Integer Linear Combinations of Catalan,Pi, and log(2) that are close to an INTEGER (but not close enough) [A FAILED ATTEMPT] By Shalosh B. Ekhad Theorem: There exist EXPLICIT integer sequences A(n), B(n), C(n), D(n) such that abs(A(n)+B(n)*Catalan+C(n)*Pi+D(n)*log(2))<= CONSTANT/max(A(n),B(n),C(n),D(n))^delta, where delta equals -0.13207062029997860310 Comment 1: Alas, since, -0.13207062029997860310, is negative, this is disappointing. Nevertheless, -0.13207062029997860310, is not that negative and this raises the hope that with other integrals it may work out to have a positive delta, like we did f\ or Dilog((a-1)/a) and Log((a-1)/a) for a integers. Comment 2: Just for comparison, the smallest empirical delta between 200 and 400 is, -0.12596134235975465484 -------------------------------------------------------------------------------- Proof of the Theorem Consider an analog of Beukers's famous integral that established the improve\ d Apery proof of the irrationality of Zeta(2) Let's call it, E(n) 1 1 / / n n n n | | x (1 - x ) y (1 - y) E(n) = | | ----------------------- dx dy | | 2 2 (n + 1) / / (x y + 1) 0 0 It is readily seen that E(n) is a linear combination with RATIONAL number co\ efficients of 1, Catalan, Pi, and log(2) E(n)=A1(n)+B1(n)*Catalan+ C1(n)*Pi+ D1(n)*log(2) This introduces four sequences of RATIONAL numbers, A1(n), B1(n), C1(n), D1(\ n) Lemma 1: The absolute value of E(n) is asymptotic (up to a constant) to n 0.059175542861351305168 x (1 - x) y (1 - y) Proof: Maximize the function, -------------------, 2 2 x y + 1 in 0<=x,y<=1 using two-variable calculus Indeed taking partial derivatives with respect to x and y, and setting it eq\ ual to zero gives that the maximum is attained at the point 4 4 {x = RootOf(_Z + 2 _Z - 1), y = RootOf(_Z + 2 _Z - 1)} x (1 - x) y (1 - y) Now plug it into, ------------------- 2 2 x y + 1 Lemma 2: The four sequences of rational numbers A1(n), B1(n), C1(n), D1(n) a\ re all asymptotic, up to a constant to n 2.3318190387053455993 Proof: Using the amazing Multivariable extension of the Almkvist-Zeilberger \ algorithm, due to Moa Apagodu and Doron Zeilberger it was discovered and proved that E(n) satisfies the following FOURTH-ORDER \ linear recurrence equation with polynomial coefficients 2 2 (43 n + 252 n + 356) (n + 1) E(n) 1/8 ----------------------------------- 2 %1 (n + 4) 4 3 2 (387 n + 3429 n + 10892 n + 14778 n + 7276) E(n + 1) - 1/4 ------------------------------------------------------- 2 %1 (n + 4) 4 3 2 (215 n + 2120 n + 7648 n + 11927 n + 6736) E(n + 2) + 1/2 ------------------------------------------------------ 2 %1 (n + 4) 4 3 2 (129 n + 1401 n + 5528 n + 9290 n + 5514) E(n + 3) - ----------------------------------------------------- + E(n + 4) = 0 2 %1 (n + 4) 2 %1 := 43 n + 166 n + 147 and in Maple format 1/8*(43*n^2+252*n+356)*(n+1)^2/(43*n^2+166*n+147)/(n+4)^2*E(n)-1/4*(387*n^4+ 3429*n^3+10892*n^2+14778*n+7276)/(43*n^2+166*n+147)/(n+4)^2*E(n+1)+1/2*(215*n^4 +2120*n^3+7648*n^2+11927*n+6736)/(43*n^2+166*n+147)/(n+4)^2*E(n+2)-(129*n^4+ 1401*n^3+5528*n^2+9290*n+5514)/(43*n^2+166*n+147)/(n+4)^2*E(n+3)+E(n+4) = 0 It follows that the four sequences of rational numbers, A1(n), B1(n), C1(n),\ D1(n) also satisy this recurrence, so let denote them all by X(n) 2 2 (43 n + 252 n + 356) (n + 1) X(n) 1/8 ----------------------------------- 2 %1 (n + 4) 4 3 2 (387 n + 3429 n + 10892 n + 14778 n + 7276) X(n + 1) - 1/4 ------------------------------------------------------- 2 %1 (n + 4) 4 3 2 (215 n + 2120 n + 7648 n + 11927 n + 6736) X(n + 2) + 1/2 ------------------------------------------------------ 2 %1 (n + 4) 4 3 2 (129 n + 1401 n + 5528 n + 9290 n + 5514) X(n + 3) - ----------------------------------------------------- + X(n + 4) = 0 2 %1 (n + 4) 2 %1 := 43 n + 166 n + 147 where X(n) stand for each of the A1(n),B1(n), C1(n),D1(n), that of course ha\ ve different initial conditions. Taking the leading term in n, gives us the constant-coefficient recurrence t\ hat approximates (up to polynomial corrections) The sequences 43*X1(n)-774*X1(n+1)+860*X1(n+2)-1032*X1(n+3)+344*X1(n+4) = 0 where X1(n) is a C-finite approximation with the same asymptotics (up to plo\ ynomial factors) to the above sequences, thanks to the Poincare lemma Solving the indicial equation 344*x^4-1032*x^3+860*x^2-774*x+43 = 0 and taking the largest root, establishes the lemma. Lemma 3: Let d(n) be the least common multiple of the first n natural number\ s. Recall that d(n) is asymptotic to exp(n) A(n)=A1(n)*d(n)^2*4^n, B(n)=B1(n)*d(n)^2*4^n, C(n)=C1(n)*d(n)^2*4^n, D(n)=D1\ (n)*d(n)^2*4^n, are INTEGER sequencs Proof: Left to the reader We are now ready to prove the theorem: A(n), B(n), C(n), D(n) are asymptotic\ to n n 2.3318190387053455993 exp(2 n) 4 d(n)^2*E(n)*4^n is asymptotic to n n 0.059175542861351305168 exp(2 n) 4 Our delta is such that (Max(A1(n),B1(n),C1(n),D1(n))*exp(2*n)*4^n)^delta=exp(2*n)*4^n/E(n) Taking logarithms of both sides, and solving for delta, establishes the disa\ ppointing theorem. QED.