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 1-4 of 4 results.

A215052 a(n) = (binomial(n,5) - floor(n/5)) / 5.

Original entry on oeis.org

1, 4, 11, 25, 50, 92, 158, 257, 400, 600, 873, 1237, 1713, 2325, 3100, 4069, 5266, 6729, 8500, 10625, 13155, 16145, 19655, 23750, 28500, 33981, 40274, 47466, 55650, 64925, 75397, 87178, 100387, 115150, 131600, 149878, 170132, 192518, 217200
Offset: 6

Views

Author

Peter Bala, Aug 01 2012

Keywords

Comments

Apparently a duplicate of A036837. - R. J. Mathar, Aug 06 2012
Not the same as A011851.
Let p be a prime. Saikia and Vogrinc have proved that (1/p)*{binomial(n,p) - floor(n/p)} is an integer sequence. The present sequence is the case p = 5. Other cases are A002620 (p = 2), A014125 (p = 3), A215053 (p = 7) and A215054 (p = 11).
There is a connection between these sequences and A178904. For a fixed prime p the o.g.f. for the sequence (1/p)*{binomial(n,p) - floor(n/p)} is a rational function of the form x^(p+1)*R(p,x)/((1-x^p)*(1-x)^p). The polynomial R(p,x) = sum {k = 0..p-1} (1/p)*{1 - (-1)^k*binomial(p-1,k)}*x^(k-1). For prime p >= 3, -R(p,x) is equal to the p-th row polynomial of A178904.

Crossrefs

Cf. A002620 (p = 2), A014125 (p = 3), A178904, A215053 (p = 7), A215054( p = 11).

Formula

a(n) = (1/5)*{binomial(n,5) - floor(n/5)}.
O.g.f.: sum {n>=0} a(n)*x^n = x^6*(1-x+x^2)/((1-x^5)*(1-x)^5) = x^6*(1 + 4*x + 11*x^2 + 25*x^3 + ...).

A215054 a(n) = 1/11*(binomial(n,11) - floor(n/11)).

Original entry on oeis.org

1, 7, 33, 124, 397, 1125, 2893, 6871, 15269, 32065, 64130, 122916, 226922, 405218, 702378, 1185263, 1952198, 3145208, 4966118, 7697483, 11729498, 17594247, 26008887, 37929627, 54618663, 77726559, 109392935, 152368731, 210163767, 287223815, 389141943
Offset: 12

Views

Author

Peter Bala, Aug 01 2012

Keywords

Comments

Let p be a prime. Saikia and Vogrinc have proved that 1/p*{binomial(n,p) - floor(n/p)} is an integer sequence. The present sequence is the case p = 11. Other cases are A002620 (p = 2), A014125 (p = 3), A215052 (p = 5) and A215053 (p = 7).

Crossrefs

A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215053(p = 7).
Partial sums of A032169.

Programs

  • Mathematica
    Table[(Binomial[n,11]-Floor[n/11])/11,{n,12,50}] (* Harvey P. Dale, Aug 06 2012 *)
  • Maxima
    A215054(n):=1/11*(binomial(n,11) - floor(n/11))$ makelist(A215054(n),n,12,30); /* Martin Ettl, Oct 25 2012 */

Formula

a(n) = 1/11*(binomial(n,11) - floor(n/11)).
O.g.f.: sum_{n>=0} a(n)*x^n = x^12*(1 - 4*x + 11*x^2 - 19*x^3 + 23*x^4 - 19*x^5 + 11*x^6 - 4*x^7 + x^8)/((1-x^11)*(1-x)^11) = x^12*(1 + 7*x + 33*x^2 + 124*x^3 + ...). The numerator polynomial 1 - 4*x + 11*x^2 - 19*x^3 + 23*x^4 - 19*x^5 + 11*x^6 - 4*x^7 + x^8 is the negative of the row generating polynomial for row 11 of A178904.

A286033 a(n) = binomial(2*n-2, n-1) + (-1)^n.

Original entry on oeis.org

0, 3, 5, 21, 69, 253, 923, 3433, 12869, 48621, 184755, 705433, 2704155, 10400601, 40116599, 155117521, 601080389, 2333606221, 9075135299, 35345263801, 137846528819, 538257874441, 2104098963719, 8233430727601, 32247603683099, 126410606437753, 495918532948103
Offset: 1

Views

Author

Peter Luschny, May 13 2017

Keywords

Comments

An odd prime p divides a((p+1)/2) which gives A163210.

Crossrefs

Programs

  • Magma
    [Binomial(2*n-2, n-1) + (-1)^n: n in [1..30]]; // G. C. Greubel, Jul 14 2024
    
  • Maple
    a := n -> binomial(2*n-2, n-1) + (-1)^n: seq(a(n), n=1..27);
  • Mathematica
    a[n_] := Binomial[2n-2, n-1] + (-1)^n; a[Range[1,27]]
  • Maxima
    a(n):=-sum((-1)^k*binomial(2*n,n-k)*(fib(2*k+1)+fib(2*k-1)),k,1,n); /* Vladimir Kruchinin, Jan 18 2025 */
  • PARI
    a(n) = binomial(2*n-2, n-1) + (-1)^n \\ David A. Corneth, May 13 2017
    
  • SageMath
    def A286033(n): return binomial(2*n-2, n-1) + (-1)^n
    [A286033(n) for n in range(1,31)] # G. C. Greubel, Jul 14 2024
    

Formula

a(n) = A000984(n-1) + A033999(n). - David A. Corneth, May 13 2017
G.f.: -1 + x/sqrt(1 - 4*x) + 1/(1 + x). - Ilya Gutkovskiy, May 13 2017
D-finite with recurrence: -(n-1)*a(n) +2*(n-1)*a(n-1) +(7*n-17)*a(n-2) +2*(2*n-7)*a(n-3)=0. - R. J. Mathar, Jan 27 2020
a(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(2*n, n-k)*A000032(2*k). - Vladimir Kruchinin, Jan 18 2025

A215053 a(n) = 1/7*( binomial(n,7) - floor(n/7) ).

Original entry on oeis.org

1, 5, 17, 47, 113, 245, 490, 919, 1634, 2778, 4546, 7198, 11074, 16611, 24363, 35022, 49443, 68671, 93971, 126861, 169148, 222968, 290828, 375653, 480836, 610292, 768516, 960645, 1192525, 1470781, 1802893, 2197276, 2663365, 3211705, 3854046
Offset: 8

Views

Author

Peter Bala, Aug 01 2012

Keywords

Comments

Not the same as A011853.
Let p be a prime. Saikia and Vogrinc have proved that 1/p*{binomial(n,p) - floor(n/p)} is an integer sequence. The present sequence is the case p = 7. Other cases are A002620 (p = 2), A014125 (p = 3), A215052 (p = 5) and A215054 (p = 11).

Crossrefs

Cf. A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215054 (p = 11).

Programs

  • Magma
    [(Binomial(n, 7)-Floor(n/7))/7: n in [8..50]]; // Vincenzo Librandi, Jun 23 2015
  • Mathematica
    Table[(Binomial[n,7]-Floor[n/7])/7,{n,8,50}] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,2,-7,21,-35,35,-21,7,-1},{1,5,17,47,113,245,490,919,1634,2778,4546,7198,11074,16611},40] (* Harvey P. Dale, Dec 23 2014 *)
  • PARI
    a(n) = (binomial(n, 7) - n\7) / 7; \\ Michel Marcus, Jan 23 2014
    

Formula

O.g.f.: sum {n>=0} a(n)*x^n = x^8*(1 - 2*x + 3*x^2 - 2*x^3 + x^4)/((1-x^7)*(1-x)^7) = x^8*(1 + 5*x + 17*x^2 + 47*x^3 + ...). The numerator polynomial 1 - 2*x + 3*x^2 - 2*x^3 + x^4 is the negative of the row generating polynomial for row 7 of A178904.
Showing 1-4 of 4 results.