%I A097344
%S A097344 1,5,29,103,887,1517,18239,63253,332839,118127,2331085,4222975,
%T A097344 100309579,184649263,1710440723,6372905521,202804884977,381240382217,
%U A097344 13667257415003,25872280345103,49119954154463,93501887462903,4103348710010689,
7846225754967739,75162749477272151
%N A097344 Numerators in binomial transform of 1/(n+1)^2.
%C A097344 Numerators in the expansion of ln((1-x)/(1-2x)) / (1-x) are 0,1,5,29,
.. - Paul Barry (pbarry(AT)wit.ie), Feb 09 2005
%C A097344 Is this identical to A097345? - Aaron Gulliver, Jul 19 2007. The answer
turns out to be No - see A134652.
%C A097344 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
%C A097344 a(n) is also the numerator of u(n+1), with u(n)=(1/n)*sum((2^k-1)/k,k=1..n)and
we have the formula : polylog(2,x/(1-x))=sum(u(n)*x^n, n=1..infinity)
on the interval [ -1/2,1/2]. [From Groux Roland (roland.groux(AT)orange.fr),
Feb 01 2009]
%H A097344 R. J. Mathar, <a href="a097345.pdf">Notes on an attempt to prove that
A097344 and A097345 are identical</a>
%e A097344 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, ...
%p A097344 f:=n->add( binomial(n,k)/(k+1)^2, k=0..n);
%o A097344 (PARI) A097344(n)=numerator(sum(k=0,n,binomial(n,k)/(k+1)^2)) \\ - M.
F. Hasler, Jan 25 2008
%Y A097344 Cf. A097345, A134652.
%Y A097344 Sequence in context: A111937 A139856 A097345 this_sequence A153076 A034700
A057721
%Y A097344 Adjacent sequences: A097341 A097342 A097343 this_sequence A097345 A097346
A097347
%K A097344 easy,nonn,frac
%O A097344 0,2
%A A097344 Paul Barry (pbarry(AT)wit.ie), Aug 06 2004
%E A097344 Edited and corrected by Daniel Glasscock (glasscock(AT)rice.edu), Jan
04 2008 and M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Jan 25
2008
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