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A005148
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Sequence of coefficients arising in connection with a rapidly converging series for Pi.
(Formerly M5290)
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10
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0, 1, 47, 2488, 138799, 7976456, 467232200, 27736348480, 1662803271215, 100442427373480, 6103747246289272, 372725876150863808, 22852464771010647496, 1405886026610765892544, 86741060172969340021952
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OFFSET
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0,3
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COMMENTS
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The paper by Newman and Shanks has an appendix by Don Zagier which eventually leads to an efficient recursive algorithm for the series itself, whereas the main paper treats each term in isolation, which is enormously slower. Using Zagier's appendix one may compute 1000 terms in 25 seconds running PARI/GP on a 500MHz Alpha. - David Broadhurst, Jun 17 2002 (see second version of PARI code here)
Conjecture: The following 2 definitions give the same sequence: (1) numbers k such that 8^m is the highest power of 2 dividing a(k), and (2) numbers k such that k has exactly (m+1) 1's in its binary representation. A018900 is the special case m=1. - Benoit Cloitre, Jun 22 2002, edited by Hugo Pfoertner, Aug 21 2021
Conjecture: There are polynomials P_k(x) such that P_k(m) = the constant term of j_m(tau)^k where j_m is modular for the Hecke group G(lambda_m), j_3 is the Klein invariant j with constant term 744, and P_k(x) = a(k+1) times a product of monic polynomials. - Barry Brent, Nov 25 2022
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REFERENCES
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F. Beukers, Letter to D. Shanks, Mar 13 1984
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 195; see Exercise 6(a).
D. Shanks, Solved and unsolved problems in number theory, Chelsea NY, 1985, p. 255-7,276
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n) = (1/24) * coefficient of x^n in Product_{k>=1} (1+x^(2k-1))^(24n).
Asymptotically (D. Zagier): a(n) = C*(64^n)/sqrt(n)*(1 - a/n + b/n^2 + ...) with C = (sqrt(Pi)/12)*Gamma(3/4)^2/Gamma(1/4)^2 = 0.0168732651....; a = 6*Gamma(3/4)^4/Gamma(1/4)^4 = 0.078300067..., b = 60*Gamma(3/4)^8/Gamma(1/4)^8 - 1/128 = 0.002405668.... - Benoit Cloitre, Jun 22 2002; numerical value of constant "a" corrected by Vaclav Kotesovec, Jul 28 2013
Alternative expressions for these constants: C = Pi^(5/2)/(6*Gamma(1/4)^4), a = 24*Pi^4/Gamma(1/4)^8, b = 960*Pi^8/Gamma(1/4)^16 - 1/128. - Vaclav Kotesovec, Jul 28 2013
A076657(n) = Sum_{i=0..n} binomial(2*n-2*i, n-i)^3 a(i) = (1/24)*binomial(2*n, n)*(16^n-binomial(2*n, n)^2) (Shanks and Beukers). - Ralf Stephan, Oct 24 2002
Expansion of ((Pi / (2 K(q)))^2 / (1 - 2*k(q)^2) - 1) / 24 in powers of (k'(q) * k(q) / 4)^2. [Borwein and Borwein, 6(a)(i)] - Michael Somos, Jul 06 2014
Expansion of ((Pi / (2 K(q)))^2 / (1 + k(q)^2) - 1) / 24 in powers of (k'(q)^-2 * k(q) / 4)^2. [Borwein and Borwein, 6(a)(ii)] - Michael Somos, Jul 06 2014
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EXAMPLE
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G.f. = x + 47*x^2 + 2488*x^3 + 138799*x^4 + 7976456*x^5 + 467232200*x^6 + ...
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MATHEMATICA
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a[n_] := a[n]=(Binomial[2n, n](16^n-Binomial[2n, n]^2))/24-Sum[Binomial[2n-2i, n-i]^3a[i], {i, 0, n-1}]
a[ n_] := If[ n < 1, 0, SeriesCoefficient[ ComposeSeries[ Series[ ((Pi / (2 EllipticK[m]))^2 / (1 - 2 m) - 1) / 24, {m, 0, n}], InverseSeries[ Series[ (1 - m) m/16, {m, 0, n}]]], {m, 0, n}]]; (* Michael Somos, Jul 06 2014 *)
a[ n_] := If[ n < 1, 0, SeriesCoefficient[ ComposeSeries[ Series[ ((Pi / (2 EllipticK[m]))^2 / (1 + m) - 1) / 24, {m, 0, n}], InverseSeries[ Series[ -(1 - m)^-2 m/16, {m, 0, n}]]], {m, 0, n}]]; (* Michael Somos, Jul 06 2014 *)
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PROG
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(PARI) {a(n) = if( n<1, 0, polcoeff( prod( k=1, (n+1)\2, 1 + x^(2*k - 1), 1 + x *O(x^n))^(24*n), n) / 24)};
(PARI) {nt=1000; a=[1]; b=[1]; d=1; e=0; g=0; print(1); for(n=2, nt, c=48*(a[n-1]+g)+128*(d-32*e); e=d; d=c; i=(n-1)\2; g=12*if(n%2==0, a[n/2]^2)+24*sum(j=1, i, a[j]*a[n-j]); h=12*if(n%2==0, b[n/2]^2)+24*sum(j=1, i, b[j]*b[n-j]); f=(c+5*h)/n^2-g; a=concat(a, f); b=concat(b, n*f); print(f))} /* Broadhurst 2002 */
(PARI) {a(n)=if(n<1, 0, va[n])} {b(n)=n*a(n)} {doit(nt)= local(c, d, e, g); va=vector(nt); va[1]=1; d=1; e=0; g=0; for(n=2, nt, c=48*(a(n-1)+g)+128*(d-32*e); e=d; d=c; g=12*if(n%2==0, a(n/2)^2)+24*sum(j=1, (n-1)\2, a(j)*a(n-j)); va[n]=(c+5*(12*if(n%2==0, b(n/2)^2)+24*sum(j=1, (n-1)\2, b(j)*b(n-j))))/n^2-g; )}; /* Michael Somos, Nov 05 2002 */
(PARI) {a(n) = local(an, cb); if( n<1, 0, an = cb = vector(n, i, binomial(2*i, i)); an[1]=1; for(j=2, n, an[j] = (cb[j]*16^j - cb[j]^3) / 24 - sum(i=1, j-1, cb[j-i]^3*an[i])); an[n])}; /* Michael Somos, Mar 09 2004 */
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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