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Search: id:A115257
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| A115257 |
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Partial sums of binomial(2n,n)^2. |
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+0
2
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| 1, 5, 41, 441, 5341, 68845, 922621, 12701245, 178338145, 2542242545, 36677022081, 534311328705, 7846771001041, 116019251361041, 1725360846921041, 25786805857871441, 387084441100423541, 5832802431123111941
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Central coefficients of number triangle A115255.
p divides all a(n) from a((p-1)/2) to a(p-1) for Gaussian primes p=7,23,31,79,167,431,479,983, ... of the form 4n+3, A002145(n), and for primes of the form 8n+7, A007522(n). - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
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FORMULA
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a(n)=sum{k=0..n, C(2k, k)^2}. a(n)=A115255(2n, n).
C(2n,n)^2 + C(2n-2,n-1)^2 + ... + C(2k,k)^2 + ... + C(2,1)^2 + C(0,0)^2, where C(2k,k)=(2k)!/(k!)^2 are the central binomial coefficients A000984[k]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
a(n) = Sum[((2k)!/(k!)^2)^2,{k,0,n}]. a(n) = Sum[A000984[k]^2,{k,0,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
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MATHEMATICA
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Table[Sum[((2k)!/(k!)^2)^2, {k, 0, n}], {n, 0, 40}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
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CROSSREFS
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Cf. A000984, A002145, A007522.
Sequence in context: A081215 A140095 A083073 this_sequence A047735 A096364 A049119
Adjacent sequences: A115254 A115255 A115256 this_sequence A115258 A115259 A115260
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KEYWORD
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easy,nonn
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), Jan 18 2006
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