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.

A079547 a(n) = ((n^6 - (n-1)^6) - (n^2 - (n-1)^2))/60.

Original entry on oeis.org

0, 1, 11, 56, 192, 517, 1183, 2408, 4488, 7809, 12859, 20240, 30680, 45045, 64351, 89776, 122672, 164577, 217227, 282568, 362768, 460229, 577599, 717784, 883960, 1079585, 1308411
Offset: 1

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Author

Xavier Acloque, Jan 22 2003

Keywords

Comments

Polynexus numbers of order 6.
A polynexus (subtractive) function is composed of two or more subtracted nexus numbers divided by an integer x. The general form of the formula is a(n)=((n^p-(n-1)^p)-(n^q-(n-1)^q))/x, where n, p, q and x are integers.
Already known: ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24, giving A006322 for n>1; ((n^4-(n-1)^4) - (n^2-(n-1)^2))/12, giving A000330; ((n^3-(n-1)^3) - (n^1-(n-1)^1))/6, giving A000217; ((n^2-(n-1)^2) - (n^1-(n-1)^1))/2, giving n; ((n^2-(n-1)^2) - (n^0-(n-1)^0))/1, giving 2*n-1. In those examples, x is equal to 1,2,6,12,24, and 3 is also possible.
Also number of monotone n-weightings of complete bipartite digraph K(3,2) if offset were 0; cf. A085464-A085465. - Goran Kilibarda, Vladeta Jovovic, Jul 01 2003
Partial sums of A037270. - J. M. Bergot, Jun 07 2012

Crossrefs

Programs

  • GAP
    List([1..30], n-> n*(6*n^4-15*n^3+20*n^2-15*n+4)/60) # G. C. Greubel, Jun 19 2019
  • Magma
    [n*(6*n^4-15*n^3+20*n^2-15*n+4)/60: n in [1..30]]; // G. C. Greubel, Jun 19 2019
    
  • Mathematica
    Table[((n^6 -(n-1)^6) - (n^2 -(n-1)^2))/60, {n, 30}] (* Bruno Berselli, Feb 13 2012 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,11,56,192,517},30] (* Harvey P. Dale, Feb 21 2023 *)
  • PARI
    a(n) = n*(6*n^4-15*n^3+20*n^2-15*n+4)/60 \\ Charles R Greathouse IV, Jan 16 2013
    
  • Sage
    [n*(6*n^4-15*n^3+20*n^2-15*n+4)/60 for n in (1..30)] # G. C. Greubel, Jun 19 2019
    

Formula

a(n+1) = Sum_{i=1..n} (i^2 + i^4)/2 = n*(2*n+1)*(n+1)*(3*n^2+3*n+4)/60. - Vladeta Jovovic, Mar 17 2006
G.f.: x^2*(x+1)*(1+4*x+x^2)/(1-x)^6. - Bruno Berselli, Feb 13 2012
a(n) = Sum_{i=1..n-1} Sum_{j=1..n-1} min(i,j)^3. - Enrique Pérez Herrero, Jan 16 2013
E.g.f.: x^2*(30 + 80*x + 45*x^2 + 6*x^3)*exp(x)/60. - G. C. Greubel, Jun 19 2019