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.

A000538 Sum of fourth powers: 0^4 + 1^4 + ... + n^4.

Original entry on oeis.org

0, 1, 17, 98, 354, 979, 2275, 4676, 8772, 15333, 25333, 39974, 60710, 89271, 127687, 178312, 243848, 327369, 432345, 562666, 722666, 917147, 1151403, 1431244, 1763020, 2153645, 2610621, 3142062, 3756718, 4463999, 5273999, 6197520, 7246096, 8432017, 9768353
Offset: 0

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Keywords

Comments

This sequence is related to A000537 by the transform a(n) = n*A000537(n) - Sum_{i=0..n-1} A000537(i). - Bruno Berselli, Apr 26 2010
A formula for the r-th successive summation of k^4, for k = 1 to n, is ((12*n^2+(12*n-5)*r+r^2)*(2*n+r)*(n+r)!)/((r+4)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
The number of four dimensional hypercubes in a 4D grid with side lengths n. This applies in general to k dimensions. That is, the number of k-dimensional hypercubes in a k-dimensional grid with side lengths n is equal to the sum of 1^k + 2^k + ... + n^k. - Alejandro Rodriguez, Oct 20 2020

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 222.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a000538 n = (3 * n * (n + 1) - 1) * (2 * n + 1) * (n + 1) * n `div` 30
    -- Reinhard Zumkeller, Nov 11 2012
    
  • Magma
    [n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30: n in [0..35]]; // Vincenzo Librandi, Apr 04 2015
  • Maple
    A000538 := n-> n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30;
  • Mathematica
    Accumulate[Range[0,40]^4] (* Harvey P. Dale, Jan 13 2011 *)
    CoefficientList[Series[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 07 2015 *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 17, 98, 354, 979}, 35] (* Jean-François Alcover, Feb 09 2016 *)
    Table[x^5/5+x^4/2+x^3/3-x/30,{x,40}] (* Harvey P. Dale, Jun 06 2021 *)
  • Maxima
    A000538(n):=n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30$
    makelist(A000538(n),n,0,30); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30 \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    concat(0, Vec(x*(1+11*x+11*x^2+x^3)/(1-x)^6 + O(x^100))) \\ Altug Alkan, Dec 07 2015
    
  • Python
    A000538_list, m = [0], [24, -36, 14, -1, 0, 0]
    for _ in range(10**2):
        for i in range(5):
            m[i+1] += m[i]
        A000538_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Python
    def A000538(n): return n*(n**2*(n*(6*n+15)+10)-1)//30 # Chai Wah Wu, Oct 03 2024
    
  • Sage
    [bernoulli_polynomial(n,5)/5 for n in range(1, 35)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30.
The preceding formula is due to al-Kachi (1394-1437). - Juri-Stepan Gerasimov, Jul 12 2009
G.f.: x*(x+1)*(1+10*x+x^2)/(1-x)^6. Simon Plouffe in his 1992 dissertation. More generally, the o.g.f. for Sum_{k=0..n} k^m is x*E(m, x)/(1-x)^(m+2), where E(m, x) is the Eulerian polynomial of degree m (cf. A008292). The e.g.f. for these o.g.f.s is: x/(1-x)^2*(exp(y/(1-x))-exp(x*y/(1-x)))/(exp(x*y/(1-x))-x*exp(y/(1-x))). - Vladeta Jovovic, May 08 2002
a(n) = Sum_{i = 1..n} J_4(i)*floor(n/i), where J_4 is A059377. - Enrique Pérez Herrero, Feb 26 2012
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 24. - Ant King, Sep 23 2013
a(n) = -Sum_{j=1..4} j*Stirling1(n+1,n+1-j)*Stirling2(n+4-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = -30*(4 + 3/cos(sqrt(7/3)*Pi/2))*Pi/7. - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + sqrt(7/12))*(n + 1/2 - sqrt(7/12))/5, see the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015

Extensions

The general V. Jovovic formula has been slightly changed after his approval by Wolfdieter Lang, Nov 03 2011