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A101102
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Fifth partial sums of cubes (A000578).
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15
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1, 13, 82, 354, 1200, 3432, 8646, 19734, 41613, 82225, 153868, 274924, 472056, 782952, 1259700, 1972884, 3016497, 4513773, 6624046, 9550750, 13550680, 18944640, 26129610, 35592570, 47926125, 63846081, 84211128, 110044792, 142559824, 183185200, 233595912
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OFFSET
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1,2
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LINKS
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FORMULA
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a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(10 + 3*n*(n+5))/20160.
This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=5. - Alexander R. Povolotsky, May 17 2008
Sum_{n>=1} 1/a(n) = -162*sqrt(21/5)*Pi*tan(sqrt(35/3)*Pi/2) - 136269/100. - Amiram Eldar, Jan 26 2022
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MATHEMATICA
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Table[Binomial[n+5, 6]*(3*n^2+15*n+10)/28, {n, 1, 30}] (* G. C. Greubel, Dec 01 2018 *)
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PROG
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(PARI) a(n)=sum(t=1, n, sum(s=1, t, sum(l=1, s, sum(j=1, l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1))))))) \\ Alexander R. Povolotsky, May 17 2008
(PARI) Vec(-x*(x^2+4*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Apr 23 2015
(Magma) [Binomial(n+5, 6)*(3*n^2+15*n+10)/28: n in [1..30]]; // G. C. Greubel, Dec 01 2018
(Sage) [binomial(n+5, 6)*(3*n^2+15*n+10)/28 for n in (1..30)] # G. C. Greubel, Dec 01 2018
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
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EXTENSIONS
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STATUS
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approved
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