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Search: id:A007750
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| A007750 |
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Nonnegative integers n such that n^2(n+1)(2n+1)^2(7n+1)/36 is a square. |
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+0
5
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| 0, 1, 7, 24, 120, 391, 1921, 6240, 30624, 99457, 488071, 1585080, 7778520, 25261831, 123968257, 402604224, 1975713600, 6416405761, 31487449351, 102259887960, 501823476024, 1629741801607, 7997688167041, 25973608937760
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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n^2(n+1)(2n+1)^2(7n+1)/36=sum(i=1,n,i^2)*sum(i=n+1,2n,i^2).
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REFERENCES
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Mentioned in a problem on p. 334 of Two-Year College Math. Jnl., Vol. 25, 1994.
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FORMULA
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G.f.: x(1+6x+x^2)/((1-x)(1-16x^2+x^4)). a(n)=16a(n-2)-a(n-4)+8.
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PROGRAM
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(PARI) a(n)=if(n<0, a(-1-n), if(n<2, n>0, 16*a(n-2)-a(n-4)+8))
(PARI) a(n)=local(w); if(n<0, 0, w=8+3*quadgen(28); n=((n+1)\2)*(-1)^(n%2); imag(w^n)+4*(real(w^n)-1)/7)
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CROSSREFS
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Cf. A007751, A007752.
Sequence in context: A050191 A021000 A129797 this_sequence A009643 A108095 A009646
Adjacent sequences: A007747 A007748 A007749 this_sequence A007751 A007752 A007753
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KEYWORD
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nonn
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AUTHOR
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John C. Hallyburton, Jr. [ hallyb(AT)vmsdev.enet.dec.com ].
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EXTENSIONS
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Edited by Michael Somos, Jul 27, 2002
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