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.

Showing 1-3 of 3 results.

A071246 a(n) = n*(n - 1)*(2*n^2 + n + 2)/6.

Original entry on oeis.org

0, 0, 4, 23, 76, 190, 400, 749, 1288, 2076, 3180, 4675, 6644, 9178, 12376, 16345, 21200, 27064, 34068, 42351, 52060, 63350, 76384, 91333, 108376, 127700, 149500, 173979, 201348, 231826, 265640, 303025, 344224, 389488, 439076, 493255, 552300, 616494
Offset: 0

Views

Author

N. J. A. Sloane, Jun 12 2002

Keywords

References

  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Crossrefs

Programs

  • Magma
    [n*(n-1)*(2*n^2+n+2)/6: n in [0..40]]; // Vincenzo Librandi, Jun 14 2011
    
  • Mathematica
    CoefficientList[Series[-(4x^2 + 3x^3 + x^4)/(x - 1)^5, {x, 0, 50}], x] (* or *) Table[n*(n - 1)*(2*n^2 + n + 2)/6, {n, 0, 50}] (* Indranil Ghosh, Apr 05 2017 *)
  • PARI
    a(n) = n*(n - 1)*(2*n^2 + n + 2)/6; \\ Indranil Ghosh, Apr 05 2017
    
  • Python
    def a(n): return n*(n - 1)*(2*n**2 + n + 2)/6 # Indranil Ghosh, Apr 05 2017
    
  • SageMath
    def A071246(n): return binomial(n,2)*(2+n+2*n^2)//3
    [A071246(n) for n in range(41)] # G. C. Greubel, Aug 07 2024

Formula

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4, a(0)=0, a(1)=0, a(2)=4, a(3)=23, a(4)=76. - Yosu Yurramendi, Sep 03 2013
From Indranil Ghosh, Apr 05 2017: (Start)
G.f.: x^2*(4 + 3*x + x^2)/(1 - x)^5.
E.g.f.: exp(x)*x^2*(4+x)*(3+2*x)/6. (End)

A277228 Convolution of the even-indexed triangular numbers (A014105) and the squares (A000290).

Original entry on oeis.org

0, 0, 3, 22, 88, 258, 623, 1316, 2520, 4476, 7491, 11946, 18304, 27118, 39039, 54824, 75344, 101592, 134691, 175902, 226632, 288442, 363055, 452364, 558440, 683540, 830115, 1000818, 1198512, 1426278, 1687423, 1985488, 2324256, 2707760, 3140291, 3626406
Offset: 0

Views

Author

Wolfdieter Lang, Oct 20 2016

Keywords

Comments

This sequence was originally proposed in a comment on A071245 by J. M. Bergot as giving the first differences. Therefore, a(n) gives the partial sums of A071245.

Crossrefs

Programs

  • Magma
    [Binomial(n+1, 3)*(4*n^2 +5*n +4)/10: n in [0..40]]; // G. C. Greubel, Oct 22 2018
  • Mathematica
    Table[(n - 1) n (n + 1) (4 n^2 + 5 n + 4)/60, {n, 0, 40}] (* Bruno Berselli, Oct 21 2016 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,0,3,22,88,258},40] (* Harvey P. Dale, Jun 04 2023 *)
  • PARI
    concat(vector(2), Vec(x^2*(1+x)*(3+x)/(1-x)^6 + O(x^50))) \\ Colin Barker, Oct 21 2016
    

Formula

O.g.f.: x^2*(1 + x)*(3 + x)/(1 - x)^6 = (x*(3 + x)/(1 - x)^3)*(x*(1 + x)/(1 - x)^3).
a(n) = Sum_{k=0..n} A014105(n-k)*A000290(k).
a(n) = binomial(n+1, 3)*(4*n^2 + 5*n + 4)/10 = (n - 1)*n*(n + 1)*(4*n^2 + 5*n + 4)/60.
a(n) = Sum_{k=0..n} A071245(k).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5. - Colin Barker, Oct 21 2016

A071252 a(n) = n*(n - 1)*(n^2 + 1)/2.

Original entry on oeis.org

0, 0, 5, 30, 102, 260, 555, 1050, 1820, 2952, 4545, 6710, 9570, 13260, 17927, 23730, 30840, 39440, 49725, 61902, 76190, 92820, 112035, 134090, 159252, 187800, 220025, 256230, 296730, 341852, 391935, 447330, 508400, 575520, 649077, 729470, 817110, 912420
Offset: 0

Views

Author

N. J. A. Sloane, Jun 12 2002

Keywords

References

  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Crossrefs

Programs

  • Magma
    [n*(n-1)*(n^2+1)/2: n in [0..40]]; // Vincenzo Librandi, Jun 14 2011
    
  • Mathematica
    f[n_] := n (n - 1) (n^2 + 1)/2 (* Or *) f[n_] := Floor[n^5/(n + 1)]/2; Array[f, 38, 0] (* Robert G. Wilson v, Apr 01 2012 *)
  • PARI
    a(n)=n*(n-1)*(n^2+1)/2; \\ Joerg Arndt, Sep 04 2013
    
  • Python
    def a(n): return  n*(n - 1)*(n**2 + 1)/2 # Indranil Ghosh, Apr 05 2017
    
  • SageMath
    def A071252(n): return binomial(n,2)*(1+n^2)
    [A071252(n) for n in range(41)] # G. C. Greubel, Aug 07 2024

Formula

a(n) = floor(n^5/(n+1))/2. - Gary Detlefs, Mar 31 2011
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) n>4, a(0)=0, a(1)=0, a(2)=5, a(3)=30, a(4)=102. - Yosu Yurramendi, Sep 03 2013
G.f.: x^2*(5+5*x+2*x^2)/(1-x)^5. - Joerg Arndt, Sep 04 2013
From Indranil Ghosh, Apr 05 2017: (Start)
a(n) = A002378(n) * A002522(n) / 2.
E.g.f.: exp(x)*x^2*(5 + 5*x + x^2)/2.
(End)
Showing 1-3 of 3 results.