A097344 Numerators in binomial transform of 1/(n+1)^2.

1, 5, 29, 103, 887, 1517, 18239, 63253, 332839, 118127, 2331085, 4222975, 100309579, 184649263, 1710440723, 6372905521, 202804884977, 381240382217, 13667257415003, 25872280345103, 49119954154463, 93501887462903, 4103348710010689, 7846225754967739, 75162749477272151

Offset: 0

Author

Paul Barry, Aug 06 2004

Keywords

• easy
• nonn
• frac

Comments

Is this identical to A097345? - Aaron Gulliver, Jul 19 2007. The answer turns out to be No - see A134652.

If the putative formula a(n)=A081528(n) sum{k=0..n, binomial(n, k)/(k+1)^2} were true, then this sequence coincides with A097345 according to Mathar's notes. However, the term n=9 in the binomial transform of 1/(n+1)^2 has the denominator 5040=A081528(9)/4=A081528(10)/5. So the formula cannot be true. - M. F. Hasler, Jan 25 2008

a(n) is also the numerator of u(n+1) with u(n) = (1/n)*Sum_{k=1..n} (2^k-1)/k and we have the formula: polylog(2,x/(1-x)) = Sum_{n>=1} u(n)*x^n on the interval [-1/2, 1/2]. - Groux Roland, Feb 01 2009

Examples

			The first values of the binomial transform of 1/(n+1)^2 are 1, 5/4, 29/18, 103/48, 887/300, 1517/360, 18239/2940, 63253/6720, 332839/22680, 118127/5040, 2331085/60984, ...
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..500
• R. J. Mathar, Notes on an attempt to prove that A097344 and A097345 are identical

Crossrefs

Cf. A097345, A134652.

Programs

Maple

f:=n->numer(add( binomial(n,k)/(k+1)^2, k=0..n));

Mathematica

Table[HypergeometricPFQ[{1, 1, -n}, {2, 2}, -1] // Numerator, {n, 0, 24}] (* Jean-François Alcover, Oct 14 2013 *)

Maxima

a(n):=if n<0 then 1 else 1/((n+1)^2)*((n)*(3*n+1)*a(n-1)-2*(n-1)*(n)*a(n-2)+1);
makelist(num(a(n),n,0,10); /* Vladimir Kruchinin, Jun 01 2016 */

PARI

A097344(n)=numerator(sum(k=0,n,binomial(n,k)/(k+1)^2)) \\ M. F. Hasler, Jan 25 2008

Python

from fractions import Fraction
A097344_list, tlist = [1], [Fraction(1,1)]
for i in range(1,100):
    for j in range(len(tlist)):
        tlist[j] *= Fraction(i,i-j)
    tlist += [Fraction(1,(i+1)**2)]
    A097344_list.append(sum(tlist).numerator) # Chai Wah Wu, Jun 04 2015

Sage

def A097344_list(size):
    R, L = [1], [1]
    inc = sqr = 1
    for i in range(1, size):
        for j in range(i):
            L[j] *= i / (i - j)
        inc += 2; sqr += inc
        L.extend(1 / sqr)
        R.append(sum(L).numerator())
    return R
A097344_list(50) # (after Chai Wah Wu) Peter Luschny, Jun 05 2016

Formula

a(n) = numerator(b(n)), b(n) = 1/((n+1)^2)*((n)*(3*n+1)*b(n-1)-2*(n-1)*(n)*b(n-2)+1). - Vladimir Kruchinin, May 31 2016

Extensions

Edited and corrected by Daniel Glasscock (glasscock(AT)rice.edu), Jan 04 2008 and M. F. Hasler, Jan 25 2008

Moved comment on numerators of a logarithmic g.f. over to A097345 - R. J. Mathar, Mar 04 2010