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A097344
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Numerators in binomial transform of 1/(n+1)^2.
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4
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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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((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]
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LINKS
| R. J. Mathar, Notes on an attempt to prove that A097344 and A097345 are identical
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EXAMPLE
| 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, ...
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MAPLE
| f:=n->add( binomial(n, k)/(k+1)^2, k=0..n);
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PROG
| (PARI) A097344(n)=numerator(sum(k=0, n, binomial(n, k)/(k+1)^2)) \\ - M. F. Hasler, Jan 25 2008
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CROSSREFS
| Cf. A097345, A134652.
Sequence in context: A205172 A139856 A097345 * A153076 A034700 A057721
Adjacent sequences: A097341 A097342 A097343 * A097345 A097346 A097347
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KEYWORD
| easy,nonn,frac
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Aug 06 2004
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EXTENSIONS
| 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
Moved comment on numerators of a logarithmic g.f. over to A097345 - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 04 2010
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