cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A250212 Second partial sums of seventh powers (A001015).

Original entry on oeis.org

1, 130, 2446, 21146, 117971, 494732, 1695036, 4992492, 13072917, 31153342, 68720938, 142120342, 278268263, 519829688, 932250488, 1613106744, 2704301673, 4407716634, 7005003334, 10882290034, 16560665275, 24733398404, 36310956980, 52474986980, 74742532605, 105041888406
Offset: 1

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Author

Luciano Ancora, Jan 18 2015

Keywords

Comments

The general formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) is the m-th Faulhaber’s polynomial.

Crossrefs

Cf. A239094 (same sequence, shifted by 1).

Programs

  • GAP
    List([1..30], n-> Binomial(n+2, 3)*(5*(n+1)^6 -25*(n+1)^4 +38*(n+ 1)^2 -12)/60); # G. C. Greubel, Aug 28 2019
  • Magma
    [(n*(n + 1)*(n + 2)*(5*n^6 + 30*n^5 + 50*n^4 -37*n^2 + 6*n + 6) / 360): n in [1..30]]; // Vincenzo Librandi, Jan 22 2015
    
  • Maple
    seq(binomial(n+2, 3)*(5*(n+1)^6 -25*(n+1)^4 +38*(n+1)^2 -12)/60, n=1..30); # G. C. Greubel, Aug 28 2019
  • Mathematica
    Accumulate[Accumulate[Range[25]^7]] (* Robert G. Wilson v, Jan 21 2015 *)
    Table[(n(n+1)(n+2)(5n^6+30n^5+50n^4-37n^2+6n+6)/360), {n, 30}] (* Vincenzo Librandi, Jan 22 2015 *)
    RecurrenceTable[{a[n]==2a[n-1]-a[n-2]+n^7, a[1]==1,a[2]==130}, a, {n,30}] (* Bruno Berselli, Jan 22 2015 *)
    LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,130,2446,21146,117971,494732,1695036,4992492,13072917,31153342},30] (* Harvey P. Dale, Jan 19 2020 *)
  • PARI
    vector(50, n, n*(n+1)*(n+2)*(5*n^6 + 30*n^5 + 50*n^4 - 37*n^2 + 6*n + 6)/360) \\ Michel Marcus, Jan 21 2015
    
  • Sage
    [binomial(n+2, 3)*(5*(n+1)^6 -25*(n+1)^4 +38*(n+1)^2 -12)/60 for n in (1..30)] # G. C. Greubel, Aug 28 2019
    

Formula

a(n) = n*(n+1)*(n+2)*(5*n^6 + 30*n^5 + 50*n^4 - 37*n^2 + 6*n + 6)/360.
a(n) = 2*a(n-1) - a(n-2) + n^7.
G.f.: x*(1 +120*x +1191*x^2 +2416*x^3 +1191*x^4 +120*x^5 +x^6)/(1-x)^10. - Georg Fischer, May 24 2019
a(n) = A239094(n+1). - Danny Rorabaugh, Apr 22 2015