Doing a=, 1, i.e. , ln(2) ---------------------------- Doing (A,B,C)=(1,1,1), The Alladi-Robinson case A sequence of Diophantine Approximations to, ln(2), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 4.6221008324542313343 and a not-yet-rigorous smaller upper bound around, 4.5977961316311489096 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) 1 /x (1 - x)\n / |---------| | \ 1 + x / | ------------ dx | 1 + x / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. (n + 1) F(n) + (-6 n - 9) F(n + 1) + (n + 2) F(n + 2) = 0 and in Maple format (n+1)*F(n)+(-6*n-9)*F(n+1)+(n+2)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(2), E(1) = -2 + 3 ln(2) and in Maple notation E(0) = ln(2), E(1) = -2 + 3 ln(2) Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 4.5469377751717949058 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 N - 6 N + 1 and in Maple format N^2-6*N+1 whose largest root let's call it alef, is, 5.8284271247461900976, and absolute value of the smaller root, let's call it bet is, 0.1715728752538099024 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = d(n), D1(n) = d(n) D(n) are integers Defining E1(n) = d(n) E(n) We have E1(n) = C1(n) + ln(2) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(n) alef and E1(n) is O of n exp(n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 1 - ------------ ln(alef) + 1 recall that alef is , 5.8284271247461900976, and bet is , 0.1715728752538099024 and delta is .27608287186263358796 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 4.6221008324542313346 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .40320060607156076656e-2 and the better delta is .27794793351635117004 that implies an irrationality measure, 1+1/delta that equals 4.5977961316311489097 ----------------------------------------- This ends this paper that took, 2.248, to generate. ---------------------------- Doing (A,B,C)=(6,8,7), The Rukhadze case A sequence of Diophantine Approximations to, ln(2), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 5.0622417785469770622 and a not-yet-rigorous smaller upper bound around, 3.9288531833114715468 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) / 6 8\n 1 |x (1 - x) | / |-----------| | | 7 | | \ (1 + x) / | -------------- dx | 1 + x | / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. -18432 (8 n + 5) (6 n + 5) (4 n + 1) (3 n + 2) (8 n + 7) (2 n + 3) (2 n + 1) 5 (8 n + 1) (1 + 3 n) (4 n + 3) (6 n + 1) (8 n + 3) (n + 1) (92815596 n 4 3 2 + 735911952 n + 2317592435 n + 3622803767 n + 2810234794 n + 865249000) F(n) + 168 (7 n + 1) (7 n + 2) (7 n + 3) (2 n + 3) (7 n + 4) (7 n + 5) 11 10 (7 n + 6) (512341393803030000 n + 6111594030912480000 n 9 8 + 32316884594583581736 n + 99790581649964348400 n 7 6 + 199464343566565915823 n + 270211940355922765844 n 5 4 + 252258533758563559774 n + 161565197325805798880 n 3 2 + 69172353402109525447 n + 18712762706641682156 n + 2850695797176012500 n + 183072808362620400) F(n + 1) + 49 (7 n + 8) (7 n + 1) (7 n + 9) (7 n + 2) (7 n + 10) (7 n + 3) (7 n + 11) (7 n + 4) 5 (7 n + 12) (7 n + 5) (7 n + 13) (7 n + 6) (n + 2) (92815596 n 4 3 2 + 271833972 n + 302100587 n + 157342214 n + 37834737 n + 3321894) F(n + 2) = 0 and in Maple format -18432*(8*n+5)*(6*n+5)*(4*n+1)*(3*n+2)*(8*n+7)*(2*n+3)*(2*n+1)*(8*n+1)*(1+3*n)* (4*n+3)*(6*n+1)*(8*n+3)*(n+1)*(92815596*n^5+735911952*n^4+2317592435*n^3+ 3622803767*n^2+2810234794*n+865249000)*F(n)+168*(7*n+1)*(7*n+2)*(7*n+3)*(2*n+3) *(7*n+4)*(7*n+5)*(7*n+6)*(512341393803030000*n^11+6111594030912480000*n^10+ 32316884594583581736*n^9+99790581649964348400*n^8+199464343566565915823*n^7+ 270211940355922765844*n^6+252258533758563559774*n^5+161565197325805798880*n^4+ 69172353402109525447*n^3+18712762706641682156*n^2+2850695797176012500*n+ 183072808362620400)*F(n+1)+49*(7*n+8)*(7*n+1)*(7*n+9)*(7*n+2)*(7*n+10)*(7*n+3)* (7*n+11)*(7*n+4)*(7*n+12)*(7*n+5)*(7*n+13)*(7*n+6)*(n+2)*(92815596*n^5+ 271833972*n^4+302100587*n^3+157342214*n^2+37834737*n+3321894)*F(n+2) = 0 The initial conditions are as follows 1660947 E(0) = ln(2), E(1) = ------- - 68464 ln(2) 35 and in Maple notation 1660947 E(0) = ln(2), E(1) = ------- - 68464 ln(2) 35 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.9160394243424431047 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 -678223072849 N - 218205656804520000 N + 1565515579392 and in Maple format -678223072849*N^2-218205656804520000*N+1565515579392 whose largest root let's call it alef, is, 321731.39715340729729, and absolute value of the smaller root, let's call it bet is, -5 0.717449585 10 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = 7 d(7 n), D1(n) = d(7 n) D(n) are integers Defining E1(n) = d(7 n) E(n) We have E1(n) = C1(n) + ln(2) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(7 n) alef and E1(n) is O of n exp(7 n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 7 - ------------ ln(alef) + 7 -5 recall that alef is , 321731.39715340729729, and bet is , 0.717449585 10 and delta is .24616949322934931760 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 5.0622417785469770625 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be 1.3976706751910628202 and the better delta is .34143056596280541299 that implies an irrationality measure, 1+1/delta that equals 3.9288531833114715469 ----------------------------------------- This ends this paper that took, 14.708, to generate. ---------------------------- Doing a=, 2, i.e. , ln(3/2) ---------------------------- Doing (A,B,C)=(1,1,1), The Alladi-Robinson case A sequence of Diophantine Approximations to, ln(3) - ln(2), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.5474705913691086066 and a not-yet-rigorous smaller upper bound around, 3.5453555543463918252 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) 1 /x (1 - x)\n / |---------| | \ 2 + x / | ------------ dx | 2 + x / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. (n + 1) F(n) + (-10 n - 15) F(n + 1) + (n + 2) F(n + 2) = 0 and in Maple format (n+1)*F(n)+(-10*n-15)*F(n+1)+(n+2)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(3) - ln(2), E(1) = -2 - 5 ln(2) + 5 ln(3) and in Maple notation E(0) = ln(3) - ln(2), E(1) = -2 - 5 ln(2) + 5 ln(3) Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.5049744302106557629 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 N - 10 N + 1 and in Maple format N^2-10*N+1 whose largest root let's call it alef, is, 9.8989794855663561964, and absolute value of the smaller root, let's call it bet is, 0.1010205144336438036 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = d(n), D1(n) = d(n) D(n) are integers Defining E1(n) = d(n) E(n) We have E1(n) = C1(n) + ln(3/2) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(n) alef and E1(n) is O of n exp(n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 1 - ------------ ln(alef) + 1 recall that alef is , 9.8989794855663561964, and bet is , 0.1010205144336438036 and delta is .39254623915503635531 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.5474705913691086065 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .77102022309226397045e-3 and the better delta is .39287242141571244114 that implies an irrationality measure, 1+1/delta that equals 3.5453555543463918251 ----------------------------------------- This ends this paper that took, 2.629, to generate. ---------------------------- Doing (A,B,C)=(6,8,7), The Rukhadze case A sequence of Diophantine Approximations to, ln(3) - ln(2), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.9903151104318804780 and a not-yet-rigorous smaller upper bound around, 3.3987437052869894553 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) / 6 8\n 1 |x (1 - x) | / |-----------| | | 7 | | \ (2 + x) / | -------------- dx | 2 + x | / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. -13824 (8 n + 5) (6 n + 5) (4 n + 1) (3 n + 2) (8 n + 7) (2 n + 3) (2 n + 1) 5 (8 n + 1) (3 n + 1) (4 n + 3) (6 n + 1) (8 n + 3) (n + 1) (2278535468 n 4 3 2 + 18065669488 n + 56893979587 n + 88937144543 n + 68991986514 n + 21243392520) F(n) + 84 (7 n + 1) (7 n + 2) (7 n + 3) (2 n + 3) (7 n + 4) 11 (7 n + 5) (7 n + 6) (890573611147656603376 n 10 9 + 10623324979195495931520 n + 56174377692744234572840 n 8 7 + 173464048762577612294064 n + 346741883519096972790015 n 6 5 + 469763293640063698303308 n + 438601410987144596861510 n 4 3 + 280959317721807705490536 n + 120318027390750752552239 n 2 + 32560040154941713519212 n + 4962715435871417451060 n + 318957068712339212400) F(n + 1) + 49 (7 n + 9) (7 n + 2) (7 n + 11) (7 n + 4) (7 n + 13) (7 n + 6) (7 n + 8) (7 n + 1) (7 n + 10) (7 n + 3) 5 4 3 (7 n + 12) (7 n + 5) (n + 2) (2278535468 n + 6672992148 n + 7416656315 n 2 + 3863868030 n + 929635577 n + 81704982) F(n + 2) = 0 and in Maple format -13824*(8*n+5)*(6*n+5)*(4*n+1)*(3*n+2)*(8*n+7)*(2*n+3)*(2*n+1)*(8*n+1)*(3*n+1)* (4*n+3)*(6*n+1)*(8*n+3)*(n+1)*(2278535468*n^5+18065669488*n^4+56893979587*n^3+ 88937144543*n^2+68991986514*n+21243392520)*F(n)+84*(7*n+1)*(7*n+2)*(7*n+3)*(2*n +3)*(7*n+4)*(7*n+5)*(7*n+6)*(890573611147656603376*n^11+10623324979195495931520 *n^10+56174377692744234572840*n^9+173464048762577612294064*n^8+ 346741883519096972790015*n^7+469763293640063698303308*n^6+ 438601410987144596861510*n^5+280959317721807705490536*n^4+ 120318027390750752552239*n^3+32560040154941713519212*n^2+4962715435871417451060 *n+318957068712339212400)*F(n+1)+49*(7*n+9)*(7*n+2)*(7*n+11)*(7*n+4)*(7*n+13)*( 7*n+6)*(7*n+8)*(7*n+1)*(7*n+10)*(7*n+3)*(7*n+12)*(7*n+5)*(n+2)*(2278535468*n^5+ 6672992148*n^4+7416656315*n^3+3863868030*n^2+929635577*n+81704982)*F(n+2) = 0 The initial conditions are as follows 13617497 E(0) = ln(3) - ln(2), E(1) = -------- + 2398920 ln(2) - 2398920 ln(3) 14 and in Maple notation 13617497 E(0) = ln(3) - ln(2), E(1) = -------- + 2398920 ln(2) - 2398920 ln(3) 14 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.3846229018340344913 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 -678223072849 N - 7725232356439662624 N + 1174136684544 and in Maple format -678223072849*N^2-7725232356439662624*N+1174136684544 8 whose largest root let's call it alef, is, 0.11390400394354787993 10 , and absolute value of the smaller root, let's call it bet is, -6 0.1519871 10 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: n n C1(n) = 7 d(7 n) 2 , D1(n) = d(7 n) 2 D(n) are integers Defining n E1(n) = d(7 n) 2 E(n) We have E1(n) = C1(n) + ln(3/2) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n n exp(7 n) 2 alef and E1(n) is O of n n exp(7 n) 2 bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 7 + ln(2) - -------------------- ln(alef) + 7 + ln(2) 8 -6 recall that alef is , 0.11390400394354787993 10 , and bet is , 0.1519871 10 and delta is .33441291739169708754 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.9903151104318804779 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be 1.3935477818642534174 and the better delta is .41688488761676955944 that implies an irrationality measure, 1+1/delta that equals 3.3987437052869894554 ----------------------------------------- This ends this paper that took, 13.925, to generate. ---------------------------- Doing a=, 3, i.e. , ln(4/3) ---------------------------- Doing (A,B,C)=(1,1,1), The Alladi-Robinson case A sequence of Diophantine Approximations to, 2 ln(2) - ln(3), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.2240532881366202850 and a not-yet-rigorous smaller upper bound around, 3.2186809591620273633 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) 1 /x (1 - x)\n / |---------| | \ 3 + x / | ------------ dx | 3 + x / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. (n + 1) F(n) + (-14 n - 21) F(n + 1) + (n + 2) F(n + 2) = 0 and in Maple format (n+1)*F(n)+(-14*n-21)*F(n+1)+(n+2)*F(n+2) = 0 The initial conditions are as follows E(0) = 2 ln(2) - ln(3), E(1) = -2 - 7 ln(3) + 14 ln(2) and in Maple notation E(0) = 2 ln(2) - ln(3), E(1) = -2 - 7 ln(3) + 14 ln(2) Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.1895842927789866521 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 N - 14 N + 1 and in Maple format N^2-14*N+1 whose largest root let's call it alef, is, 13.928203230275509174, and absolute value of the smaller root, let's call it bet is, 0.0717967697244908260 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = d(n), D1(n) = d(n) D(n) are integers Defining E1(n) = d(n) E(n) We have E1(n) = C1(n) + ln(4/3) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(n) alef and E1(n) is O of n exp(n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 1 - ------------ ln(alef) + 1 recall that alef is , 13.928203230275509174, and bet is , 0.0717967697244908260 and delta is .44962951442491313526 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.2240532881366202850 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .27271833625997793011e-2 and the better delta is .45071825035073522272 that implies an irrationality measure, 1+1/delta that equals 3.2186809591620273633 ----------------------------------------- This ends this paper that took, 2.721, to generate. ---------------------------- Doing (A,B,C)=(6,8,7), The Rukhadze case A sequence of Diophantine Approximations to, 2 ln(2) - ln(3), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.6544022768844813116 and a not-yet-rigorous smaller upper bound around, 3.2073918575231816483 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) / 6 8\n 1 |x (1 - x) | / |-----------| | | 7 | | \ (3 + x) / | -------------- dx | 3 + x | / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. -12288 (8 n + 7) (6 n + 5) (4 n + 3) (3 n + 2) (8 n + 5) (2 n + 3) (2 n + 1) 5 (8 n + 3) (3 n + 1) (4 n + 1) (6 n + 1) (8 n + 1) (n + 1) (17785696396 n 4 3 2 + 141015704192 n + 444099277115 n + 694223591707 n + 538541890974 n + 165825655800) F(n) + 56 (7 n + 5) (7 n + 3) (7 n + 1) (2 n + 3) 11 (7 n + 6) (7 n + 4) (7 n + 2) (107411492883705888657520 n 10 9 + 1281268925737295742813120 n + 6775149185436440434268456 n 8 7 + 20921508524798692079465040 n + 41821088033076937117611063 n 6 5 + 56660033312825556667392864 n + 52903045482407366998387154 n 4 3 + 33890085839556129224339940 n + 14513982349071266289345787 n 2 + 3928072965201261367817916 n + 598786020676145897892900 n + 38492234744580799988400) F(n + 1) + 49 (7 n + 8) (7 n + 1) (7 n + 9) (7 n + 2) (7 n + 10) (7 n + 3) (7 n + 11) (7 n + 4) (7 n + 12) (7 n + 5) 5 4 (7 n + 13) (7 n + 6) (n + 2) (17785696396 n + 52087222212 n 3 2 + 57893424307 n + 30163021554 n + 7258204117 n + 638087214) F(n + 2) = 0 and in Maple format -12288*(8*n+7)*(6*n+5)*(4*n+3)*(3*n+2)*(8*n+5)*(2*n+3)*(2*n+1)*(8*n+3)*(3*n+1)* (4*n+1)*(6*n+1)*(8*n+1)*(n+1)*(17785696396*n^5+141015704192*n^4+444099277115*n^ 3+694223591707*n^2+538541890974*n+165825655800)*F(n)+56*(7*n+5)*(7*n+3)*(7*n+1) *(2*n+3)*(7*n+6)*(7*n+4)*(7*n+2)*(107411492883705888657520*n^11+ 1281268925737295742813120*n^10+6775149185436440434268456*n^9+ 20921508524798692079465040*n^8+41821088033076937117611063*n^7+ 56660033312825556667392864*n^6+52903045482407366998387154*n^5+ 33890085839556129224339940*n^4+14513982349071266289345787*n^3+ 3928072965201261367817916*n^2+598786020676145897892900*n+ 38492234744580799988400)*F(n+1)+49*(7*n+8)*(7*n+1)*(7*n+9)*(7*n+2)*(7*n+10)*(7* n+3)*(7*n+11)*(7*n+4)*(7*n+12)*(7*n+5)*(7*n+13)*(7*n+6)*(n+2)*(17785696396*n^5+ 52087222212*n^4+57893424307*n^3+30163021554*n^2+7258204117*n+638087214)*F(n+2) = 0 The initial conditions are as follows 106347869 E(0) = 2 ln(2) - ln(3), E(1) = --------- + 24644768 ln(3) - 49289536 ln(2) 15 and in Maple notation 106347869 E(0) = 2 ln(2) - ln(3), E(1) = --------- + 24644768 ln(3) - 49289536 ln(2) 15 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.2008221731658347952 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 -678223072849 N - 79576739523177722560 N + 1043677052928 and in Maple format -678223072849*N^2-79576739523177722560*N+1043677052928 9 whose largest root let's call it alef, is, 0.11733121845722689744 10 , -7 and absolute value of the smaller root, let's call it bet is, 0.13115 10 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: n n C1(n) = 7 d(7 n) 3 , D1(n) = d(7 n) 3 D(n) are integers Defining n E1(n) = d(7 n) 3 E(n) We have E1(n) = C1(n) + ln(4/3) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n n exp(7 n) 3 alef and E1(n) is O of n n exp(7 n) 3 bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 7 + ln(3) - -------------------- ln(alef) + 7 + ln(3) 9 -7 recall that alef is , 0.11733121845722689744 10 , and bet is , 0.13115 10 and delta is .37673264851690739467 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.6544022768844813117 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be 1.4007816485893661062 and the better delta is .45302332551052194524 that implies an irrationality measure, 1+1/delta that equals 3.2073918575231816485 ----------------------------------------- This ends this paper that took, 14.329, to generate. ---------------------------- Doing a=, 4, i.e. , ln(5/4) ---------------------------- Doing (A,B,C)=(1,1,1), The Alladi-Robinson case A sequence of Diophantine Approximations to, ln(5) - 2 ln(2), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.0597312482455252048 and a not-yet-rigorous smaller upper bound around, 3.0567301785332553673 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) 1 /x (1 - x)\n / |---------| | \ 4 + x / | ------------ dx | 4 + x / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. (n + 1) F(n) + (-18 n - 27) F(n + 1) + (n + 2) F(n + 2) = 0 and in Maple format (n+1)*F(n)+(-18*n-27)*F(n+1)+(n+2)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(5) - 2 ln(2), E(1) = -2 - 18 ln(2) + 9 ln(5) and in Maple notation E(0) = ln(5) - 2 ln(2), E(1) = -2 - 18 ln(2) + 9 ln(5) Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.0361403626623286862 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 N - 18 N + 1 and in Maple format N^2-18*N+1 whose largest root let's call it alef, is, 17.944271909999158786, and absolute value of the smaller root, let's call it bet is, 0.0557280900008412144 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = d(n), D1(n) = d(n) D(n) are integers Defining E1(n) = d(n) E(n) We have E1(n) = C1(n) + ln(5/4) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(n) alef and E1(n) is O of n exp(n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 1 - ------------ ln(alef) + 1 recall that alef is , 17.944271909999158786, and bet is , 0.0557280900008412144 and delta is .48550023254334657546 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.0597312482455252048 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .18529053456339762159e-2 and the better delta is .48620864828907403168 that implies an irrationality measure, 1+1/delta that equals 3.0567301785332553674 ----------------------------------------- This ends this paper that took, 2.773, to generate. ---------------------------- Doing (A,B,C)=(6,8,7), The Rukhadze case A sequence of Diophantine Approximations to, ln(5) - 2 ln(2), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.4799504980542251212 and a not-yet-rigorous smaller upper bound around, 3.1064931197345016785 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) / 6 8\n 1 |x (1 - x) | / |-----------| | | 7 | | \ (4 + x) / | -------------- dx | 4 + x | / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. -11520 (8 n + 7) (6 n + 5) (4 n + 3) (3 n + 2) (8 n + 5) (2 n + 3) (2 n + 1) 5 (8 n + 3) (3 n + 1) (4 n + 1) (6 n + 1) (8 n + 1) (n + 1) (81527988492 n 4 3 2 + 646401965376 n + 2035707591131 n + 3182260214891 n + 2468639660014 n + 760138582840) F(n) + 42 (7 n + 1) (7 n + 2) (7 n + 3) (2 n + 3) 11 (7 n + 4) (7 n + 5) (7 n + 6) (3745884524740886197174896 n 10 9 + 44683119355057569297185472 n + 236277500006364906882193704 n 8 7 + 729621606156898458211588560 n + 1458485943002444838346322135 n 6 5 + 1976001527582107888574021072 n + 1844999889613585620361944226 n 4 3 + 1181941157378937487005078260 n + 506198160396156680061441739 n 2 + 137002719573514632533753276 n + 20885488293103003289472740 n + 1342709265613597218682800) F(n + 1) + 49 (7 n + 12) (7 n + 5) (7 n + 10) (7 n + 3) (7 n + 8) (7 n + 1) (7 n + 13) (7 n + 6) (7 n + 11) (7 n + 4) 5 4 (7 n + 9) (7 n + 2) (n + 2) (81527988492 n + 238762022916 n 3 2 + 265379614547 n + 138269348834 n + 33274084581 n + 2925523470) F(n + 2) = 0 and in Maple format -11520*(8*n+7)*(6*n+5)*(4*n+3)*(3*n+2)*(8*n+5)*(2*n+3)*(2*n+1)*(8*n+3)*(3*n+1)* (4*n+1)*(6*n+1)*(8*n+1)*(n+1)*(81527988492*n^5+646401965376*n^4+2035707591131*n ^3+3182260214891*n^2+2468639660014*n+760138582840)*F(n)+42*(7*n+1)*(7*n+2)*(7*n +3)*(2*n+3)*(7*n+4)*(7*n+5)*(7*n+6)*(3745884524740886197174896*n^11+ 44683119355057569297185472*n^10+236277500006364906882193704*n^9+ 729621606156898458211588560*n^8+1458485943002444838346322135*n^7+ 1976001527582107888574021072*n^6+1844999889613585620361944226*n^5+ 1181941157378937487005078260*n^4+506198160396156680061441739*n^3+ 137002719573514632533753276*n^2+20885488293103003289472740*n+ 1342709265613597218682800)*F(n+1)+49*(7*n+12)*(7*n+5)*(7*n+10)*(7*n+3)*(7*n+8)* (7*n+1)*(7*n+13)*(7*n+6)*(7*n+11)*(7*n+4)*(7*n+9)*(7*n+2)*(n+2)*(81527988492*n^ 5+238762022916*n^4+265379614547*n^3+138269348834*n^2+33274084581*n+2925523470)* F(n+2) = 0 The initial conditions are as follows 877657041 E(0) = ln(5) - 2 ln(2), E(1) = --------- + 280939280 ln(2) - 140469640 ln(5) 28 and in Maple notation 877657041 E(0) = ln(5) - 2 ln(2), E(1) = --------- + 280939280 ln(2) - 140469640 ln(5) 28 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.1003619379286928403 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 -678223072849 N - 454062027466024136208 N + 978447237120 and in Maple format -678223072849*N^2-454062027466024136208*N+978447237120 9 whose largest root let's call it alef, is, 0.66948773293520314272 10 , -8 and absolute value of the smaller root, let's call it bet is, 0.214 10 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: n n C1(n) = 7 d(7 n) 4 , D1(n) = d(7 n) 4 D(n) are integers Defining n E1(n) = d(7 n) 4 E(n) We have E1(n) = C1(n) + ln(5/4) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n n exp(7 n) 4 alef and E1(n) is O of n n exp(7 n) 4 bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 7 + 2 ln(2) - ---------------------- ln(alef) + 7 + 2 ln(2) 9 -8 recall that alef is , 0.66948773293520314272 10 , and bet is , 0.214 10 and delta is .40323385518565886929 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.4799504980542251211 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be 1.3916671651329957702 and the better delta is .47472265189550586673 that implies an irrationality measure, 1+1/delta that equals 3.1064931197345016786 ----------------------------------------- This ends this paper that took, 15.359, to generate. ---------------------------- Doing a=, 5, i.e. , ln(6/5) ---------------------------- Doing (A,B,C)=(1,1,1), The Alladi-Robinson case A sequence of Diophantine Approximations to, ln(6) - ln(5), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 2.9574096761096127808 and a not-yet-rigorous smaller upper bound around, 2.9563931275052114054 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) 1 /x (1 - x)\n / |---------| | \ 5 + x / | ------------ dx | 5 + x / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. (n + 1) F(n) + (-22 n - 33) F(n + 1) + (n + 2) F(n + 2) = 0 and in Maple format (n+1)*F(n)+(-22*n-33)*F(n+1)+(n+2)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(2) + ln(3) - ln(5), E(1) = -2 - 11 ln(5) + 11 ln(2) + 11 ln(3) and in Maple notation E(0) = ln(2) + ln(3) - ln(5), E(1) = -2 - 11 ln(5) + 11 ln(2) + 11 ln(3) Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 2.9332178025782929231 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 N - 22 N + 1 and in Maple format N^2-22*N+1 whose largest root let's call it alef, is, 21.954451150103322269, and absolute value of the smaller root, let's call it bet is, 0.045548849896677731 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = d(n), D1(n) = d(n) D(n) are integers Defining E1(n) = d(n) E(n) We have E1(n) = C1(n) + ln(6/5) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(n) alef and E1(n) is O of n exp(n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 1 - ------------ ln(alef) + 1 recall that alef is , 21.954451150103322269, and bet is , 0.045548849896677731 and delta is .51087925650164258413 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 2.9574096761096127807 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .71828723374087596831e-3 and the better delta is .51114471112214445138 that implies an irrationality measure, 1+1/delta that equals 2.9563931275052114053 ----------------------------------------- This ends this paper that took, 3.086, to generate. ---------------------------- Doing (A,B,C)=(6,8,7), The Rukhadze case A sequence of Diophantine Approximations to, ln(6) - ln(5), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.4018949045623248333 and a not-yet-rigorous smaller upper bound around, 3.0652827262611494098 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) / 6 8\n 1 |x (1 - x) | / |-----------| | | 7 | | \ (5 + x) / | -------------- dx | 5 + x | / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. -55296 (8 n + 5) (6 n + 5) (4 n + 1) (3 n + 2) (8 n + 7) (2 n + 3) (2 n + 1) 5 (8 n + 1) (3 n + 1) (4 n + 3) (6 n + 1) (8 n + 3) (n + 1) (273789319852 n 4 3 2 + 2170761504176 n + 6836359806371 n + 10686751350991 n + 8290269875394 n + 2552731709640) F(n) + 168 (7 n + 5) (7 n + 3) (7 n + 1) (2 n + 3) (7 n + 6) (7 n + 4) (7 n + 2) ( 11 10 63234091110265992002231536 n + 754293140548780797643350912 n 9 8 + 3988588200584873487284675624 n + 12316721341869281905905096240 n 7 6 + 24620722783618171246005036255 n + 33357047278350810336311717532 n 5 4 + 31145785762477583794055601206 n + 19952741676947790527627859480 n 3 2 + 8545406084524794480343438879 n + 2312859254596305392186007276 n + 352595186598073584378490740 n + 22668968880637198456402800) F(n + 1) + 245 (7 n + 10) (7 n + 3) (7 n + 13) (7 n + 6) (7 n + 9) (7 n + 2) (7 n + 12) (7 n + 5) (7 n + 8) (7 n + 1) (7 n + 11) (7 n + 4) (n + 2) ( 5 4 3 2 273789319852 n + 801814904916 n + 891206988187 n + 464347758414 n + 111747175081 n + 9825563190) F(n + 2) = 0 and in Maple format -55296*(8*n+5)*(6*n+5)*(4*n+1)*(3*n+2)*(8*n+7)*(2*n+3)*(2*n+1)*(8*n+1)*(3*n+1)* (4*n+3)*(6*n+1)*(8*n+3)*(n+1)*(273789319852*n^5+2170761504176*n^4+6836359806371 *n^3+10686751350991*n^2+8290269875394*n+2552731709640)*F(n)+168*(7*n+5)*(7*n+3) *(7*n+1)*(2*n+3)*(7*n+6)*(7*n+4)*(7*n+2)*(63234091110265992002231536*n^11+ 754293140548780797643350912*n^10+3988588200584873487284675624*n^9+ 12316721341869281905905096240*n^8+24620722783618171246005036255*n^7+ 33357047278350810336311717532*n^6+31145785762477583794055601206*n^5+ 19952741676947790527627859480*n^4+8545406084524794480343438879*n^3+ 2312859254596305392186007276*n^2+352595186598073584378490740*n+ 22668968880637198456402800)*F(n+1)+245*(7*n+10)*(7*n+3)*(7*n+13)*(7*n+6)*(7*n+9 )*(7*n+2)*(7*n+12)*(7*n+5)*(7*n+8)*(7*n+1)*(7*n+11)*(7*n+4)*(n+2)*(273789319852 *n^5+801814904916*n^4+891206988187*n^3+464347758414*n^2+111747175081*n+ 9825563190)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(2) + ln(3) - ln(5), 3602706503 E(1) = ---------- + 564576528 ln(5) - 564576528 ln(2) - 564576528 ln(3) 35 and in Maple notation E(0) = ln(2) + ln(3) - ln(5), 3602706503 E(1) = ---------- + 564576528 ln(5) - 564576528 ln(2) - 564576528 ln(3) 35 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 3.0359472917639016186 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 -3391115364245 N - 9129821681582975295552 N + 4696546738176 and in Maple format -3391115364245*N^2-9129821681582975295552*N+4696546738176 10 whose largest root let's call it alef, is, 0.26922769357378215836 10 , -9 and absolute value of the smaller root, let's call it bet is, 0.6 10 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: n n C1(n) = 7 d(7 n) 5 , D1(n) = d(7 n) 5 D(n) are integers Defining n E1(n) = d(7 n) 5 E(n) We have E1(n) = C1(n) + ln(6/5) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n n exp(7 n) 5 alef and E1(n) is O of n n exp(7 n) 5 bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 7 + ln(5) - -------------------- ln(alef) + 7 + ln(5) 10 -9 recall that alef is , 0.26922769357378215836 10 , and bet is , 0.6 10 and delta is .41633794971650551111 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.4018949045623248333 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be 1.3863687335513902639 and the better delta is .48419520837727314006 that implies an irrationality measure, 1+1/delta that equals 3.0652827262611494097 ----------------------------------------- This ends this paper that took, 16.140, to generate. ---------------------------- Doing a=, 6, i.e. , ln(7/6) ---------------------------- Doing (A,B,C)=(1,1,1), The Alladi-Robinson case A sequence of Diophantine Approximations to, ln(7) - ln(6), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 2.8862836267345560541 and a not-yet-rigorous smaller upper bound around, 2.8853838483466796686 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) 1 /x (1 - x)\n / |---------| | \ 6 + x / | ------------ dx | 6 + x / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. (n + 1) F(n) + (-26 n - 39) F(n + 1) + (n + 2) F(n + 2) = 0 and in Maple format (n+1)*F(n)+(-26*n-39)*F(n+1)+(n+2)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(7) - ln(2) - ln(3), E(1) = -2 - 13 ln(2) - 13 ln(3) + 13 ln(7) and in Maple notation E(0) = ln(7) - ln(2) - ln(3), E(1) = -2 - 13 ln(2) - 13 ln(3) + 13 ln(7) Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 2.8671031907470747444 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 N - 26 N + 1 and in Maple format N^2-26*N+1 whose largest root let's call it alef, is, 25.961481396815720462, and absolute value of the smaller root, let's call it bet is, 0.038518603184279538 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: C1(n) = d(n), D1(n) = d(n) D(n) are integers Defining E1(n) = d(n) E(n) We have E1(n) = C1(n) + ln(7/6) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n exp(n) alef and E1(n) is O of n exp(n) bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 1 - ------------ ln(alef) + 1 recall that alef is , 25.961481396815720462, and bet is , 0.038518603184279538 and delta is .53014296780551085240 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 2.8862836267345560541 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be .70370272138065513647e-3 and the better delta is .53039597261688353023 that implies an irrationality measure, 1+1/delta that equals 2.8853838483466796686 ----------------------------------------- This ends this paper that took, 3.063, to generate. ---------------------------- Doing (A,B,C)=(6,8,7), The Rukhadze case A sequence of Diophantine Approximations to, ln(7) - ln(6), that implies its irrationality With a fully rigorous crude upper bound on the irrationality measure of, 3.2245831645621245282 and a not-yet-rigorous smaller upper bound around, 2.9372336371610440332 By Shalosh B. Ekhad Consider the following Rukhazde-type integral, let's call it E(n) / 6 8\n 1 |x (1 - x) | / |-----------| | | 7 | | \ (6 + x) / | -------------- dx | 6 + x | / 0 It is readily seen that E(n) can be written as E(n)=C(n) + D(n)*log((a+1)/a) For certain sequences of RATIONAL numbers C(n), D(n) It follows from the amazing Almkvist-Zeilberger algorithm that E(n), and hen\ ce C(n) and D(n) satisfy the following second-order linear recurrence equation with polynomial coefficients. -1536 (8 n + 7) (6 n + 5) (4 n + 3) (3 n + 2) (8 n + 5) (2 n + 3) (2 n + 1) 5 (8 n + 3) (3 n + 1) (4 n + 1) (6 n + 1) (8 n + 1) (n + 1) (749137234796 n 4 3 2 + 5939594575312 n + 18705513937555 n + 29240899241927 n + 22683717673914 n + 6984761116200) F(n) + 4 (7 n + 5) (7 n + 3) (7 n + 1) 11 (2 n + 3) (7 n + 6) (7 n + 4) (7 n + 2) (661980470447577723735197680 n 10 9 + 7896487069320504444623623680 n + 41755439719239172232392907816 n 8 7 + 128940490882255485006564462960 n + 257748201675105370006510623183 n 6 5 + 349207419543640014320984736324 n + 326059348671367301946219322094 n 4 3 + 208882524784930301795807859120 n + 89461327532531377732346398807 n 2 + 24213431938386113653536991596 n + 3691392653912439516298910100 n + 237331817218266085197212400) F(n + 1) + 7 (7 n + 10) (7 n + 3) (7 n + 13) (7 n + 6) (7 n + 9) (7 n + 2) (7 n + 12) (7 n + 5) (7 n + 8) 5 4 (7 n + 1) (7 n + 11) (7 n + 4) (n + 2) (749137234796 n + 2193908401332 n 3 2 + 2438507984267 n + 1270552533174 n + 305768875457 n + 26886087174) F(n + 2) = 0 and in Maple format -1536*(8*n+7)*(6*n+5)*(4*n+3)*(3*n+2)*(8*n+5)*(2*n+3)*(2*n+1)*(8*n+3)*(3*n+1)*( 4*n+1)*(6*n+1)*(8*n+1)*(n+1)*(749137234796*n^5+5939594575312*n^4+18705513937555 *n^3+29240899241927*n^2+22683717673914*n+6984761116200)*F(n)+4*(7*n+5)*(7*n+3)* (7*n+1)*(2*n+3)*(7*n+6)*(7*n+4)*(7*n+2)*(661980470447577723735197680*n^11+ 7896487069320504444623623680*n^10+41755439719239172232392907816*n^9+ 128940490882255485006564462960*n^8+257748201675105370006510623183*n^7+ 349207419543640014320984736324*n^6+326059348671367301946219322094*n^5+ 208882524784930301795807859120*n^4+89461327532531377732346398807*n^3+ 24213431938386113653536991596*n^2+3691392653912439516298910100*n+ 237331817218266085197212400)*F(n+1)+7*(7*n+10)*(7*n+3)*(7*n+13)*(7*n+6)*(7*n+9) *(7*n+2)*(7*n+12)*(7*n+5)*(7*n+8)*(7*n+1)*(7*n+11)*(7*n+4)*(n+2)*(749137234796* n^5+2193908401332*n^4+2438507984267*n^3+1270552533174*n^2+305768875457*n+ 26886087174)*F(n+2) = 0 The initial conditions are as follows E(0) = ln(7) - ln(2) - ln(3), 58253188877 E(1) = ----------- + 1799512904 ln(2) + 1799512904 ln(3) - 1799512904 ln(7) 210 and in Maple notation E(0) = ln(7) - ln(2) - ln(3), 58253188877 E(1) = ----------- + 1799512904 ln(2) + 1799512904 ln(3) - 1799512904 ln(7) 210 Using this recurrence, we can compute many values, and use them to ESTIMATE \ the implied bound irrationality measure. Using the values from n=, 990, to M=, 1000, and taking the largest value yields the following estimate 2.9883439667370171325 We can use the Poincare lemma to find the asymptotic behavior of the exponen\ tially growing C(n), D(n) and the exponentially decaying E(n). The constant-coefficient approximation to the above recurrence has indicial \ polynomial 2 -96889010407 N - 831691036037157523360 N + 130459631616 and in Maple format -96889010407*N^2-831691036037157523360*N+130459631616 10 whose largest root let's call it alef, is, 0.85839563490584462501 10 , -9 and absolute value of the smaller root, let's call it bet is, 0.1 10 It follows that, up to polynomial correction that will get out in the wash, \ C(n) and D(n) are OMEGA(alef^n) and E(n) is O(bet^n) We now need a divisibility lemma that is left to the reader Let d(n)=lcm(1..n) Lemma: n n C1(n) = 7 d(7 n) 6 , D1(n) = d(7 n) 6 D(n) are integers Defining n E1(n) = d(7 n) 6 E(n) We have E1(n) = C1(n) + ln(7/6) D1(n) where C1(n) and D1(n) are INTEGERS Since, famously, d(n)=OMEGA(e^n), we have C1(n) and D1(n) are OMEGA of n n exp(7 n) 6 alef and E1(n) is O of n n exp(7 n) 6 bet It follows that E1(n)=max(C1(n),D1(n))^(-delta) where delta equals ln(bet) + 7 + ln(6) - -------------------- ln(alef) + 7 + ln(6) 10 -9 recall that alef is , 0.85839563490584462501 10 , and bet is , 0.1 10 and delta is .44952241657229067461 implying the rigorous, but crude upper bound for the irrationality measure, \ 1+1/delta that equals 3.2245831645621245282 confirming the first part of the statement in the title We now need a harder lemma, that we confess that we can't do, but you the re\ ader may be able to Harder Lemma: There exists a rigorously proved constant K1 such that gcd(C1(n),D1(n))=O(exp(K1*n)) Once we find such a constant K1 the delta can be improved to By looking at the smallest value of log(gcd(C1(n),D1(n))/n for n between n=, 500, and , 1000, we estimate that K1 can be taken to be 1.3925209770417752458 and the better delta is .51619999819199354031 that implies an irrationality measure, 1+1/delta that equals 2.9372336371610440332 ----------------------------------------- This ends this paper that took, 15.351, to generate. ----------------------------