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Search: id:A110494
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| A110494 |
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Least k such that prime(n)^2 divides binomial(2k,k). |
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+0
2
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| 3, 5, 13, 25, 61, 85, 145, 181, 265, 421, 481, 685, 841, 925, 1105, 1405, 1741, 1861, 2245, 2521, 2665, 3121, 3445, 3961, 4705, 5101, 5305, 5725, 5941, 6385, 8065, 8581, 9385, 9661, 11101, 11401, 12325, 13285, 13945, 14965
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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For prime p > sqrt(2n), p^2 does not divide binomial(2n,n).
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FORMULA
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a(n)=(prime(n)^2+1)/2 for n>1
a(n)=A066885(n), n>1. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 18 2008]
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MATHEMATICA
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t=Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>1&]; If[s=={}, 0, s[[ -1, 1]]], {n, 100}]; Table[p=Prime[i]; First[Flatten[Position[t, p]]], {i, PrimePi[Max[t]]}]
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CROSSREFS
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Cf. A110493 (largest prime p such that p^2 divides binomial(2n, n)).
Sequence in context: A026766 A026709 A082010 this_sequence A098615 A026720 A026003
Adjacent sequences: A110491 A110492 A110493 this_sequence A110495 A110496 A110497
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KEYWORD
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nonn,new
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Jul 22 2005
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