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.

A000540 Sum of 6th powers: 0^6 + 1^6 + 2^6 + ... + n^6.

Original entry on oeis.org

0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, 1978405, 3749966, 6735950, 11562759, 19092295, 30482920, 47260136, 71397705, 105409929, 152455810, 216455810, 302221931, 415601835, 563637724, 754740700, 998881325, 1307797101, 1695217590
Offset: 0

Views

Author

Keywords

Comments

This sequence is related to A000539 by a(n) = n*A000539(n)-sum(A000539(i), i=0..n-1). - Bruno Berselli, Apr 26 2010

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.
  • J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, (2008), p. 289.
  • 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

Row 6 of array A103438.
Partial sums of A001014.

Programs

  • Haskell
    a000540 n = a000540_list !! n
    a000540_list = scanl1 (+) a001014_list -- Reinhard Zumkeller, Dec 04 2011
    
  • Magma
    [n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015
  • Maple
    a:=n->sum (j^6,j=0..n): seq(a(n),n=0..27); # Zerinvary Lajos, Jun 27 2007
    A000540:=(z+1)*(z**4+56*z**3+246*z**2+56*z+1)/(z-1)**8; # g.f. by Simon Plouffe in his 1992 dissertation, without the leading 0.
    A000540 := proc(n) n^7/7+n^6/2+n^5/2-n^3/6+n/42 ; end proc: # R. J. Mathar
  • Mathematica
    Accumulate[Range[0,30]^6] (* Harvey P. Dale, Jul 30 2009 *)
    LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 1, 65, 794, 4890, 20515, 67171, 184820}, 31] (* Jean-François Alcover, Feb 09 2016 *)
  • PARI
    a(n)=n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42 \\ Edward Jiang, Sep 10 2014
    
  • PARI
    a(n)=sum(i=1, n, i^6); \\ Michel Marcus, Sep 11 2014
    
  • Python
    A000540_list, m = [0], [720, -1800, 1560, -540, 62, -1, 0, 0]
    for _ in range(10**2):
        for i in range(7):
            m[i+1] += m[i]
        A000540_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Sage
    [bernoulli_polynomial(n,7)/7 for n in range(1, 29)]# Zerinvary Lajos, May 17 2009
    

Formula

a(n) = n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^6). - Alexander Adamchuk, Oct 26 2004
G.f.: A(x) = 3*x/7*G(0); with G(k) = 1 + 2/(k+1+(k+1)/(2*k^2 + 4*k + 1 + 2*(k+1)^2/(3*k + 2 - 9*x*(k+1)*(k+2)^4*(k+3)*(2*k+5)/(3*x*(k+2)^4*(k+3)*(2*k+5)+(k+1)*(2*k+3)/G(k+1))))); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
G.f.: x*(1+x)*(x^4 + 56*x^3 + 246*x^2 + 56*x + 1) / (x-1)^8 . - R. J. Mathar, Aug 07 2012
a(n) = Sum_{i=1..n} J_6(i)*floor(n/i), where J_6 is A069091. - Enrique Pérez Herrero, Mar 09 2013
a(n) = 7*a(n-1) - 21* a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + 720. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..6} j*Stirling1(n+1,n+1-j)*Stirling2(n+6-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 84*Pi*(8*cos(sqrt((sqrt(93) + 9)/6)*Pi) + 15*cos(sqrt((sqrt(93) + 9)/6)*Pi/2) * cosh(sqrt((sqrt(93) - 9)/6)*Pi/2) + 8*cosh(sqrt((sqrt(93) - 9)/6)*Pi) - 7*sqrt(3)*sin(sqrt((sqrt(93) + 9)/6)*Pi/2) * sinh(sqrt((sqrt(93) - 9)/6)*Pi/2)) / (31*(cos(sqrt((sqrt(93) + 9)/6)*Pi) + cosh(sqrt((sqrt(93) - 9)/6)*Pi))) = 0.985708051237101247832970793342271511... . - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + z)*(n + 1/2 - z)*(n + 1/2 + zbar)*(n + 1/2 - zbar)/7, with I^2 = -1 and z = 2^(-3/2)*3^(-1/4)*(sqrt(sqrt(31) + 3*sqrt(3)) + I*sqrt(sqrt(31) - 3*sqrt(3))), and zbar is the complex conjugate of z. See the Graham et al. reference, eq. (6.98), pp. 288-289 (with n -> n+1). (There was a typo in the first edition, which was corrected in the second edition.) - Wolfdieter Lang, Apr 03 2015
a(n+2) = 36*A086020(n+1) + 24*A005585(n+1) + A000330(n+2). - Yasser Arath Chavez Reyes, Apr 16 2024