A134652 -id:A134652 - cp's OEIS frontend

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

Paul Barry, Aug 06 2004

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

			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, ...

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

Cf. A097345, A134652.

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

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

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 */

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

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

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

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

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

A097345 Numerators of the partial sums of the binomial transform of 1/(n+1).

Original entry on oeis.org

Offset: 0

Paul Barry, Aug 06 2004

Numerators in the expansion of log((1-x)/(1-2x)) / (1-x) are 0,1,5,29,.. - Paul Barry, Feb 09 2005
Is this identical to A097344? - Aaron Gulliver, Jul 19 2007. The answer turns out to be No - see A134652.
From n=9 on, the putative formula a(n)=A003418(n+1)*sum{k=0..n, (2^(k+1)-1)/(k+1)} is false. The least n for which a(n) is different from A097344(n) is n=59, then they agree again until n=1519. - M. F. Hasler, Jan 25 2008

R. J. Mathar, Notes on an attempt to prove that A097344 and A097345 are identical

Cf. A097344, A134652.

Table[ Sum[(2^(k+1)-1)/(k+1), {k, 0, n}] // Numerator, {n, 0, 21}] (* Jean-François Alcover, Oct 14 2013, after Pari *)

A097345(n) = numerator(sum(k=0,n,(2^(k+1)-1)/(k+1)))

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

Moved comment concerning numerators of the logarithm from A097344 to here where it is correct - R. J. Mathar, Mar 04 2010

cp's OEIS Frontend

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

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