    (1 - z)    |-----------------------------|
     |                                   \        x y z - z + 1        /
     |   ---------------------------------------------------------------- dx dy
     |                                         4/3
     |                          (x y z - z + 1)
    /
      0

    dz

                                   divided by

               1    1    1
              /    /    /   1/3        2/3  2/3        2/3
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        1/3
            /    /    /          (x y z - z + 1)
              0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



                                  Let E(n) be



                             3                              2
PP(1/3, 2/3, 0, 3, {1}, 3, n)  PP(2/3, 1, 0, 3/2, {0}, 1, n)

                                     /3 n\
                                     |---|
                                  /  \ 2 /
    PP(0, 1/3, 0, 1, {0}, 1, n)  /  3
                                /



where LCM(1..m) is the least-common-multiple of the first m integers, and fo\
    r 0<alpha<beta<1, a,b, an integer M and a subset C of {0,...M-1}

 PP(alpha,beta,a,b,C,M,n) is the product of all r primes between a*n and b*n\
    , such that of you take them mod M it belongs to C

            and the fractional part of n/p is between alpha and beta



Then there exists a sequence G(n) of integers that is asymptotically negligi\
    ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity

                                   such that

           A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers



                    For example, G(n) for n=, 3000, equals

    6               3      2      5      2      4            2
 (2)   (3)  (5)  (7)   (11)   (13)   (17)   (19)   (23)  (29)   (31)  (37)

               3      2            2      2      2      3      3      2      2
     (41)  (43)   (47)   (59)  (61)   (67)   (71)   (73)   (79)   (83)   (89)

           3
     (9001)   (3001)

                           whose log divided by n is



                                 0.06524417399



                                Indeed peanuts.



Using standard asymptotics derived from the Prime Number Theorem it is readi\
    ly seen that the limit of

           log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 3



                          that equals in decimals, 3.



Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980,

    to , 3001, are



[2.975884195, 2.974027945, 2.971904962, 2.975628937, 2.976200312, 2.973526335,

    2.975011194, 2.970661922, 2.975303482, 2.973738878, 2.976306081,

    2.966623074, 2.975530896, 2.974664264, 2.969223665, 2.968069706,

    2.970779304, 2.965391706, 2.964454231, 2.964292422, 2.974442783]



                     Multiplying F(n) by E(n)*G(n) we get



                        E(n) F(n) G(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - 3
                      where , delta = --------------------
                                                  1/2
                                      ln(17 + 12 2   ) + 3



                      that in decimals is, , 0.08052943118



                      As you can see, delta is  positive



               We leave it to the reader to fill-in the details.





                  --------------------------------------------





For GBCZ3ck(0, 1/5, 0, 3/5, 2/5), not yet identified, possibly a first irrat\
    ionality proof  of that constant

TheoremZ3(0,1/5,0,3/5,2/5,K,[[5, 5/2], [5, 1], [[4/5, 1, 0, 5/4, {0}, 1]$2, \
    [2/5, 4/5, 0, 5/2, {0}, 1]]]);



                  --------------------------------------------





                  --------------------------------------------





For the following we do NOT (yet) have a proposed INTEGERating factor, hence\
     the last argument is 0.



                       ---------------------------------



For GBCZ3ck(0, 1/2, 1/2, 1/3, 1/6), not yet identified, possibly a first irr\
    ationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



               1    1    1
              /    /    /   1/2        1/2  1/6        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        4/3
            /    /    /          (x y z - z + 1)
              0    0    0

                                  divided by

               1    1    1
              /    /    /   1/2        1/2  1/6        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        1/3
            /    /    /          (x y z - z + 1)
              0    0    0



that equals, up to, 200,  digits , 1.8898985334508068501436739741573136333283\
    223724632627780021711026729282558082529351932369559874264425705901923252\
    864406348326733117975795915185609964050744756296311730361049477303631253\
    110366298796739



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

that is approximately , 2.096647475,

     yielding an irrationality measure that is approximately , 4.934740614



                 We hope that the reader can find  nu exactly.

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                        2                       2        2
      81 (2 n + 5) (18 n  + 84 n + 97) (3 + 2 n)  (1 + n)  X(n)
---------------------------------------------------------------------- - 9
     2
(18 n  + 48 n + 31) (11 + 6 n) (17 + 6 n) (8 + 3 n) (13 + 6 n) (3 + n)

                             4          3          2
    (2 n + 5) (n + 2) (3672 n  + 26928 n  + 71838 n  + 82390 n + 34165)

               /       2
    X(1 + n)  /  ((18 n  + 48 n + 31) (17 + 6 n) (8 + 3 n) (13 + 6 n) (3 + n))
             /

     + X(n + 2) = 0



                       Subject to the initial conditions



                                               279
                              A(0) = 0, A(1) = ---
                                               22



                                              1035
                             B(0) = 1, B(1) = ----
                                              154



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1
  /    /
 |    |
 |    |
 |    |
 |    |
/    /
  0    0

       1  1/2        1/2  1/6        1/2 /x (1 - x) y (1 - y) z (1 - z)\n
      /  x    (1 - x)    y    (1 - z)    |-----------------------------|
     |                                   \        x y z - z + 1        /
     |   ---------------------------------------------------------------- dx dy
     |                                         4/3
     |                          (x y z - z + 1)
    /
      0

    dz

                                   divided by

               1    1    1
              /    /    /   1/2        1/2  1/6        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        1/3
            /    /    /          (x y z - z + 1)
              0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



Lemma TWO: There exists a sequence of rational numbers, whose prime factoriz\
    ations consists of small primes, that hopefully

can be described (and proved) explicity, that we leave to the expert reader \
    such that

               A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers



Furthermore there exists a contant, nu, that hopefully the learned reader ca\

    n determine such that E(n) is OMEGA of , exp(nu n)



         The empircal values of nu for E(n) from, 2980, to , 3001, are



[2.081514887, 2.080739984, 2.084174675, 2.085208522, 2.083483904, 2.089968288,

    2.086373880, 2.091866290, 2.090481063, 2.092514660, 2.094317631,

    2.096647475, 2.094613209, 2.095063963, 2.088679239, 2.089341681,

    2.092283925, 2.091056826, 2.090934340, 2.090645505, 2.088435724]



                        Multiplying F(n) by E(n) we get



                          E(n) F(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - nu
                      where , delta = ---------------------
                                                  1/2
                                      ln(17 + 12 2   ) + nu



Using the above values of nu for E(n) from, 2980, to , 3001,

    the estimated deltas are



[0.2575311366, 0.2577049546, 0.2569348864, 0.2567032793, 0.2570896836,

    0.2556380753, 0.2564423129, 0.2552138184, 0.2555234274, 0.2550689552,

    0.2546662989, 0.2541463588, 0.2546003123, 0.2544996966, 0.2559263775,

    0.2557782025, 0.2551205038, 0.2553947213, 0.2554220996, 0.2554866653,

    0.2559808558]



                     As you can see, they are all positive



               We leave it to the reader to fill-in the details.







                       ---------------------------------



For GBCZ3ck(0, 1/2, 1/2, 1/6, 1/3), not yet identified, possibly a first irr\
    ationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



               1    1    1
              /    /    /   1/2        1/2  1/3        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        7/6
            /    /    /          (x y z - z + 1)
              0    0    0

                                  divided by

               1    1    1
              /    /    /   1/2        1/2  1/3        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        1/6
            /    /    /          (x y z - z + 1)
              0    0    0



that equals, up to, 200,  digits , 1.7329598953601183808997205232912873697230\
    554865727420659161705234401481508701678348764052017260739255480722229761\
    954155104621252903038960562043090227507429720104710550314622726109821970\
    185114206316428



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

that is approximately , 2.110035384,

     yielding an irrationality measure that is approximately , 4.981415128



                 We hope that the reader can find  nu exactly.

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                2          2
      81 (1 + n)  (3 + 2 n)  (2 n + 5) (12 n + 29) X(n)
-------------------------------------------------------------
(12 n + 17) (6 n + 7) (13 + 6 n) (17 + 6 n) (7 + 3 n) (3 + n)

                                              2
       18 (2 n + 5) (n + 2) (12 n + 23) (102 n  + 391 n + 349) X(1 + n)
     - ----------------------------------------------------------------
             (12 n + 17) (13 + 6 n) (17 + 6 n) (7 + 3 n) (3 + n)

     + X(n + 2) = 0



                       Subject to the initial conditions



                                               153
                              A(0) = 0, A(1) = ---
                                               10



                                               495
                              B(0) = 1, B(1) = ---
                                               56



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1
  /    /
 |    |
 |    |
 |    |
 |    |
/    /
  0    0

       1  1/2        1/2  1/3        1/2 /x (1 - x) y (1 - y) z (1 - z)\n
      /  x    (1 - x)    y    (1 - z)    |-----------------------------|
     |                                   \        x y z - z + 1        /
     |   ---------------------------------------------------------------- dx dy
     |                                         7/6
     |                          (x y z - z + 1)
    /
      0

    dz

                                   divided by

               1    1    1
              /    /    /   1/2        1/2  1/3        1/2
             |    |    |   x    (1 - x)    y    (1 - z)
             |    |    |   ------------------------------- dx dy dz
             |    |    |                        1/6
            /    /    /          (x y z - z + 1)
              0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



Lemma TWO: There exists a sequence of rational numbers, whose prime factoriz\
    ations consists of small primes, that hopefully

can be described (and proved) explicity, that we leave to the expert reader \
    such that

               A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers



Furthermore there exists a contant, nu, that hopefully the learned reader ca\

    n determine such that E(n) is OMEGA of , exp(nu n)



         The empircal values of nu for E(n) from, 2980, to , 3001, are



[2.107523848, 2.104542106, 2.103575915, 2.104751941, 2.110035384, 2.109044802,

    2.109119661, 2.105400143, 2.104737251, 2.107254179, 2.108346543,

    2.104761692, 2.103512285, 2.102775019, 2.106472193, 2.103952204,

    2.101398002, 2.105522691, 2.099972733, 2.102646132, 2.105954844]



                        Multiplying F(n) by E(n) we get



                          E(n) F(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - nu
                      where , delta = ---------------------
                                                  1/2
                                      ln(17 + 12 2   ) + nu



Using the above values of nu for E(n) from, 2980, to , 3001,

    the estimated deltas are



[0.2517248215, 0.2523877516, 0.2526027153, 0.2523410761, 0.2511669765,

    0.2513869384, 0.2513703130, 0.2521969125, 0.2523443436, 0.2517847481,

    0.2515420354, 0.2523389071, 0.2526168746, 0.2527809592, 0.2519585556,

    0.2525189876, 0.2530875406, 0.2521696609, 0.2534050230, 0.2528096484,

    0.2520735703]



                     As you can see, they are all positive



               We leave it to the reader to fill-in the details.





                       ---------------------------------





For GBCZ3ck( 1/3, 0, 2/3, 1/2, 5/6), not yet identified, possibly a first ir\
    rationality proof  of that constant

                     TheoremZ3( 1/3, 0, 2/3, 1/2, 5/6,K,0);



                       ---------------------------------





For GBCZ3ck(  1/7, 0, 2/7, 3/7, 4/7), not yet identified, possibly a first i\
    rrationality proof  of that constant

                    TheoremZ3(  1/7, 0, 2/7, 3/7, 4/7,K,0);



                       ---------------------------------



For GBCZ3ck(  1/7, 0, 2/7, 5/7, 3/7) ,not yet identified, possibly a first i\
    rrationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



          1    1    1
         /    /    /         2/7  3/7        1/7  1/7        2/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                             12/7
       /    /    /               (x y z - z + 1)
         0    0    0

                                  divided by

          1    1    1
         /    /    /         2/7  3/7        1/7  1/7        2/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              5/7
       /    /    /                (x y z - z + 1)
         0    0    0



that equals, up to, 200,  digits , 3.5993953413268197448632982394831514536182\
    213567125450319658661740216445555035415742833854487065841057874766474599\
    491604667869585153678222892786433230639342738737353218135288625529218811\
    492264195781451



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

that is approximately , 3.371014220,

     yielding an irrationality measure that is approximately , 45.64333800



                 We hope that the reader can find  nu exactly.

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



                 2                                                2
49 (n + 2) (588 n  + 2667 n + 3019) (10 + 7 n) (4 + 7 n) (9 + 7 n)  (1 + n)

                     /                  2
    (8 + 7 n) X(n)  /  ((7 n + 6) (588 n  + 1491 n + 940) (18 + 7 n) (17 + 7 n)
                   /

                                                                        5
    (13 + 7 n) (19 + 7 n) (11 + 7 n) (12 + 7 n)) - 14 (n + 2) (3428628 n

                 4             3              2
     + 29510691 n  + 99744743 n  + 165467561 n  + 134797705 n + 43175200)

               /        2
    X(1 + n)  /  ((588 n  + 1491 n + 940) (18 + 7 n) (17 + 7 n) (13 + 7 n)
             /

    (19 + 7 n)) + X(n + 2) = 0



                       Subject to the initial conditions



                                               329
                              A(0) = 0, A(1) = ---
                                               66



                                              1379
                             B(0) = 1, B(1) = ----
                                              990



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1    1
  /    /    /
 |    |    |          2/7  3/7        1/7  1/7        2/7
 |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
 |    |    |
/    /    /
  0    0    0

    /x (1 - x) y (1 - y) z (1 - z)\n   /                12/7
    |-----------------------------|   /  (x y z - z + 1)     dx dy dz
    \        x y z - z + 1        /  /

                                   divided by

          1    1    1
         /    /    /         2/7  3/7        1/7  1/7        2/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              5/7
       /    /    /                (x y z - z + 1)
         0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



Lemma TWO: There exists a sequence of rational numbers, whose prime factoriz\
    ations consists of small primes, that hopefully

can be described (and proved) explicity, that we leave to the expert reader \
    such that

               A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers



Furthermore there exists a contant, nu, that hopefully the learned reader ca\

    n determine such that E(n) is OMEGA of , exp(nu n)



         The empircal values of nu for E(n) from, 2980, to , 3001, are



[3.364788642, 3.361473865, 3.360365874, 3.356346770, 3.352982900, 3.356679591,

    3.360384463, 3.361189883, 3.361151338, 3.364914246, 3.366327897,

    3.364084374, 3.368695277, 3.367402563, 3.371014220, 3.369223831,

    3.369153599, 3.370336284, 3.368925290, 3.370110328, 3.366929172]



                        Multiplying F(n) by E(n) we get



                          E(n) F(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - nu
                      where , delta = ---------------------
                                                  1/2
                                      ln(17 + 12 2   ) + nu



Using the above values of nu for E(n) from, 2980, to , 3001,

    the estimated deltas are



[0.02332352767, 0.02381606506, 0.02398080540, 0.02457882638, 0.02507988931,

    0.02452927788, 0.02397804108, 0.02385828354, 0.02386401414, 0.02330487370,

    0.02309497334, 0.02342813407, 0.02274365510, 0.02293546343, 0.02239975873,

    0.02266525084, 0.02267566817, 0.02250027187, 0.02270953412, 0.02253377728,

    0.02300572151]



                     As you can see, they are all positive



               We leave it to the reader to fill-in the details.





                       ---------------------------------





For GBCZ3ck(  1/7, 0, 3/7, 4/7, 5/7) , not yet identified, possibly a first \
    irrationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



          1    1    1
         /    /    /         3/7  5/7        1/7  1/7        3/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                             11/7
       /    /    /               (x y z - z + 1)
         0    0    0

                                  divided by

          1    1    1
         /    /    /         3/7  5/7        1/7  1/7        3/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              4/7
       /    /    /                (x y z - z + 1)
         0    0    0



that equals, up to, 200,  digits , 2.5236091052230484525995911611935256831550\
    105995318593608418229761467623331438115345593424006491383964641052000249\
    077346655606691670601910267684011893116853935029860990326629944935086622\
    314700618043815



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

that is approximately , 3.396422246,

     yielding an irrationality measure that is approximately , 54.62829370



                 We hope that the reader can find  nu exactly.

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



         2           2                 2                       /
(8 + 7 n)  (10 + 7 n)  (6 + 7 n) (147 n  + 672 n + 769) X(n)  /  ((7 n + 9)
                                                             /

          2
    (147 n  + 378 n + 244) (20 + 7 n) (19 + 7 n) (13 + 7 n) (16 + 7 n)) -

                         4           3            2
    (14 n + 25) (122451 n  + 874650 n  + 2282273 n  + 2572850 n + 1060356)

               /        2
    X(1 + n)  /  ((147 n  + 378 n + 244) (20 + 7 n) (19 + 7 n) (16 + 7 n))
             /

     + X(2 + n) = 0



                       Subject to the initial conditions



                                              1708
                             A(0) = 0, A(1) = ----
                                              195



                                              2036
                             B(0) = 1, B(1) = ----
                                              585



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1    1
  /    /    /
 |    |    |          3/7  5/7        1/7  1/7        3/7
 |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
 |    |    |
/    /    /
  0    0    0

    /x (1 - x) y (1 - y) z (1 - z)\n   /                11/7
    |-----------------------------|   /  (x y z - z + 1)     dx dy dz
    \        x y z - z + 1        /  /

                                   divided by

          1    1    1
         /    /    /         3/7  5/7        1/7  1/7        3/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              4/7
       /    /    /                (x y z - z + 1)
         0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



Lemma TWO: There exists a sequence of rational numbers, whose prime factoriz\
    ations consists of small primes, that hopefully

can be described (and proved) explicity, that we leave to the expert reader \
    such that

               A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers



Furthermore there exists a contant, nu, that hopefully the learned reader ca\

    n determine such that E(n) is OMEGA of , exp(nu n)



         The empircal values of nu for E(n) from, 2980, to , 3001, are



[3.396422246, 3.389726753, 3.385295348, 3.385044934, 3.391203425, 3.383931088,

    3.385181240, 3.384476480, 3.379758512, 3.377400489, 3.380043368,

    3.380383996, 3.384513517, 3.385575693, 3.390475368, 3.389080734,

    3.391124834, 3.389846874, 3.390441005, 3.384979090, 3.386869380]



                        Multiplying F(n) by E(n) we get



                          E(n) F(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - nu
                      where , delta = ---------------------
                                                  1/2
                                      ln(17 + 12 2   ) + nu



Using the above values of nu for E(n) from, 2980, to , 3001,

    the estimated deltas are



[0.01864687334, 0.01963315316, 0.02028697243, 0.02032394409, 0.01941546782,

    0.02048842719, 0.02030381924, 0.02040788181, 0.02110506870, 0.02145387732,

    0.02106294774, 0.02101258446, 0.02040241252, 0.02024558486, 0.01952278358,

    0.01972841605, 0.01942705106, 0.01961544191, 0.01952784925, 0.02033366589,

    0.02005464027]



                     As you can see, they are all positive



               We leave it to the reader to fill-in the details.





                       ---------------------------------





For GBCZ3ck(  1/7, 0, 4/7, 2/7, 5/7) , not yet identified, possibly a first \
    irrationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



          1    1    1
         /    /    /         4/7  5/7        1/7  1/7        4/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              9/7
       /    /    /                (x y z - z + 1)
         0    0    0

                                  divided by

          1    1    1
         /    /    /         4/7  5/7        1/7  1/7        4/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              2/7
       /    /    /                (x y z - z + 1)
         0    0    0



that equals, up to, 200,  digits , 1.9977352760675041918208007115520369755004\
    136874075820196753684235084097123356644076603835730011981627813679871637\
    850205967169713829685642732482955777481203898000356474282299106590914852\
    767495225917755



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

that is approximately , 3.368943986,

     yielding an irrationality measure that is approximately , 45.03974698



                 We hope that the reader can find  nu exactly.

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



            2                                           2
49 (8 + 7 n)  (1 + n) (11 + 7 n) (12 + 7 n) (2 + n) (7 n  + 34 n + 41) X(n)

       /                 2
      /  ((7 n + 10) (7 n  + 20 n + 14) (20 + 7 n) (19 + 7 n) (13 + 7 n)
     /

                                                         2
    (16 + 7 n) (17 + 7 n)) - 7 (14 n + 27) (2 + n) (119 n  + 459 n + 384)

         2                           /      2
    (49 n  + 189 n + 178) X(1 + n)  /  ((7 n  + 20 n + 14) (20 + 7 n)
                                   /

    (19 + 7 n) (16 + 7 n) (17 + 7 n)) + X(2 + n) = 0



                       Subject to the initial conditions



                                               539
                              A(0) = 0, A(1) = ---
                                               39



                                              1351
                             B(0) = 1, B(1) = ----
                                              195



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1    1
  /    /    /
 |    |    |          4/7  5/7        1/7  1/7        4/7
 |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
 |    |    |
/    /    /
  0    0    0

    /x (1 - x) y (1 - y) z (1 - z)\n   /                9/7
    |-----------------------------|   /  (x y z - z + 1)    dx dy dz
    \        x y z - z + 1        /  /

                                   divided by

          1    1    1
         /    /    /         4/7  5/7        1/7  1/7        4/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              2/7
       /    /    /                (x y z - z + 1)
         0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



Lemma TWO: There exists a sequence of rational numbers, whose prime factoriz\
    ations consists of small primes, that hopefully

can be described (and proved) explicity, that we leave to the expert reader \
    such that

               A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers



Furthermore there exists a contant, nu, that hopefully the learned reader ca\

    n determine such that E(n) is OMEGA of , exp(nu n)



         The empircal values of nu for E(n) from, 2980, to , 3001, are



[3.359283380, 3.359612293, 3.359984637, 3.362693891, 3.362901186, 3.367760507,

    3.368943986, 3.363504068, 3.367347157, 3.368233506, 3.353909809,

    3.358199987, 3.359050060, 3.361276412, 3.360063627, 3.362854697,

    3.354926077, 3.356266842, 3.350627864, 3.353359845, 3.354104996]



                        Multiplying F(n) by E(n) we get



                          E(n) F(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - nu
                      where , delta = ---------------------
                                                  1/2
                                      ln(17 + 12 2   ) + nu



Using the above values of nu for E(n) from, 2980, to , 3001,

    the estimated deltas are



[0.02414180596, 0.02409288100, 0.02403750144, 0.02363472823, 0.02360392361,

    0.02288234576, 0.02270676079, 0.02351434421, 0.02294368598, 0.02281216278,

    0.02494177331, 0.02430299093, 0.02417651454, 0.02384541924, 0.02402575385,

    0.02361083185, 0.02479038496, 0.02459072632, 0.02543097383, 0.02502371735,

    0.02491269381]



                     As you can see, they are all positive



               We leave it to the reader to fill-in the details.





                       ---------------------------------





For GBCZ3ck(  2/7, 0, 3/7, 4/7, 5/7) , not yet identified, possibly a first \
    irrationality proof  of that constant



              ----------------------------------------------------



               Sketch of an Irrationality Proof of the constant



          1    1    1
         /    /    /         3/7  5/7        2/7  2/7        3/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                             11/7
       /    /    /               (x y z - z + 1)
         0    0    0

                                  divided by

          1    1    1
         /    /    /         3/7  5/7        2/7  2/7        3/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              4/7
       /    /    /                (x y z - z + 1)
         0    0    0



that equals, up to, 200,  digits , 2.7254385646253748757272516281824182504031\
    273637689522552731207740575005781169771155240217346544670136738932924232\
    537866609194553124551251554008825769181653031499948435690512085871105020\
    687365533402650



                              By Shalosh B. Ekhad



              Theorem: The constant of the title , let's all it C

                                                              1/2
                                                  ln(17 + 12 2   ) + nu
is irrational, with an irrationality measure, 1 + ---------------------,
                                                              1/2
                                                  ln(17 + 12 2   ) - nu

    for a certain number nu

that is approximately , 3.344526404,

     yielding an irrationality measure that is approximately , 38.96263912



                 We hope that the reader can find  nu exactly.

                              We need two  lemmas



Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\
    fy the second-order recurrence



           2                                   2
  (9 + 7 n)  (1 + n) (10 + 7 n) (6 + 7 n) (28 n  + 125 n + 139) X(n)
---------------------------------------------------------------------- - 2
               2
(7 n + 8) (28 n  + 69 n + 42) (3 + n) (19 + 7 n) (13 + 7 n) (15 + 7 n)

    (n + 2)

             5            4            3            2
    (163268 n  + 1416933 n  + 4832478 n  + 8087135 n  + 6633948 n + 2132286)

               /       2
    X(1 + n)  /  ((28 n  + 69 n + 42) (3 + n) (19 + 7 n) (13 + 7 n) (15 + 7 n))
             /

     + X(n + 2) = 0



                       Subject to the initial conditions



                             A(0) = 0, A(1) = 21/2



                                               309
                              B(0) = 1, B(1) = ---
                                               80



            A(n)
      Then, ----, approximates the constant of the title, let's call it C
            B(n)



                                               /       1       \n
               with an error that is OMEGA of, |---------------|
                                               |          1/2 2|
                                               \(17 + 12 2   ) /



                   Proof, consider the Beukers-type integral



   1    1    1
  /    /    /
 |    |    |          3/7  5/7        2/7  2/7        3/7
 |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
 |    |    |
/    /    /
  0    0    0

    /x (1 - x) y (1 - y) z (1 - z)\n   /                11/7
    |-----------------------------|   /  (x y z - z + 1)     dx dy dz
    \        x y z - z + 1        /  /

                                   divided by

          1    1    1
         /    /    /         3/7  5/7        2/7  2/7        3/7
        |    |    |   (1 - x)    y    (1 - y)    z    (1 - z)
        |    |    |   ------------------------------------------ dx dy dz
        |    |    |                              4/7
       /    /    /                (x y z - z + 1)
         0    0    0



               Then , F(0) = B(0) C - A(0), F(1) = B(1) C - A(1)



and F(n) also satisfies the above recurrence, thanks to the amazing Mathemat\
    ica package HolonomicFunctions of Christoph Koutschan



                          Hence, F(n) = B(n) C - A(n)



                                                              1
By a simple bound of the integrand,  F(n) is OMEGA of, ---------------,
                                                                 1/2 n
                                                       (17 + 12 2   )

     and  by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of,

              1/2 n
    (17 + 12 2   )



                                                               1
Dividing by B(n) gives that A(n)/B(n)-C is OMEGA of , -------------------,
                                                                1/2 (2 n)
                                                      (17 + 12 2   )

    QED.



we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\
    ied by another sequence of rational numbers

            E(n) such that both A(n)E(n) and B(n)E(n) are integers



Lemma TWO: There exists a sequence of rational numbers, whose prime factoriz\
    ations consists of small primes, that hopefully

can be described (and proved) explicity, that we leave to the expert reader \
    such that

               A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers



Furthermore there exists a contant, nu, that hopefully the learned reader ca\

    n determine such that E(n) is OMEGA of , exp(nu n)



         The empircal values of nu for E(n) from, 2980, to , 3001, are



[3.338351605, 3.333131097, 3.338530792, 3.333915710, 3.338335989, 3.340711813,

    3.332827638, 3.330214069, 3.338554277, 3.336098008, 3.333411497,

    3.339843874, 3.339545663, 3.336874642, 3.337087954, 3.332966702,

    3.340572835, 3.343196710, 3.344526404, 3.339919794, 3.343777643]



                        Multiplying F(n) by E(n) we get



                          E(n) F(n) = B1(n) C - A1(n)



                             and this implies that



                       |      A1(n) |        CONSTANT
                       | -C + ----- | <= ----------------
                       |      B1(n) |         (1 + delta)
                                         B1(n)



                                                  1/2
                                      ln(17 + 12 2   ) - nu
                      where , delta = ---------------------
                                                  1/2
                                      ln(17 + 12 2   ) + nu



Using the above values of nu for E(n) from, 2980, to , 3001,

    the estimated deltas are



[0.02726499754, 0.02804691006, 0.02723818054, 0.02792931701, 0.02726733468,

    0.02691188273, 0.02809239788, 0.02848433264, 0.02723466590, 0.02760238880,

    0.02800488232, 0.02704170836, 0.02708632210, 0.02748609209, 0.02745415439,

    0.02807155200, 0.02693266874, 0.02654037523, 0.02634168812, 0.02703035100,

    0.02645356091]



                     As you can see, they are all positive



               We leave it to the reader to fill-in the details.





                       ---------------------------------





                       ---------------------------------



                         This took, 77.737, seconds .