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 1-5 of 5 results.

A046662 Sum of mistyped version of binomial coefficients.

Original entry on oeis.org

1, 2, 7, 52, 749, 17686, 614227, 29354312, 1844279257, 147273109354, 14561325802271, 1745720380045852, 249461639720702917, 41886684733511640062, 8164388189339113521259, 1828191138807263097870256, 466057478369217965809683377, 134193343258948416556377786322
Offset: 0

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Author

Keywords

Comments

Origin of the name of this sequence: Binomial coefficients are n!/((n-k)!*k!) but if parentheses are omitted in the denominator, the formula might result in n!/(n-k)!*k! = n!*k!/(n-k)! and the sum giving a(n) instead of 2^n. If k! is forgotten altogether, one gets A000522. - Olivier Gérard, Mar 05 2024
Binomial transform of (n!)^2. - Peter Luschny, May 31 2014

Crossrefs

Cf. A000522 (Total number of ordered k-subsets of [1,n], k=0..n.)

Programs

  • Mathematica
    Table[Sum[(n!k!)/(n-k)!,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Sep 29 2012 *)

Formula

a(n) = Sum_{k=0..n} n!*k!/(n-k)!.
E.g.f.: exp(x)*F(x), with F(x) = Sum_{k>=0} k!*x^k. - Ralf Stephan, Apr 02 2004
a(n) = n^2*a(n - 1) - n*(n - 1)*a(n - 2) + 1. - Vladeta Jovovic, Jul 15 2004
From Peter Bala, Nov 26 2017: (Start)
a(k) == a(0) (mod k) for all k (by the inhomogeneous recurrence equation).
More generally, a(n+k) = a(n) (mod k) for all n and k (by an induction argument on n). It follows that for each positive integer k, the sequence a(n) (mod k) is periodic, with the exact period dividing k. For example, modulo 10 the sequence becomes 1, 2, 7, 2, 9, 6, 7, 2, 7, 4, 1, 2, 7, 2, 9, 6, 7, 2, 7, 4, ... with exact period 10. (End)
G.f.: Sum_{k>=0} (k!)^2*x^k/(1 - x)^(k+1). - Ilya Gutkovskiy, Apr 12 2019
a(n) ~ (n!)^2. - Vaclav Kotesovec, May 03 2021
a(n) = 3F0(1,1,-n;;-1). - R. J. Mathar, Jun 26 2024

Extensions

Corrected and extended by Harvey P. Dale, Sep 29 2012

A323280 a(n) = Sum_{k=0..n} binomial(n,k) * k^(2*k).

Original entry on oeis.org

1, 2, 19, 781, 68553, 10100761, 2236373953, 693667946945, 286962262702657, 152652510206521921, 101513694573289791441, 82511051259976074269425, 80480313356721971865934369, 92773167329045961244649105633, 124768226258051318899374299271601
Offset: 0

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Author

Seiichi Manyama, Jan 12 2019

Keywords

Crossrefs

Programs

  • Mathematica
    Table[1 + Sum[Binomial[n, k]*k^(2*k), {k, 1, n}], {n, 0, 15}] (* Vaclav Kotesovec, May 31 2019 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*k^(2*k));
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(sum(k=0, N, (k^2*x)^k/(1-x)^(k+1))) \\ Seiichi Manyama, Jul 04 2022
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(x)*sum(k=0, N, (k^2*x)^k/k!))) \\ Seiichi Manyama, Jul 04 2022

Formula

a(n) ~ n^(2*n). - Vaclav Kotesovec, May 31 2019
From Seiichi Manyama, Jul 04 2022: (Start)
G.f.: Sum_{k>=0} (k^2 * x)^k/(1 - x)^(k+1).
E.g.f.: exp(x) * Sum_{k>=0} (k^2 * x)^k/k!. (End)

A177840 Consider the n pairs (1,2), ..., (2n-1,2n); a(n) is the number of permutations of [ 2n ] with no two fixed points for any pair.

Original entry on oeis.org

1, 1, 21, 653, 37577, 3434169, 457819549, 83900098309, 20238575173137, 6217167231292913, 2369809434953636261, 1097587512530348834301, 607119566298408076479961, 395312612701784187384578473, 299298318246814086742418737197, 260721599469397754183307347278709
Offset: 0

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Author

Paul Weisenhorn, May 14 2010

Keywords

Comments

Inverse binomial transform of (2n)!. - Peter Luschny, May 31 2014
Also, the number of permutations of [2n] with no two cycle (2i-1,2i) for any pair. The number of permutation where no such pair is exchanged or fixed pointwise is A116218. - Aaron Meyerowitz, Jul 22 2023

Examples

			a(2) = 21, because there are 4! = 24 permutations of [ 4 ], only 3 of them have pairs with 2 fixed points: [1,2,3,4], [1,2,4,3], [2,1,3,4].
a(3) = A(3,0) = 653, A(3,1) = 63, A(3,2) = 3, A(3,4) = 1, sum = 720 = 6!.
		

Crossrefs

Programs

  • Maple
    f:= proc(n) option remember;
          `if`(n<2, 1-n, (n-1) *(f(n-1)+f(n-2)))
        end:
    a:= n-> add(binomial(n,j) *2^j *f(2*n-j), j=0..n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Sep 06 2011
  • Mathematica
    f[n_] := f[n] = If[n<2, 1-n, (n-1)*(f[n-1]+f[n-2])]; a[n_] := Sum[Binomial[ n, j]*2^j*f[2*n-j], {j, 0, n}]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Feb 25 2017, after Alois P. Heinz *)

Formula

a(n) = Sum_{j=0..n} C(n,j) * 2^j * f(2*n-j), where f(n) is the number of permutations of [n] with no fixed-points (A000166).
a(n) = A(n,0), with A(n,s) = number of permutations of [2n] with exactly s pairs with 2 fixed points:
A(n,s) = (n!/s!) * Sum_{j=0..n-s} 1/(j!*(n-s-j)!) * 2^j * f(2*(n-s)-j).
A(n,n) = 1, A(n,n-1) = n, A(n,n-2) = 21*n!/(2*(n-2)!).
Sum_{s=0..n} A(n,s) = (2*n)!.
a(n) = Sum_{j=0..n} C(n,j)*(2*n-2*j)!*(-1)^j. - Tani Akinari, Feb 01 2015
A(n,s) = Sum_{j=s..n} C(n,j)*C(j,s)*(2*n-2*j)!*(-1)^(j-s). - Tani Akinari, Feb 01 2015
From Peter Bala, Mar 07 2015: (Start)
a(n) = Integral_{x = 0..oo} (x^2 - 1)^n*exp(-x) dx.
For n >= 1, Integral_{x = 0..1} (x^2 - 1)^n*exp(x) dx = A007060(n)*e - a(n). Hence lim_{n->oo} a(n)/A007060(n) = e.
O.g.f. with a(0) := 1: Sum_{k >= 0} (2*k)!*x^k/(1 + x)^(k + 1) = 1 + x + 21*x^2 + 653*x^3 + ....
a(n) = 2*n*(2*n - 1)*a(n-1) + 4*n*(n - 1)*a(n-2) + (-1)^n, with initial conditions a(0) = 1, a(1) = 1.
Homogeneous recurrence: a(n) = (4*n^2 - 2*n - 1)*a(n-1) + 2*(n - 1)*(4*n - 3)*a(n-2) + 4*(n - 1)*(n - 2)*a(n-3), with initial conditions a(0) = 1, a(1) = 1 and a(2) = 21. Cf. A064570. (End)
a(n) ~ sqrt(Pi) * 2^(2*n+1) * n^(2*n + 1/2) / exp(2*n). - Vaclav Kotesovec, Mar 10 2015
a(n) = (2*n)!*hypergeom([],[1/2-n],1/4)+(-1)^n*(1-hypergeom([1],[1/2,n+1],1/4)). - Peter Luschny, Mar 15 2015

Extensions

b-file changed to a-file by N. J. A. Sloane, Oct 05 2010
Edited by Alois P. Heinz, Sep 06 2011
a(0)=1 prepended by Alois P. Heinz, Jul 23 2023

A229464 Binomial transform of (2*n + 1)!.

Original entry on oeis.org

1, 7, 133, 5419, 383785, 41782831, 6472067437, 1352114646163, 366325440650449, 124893891684358615, 52323557348796456661, 26420766706149889279867, 15824833185409769038803193, 11092546337733020334329204479, 8995627147680234199615065312445
Offset: 0

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Author

Peter Bala, Sep 25 2013

Keywords

Comments

Companion sequence to A064570.

Examples

			a(3) = 1*1! + 3*3! + 3*5! + 1*7! = 5419.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] * (2*k+1)!, {k, 0, n}], {n, 0, 15}] (* Vaclav Kotesovec, Oct 30 2017 *)

Formula

a(n) = Sum_{k = 0..n} binomial(n,k)*(2*k + 1)!.
Clearly a(n) is always odd; indeed, for n >= 1, a(n) = 1 + 6*n*b(n-1), where b(n) = [1, 11, 301, 15991, 1392761, ...] is the binomial transform of A051618.
a(n) = Integral_{x >= 0} x*(1 + x^2)^n*exp(-x) dx.
a(n) = (2*n + 1)*A064570(n) - 2*n*A064570(n-1).
Recurrence equation: a(n) = 1 + 2*n*(2*n + 1)*a(n-1) - 2*n*(2*n - 2)*a(n-2) with a(0) = 1 and a(1) = 7.
O.g.f.: Sum_{k >= 0} (2*k + 1)!*x^k/(1 - x)^(k + 1) = 1 + 7*x + 133*x^2 + 5419*x^3 + ....
a(n) ~ sqrt(Pi) * 2^(2*n + 2) * n^(2*n + 3/2) / exp(2*n). - Vaclav Kotesovec, Oct 30 2017
From Peter Bala, Nov 26 2017: (Start)
E.g.f.: exp(x)*Sum_{n >= 0} A000407(n)*x^n.
a(k) = a(0) (mod k) for all k (by the inhomogeneous recurrence equation).
More generally a(n+k) = a(n) (mod k) for all n and k (by an induction argument on n).
It follows that for each positive integer k, the sequence a(n) (mod k) is periodic, with the exact period dividing k. For example, modulo 10 the sequence becomes 1, 7, 3, 9, 5, 1, 7, 3, 9, 5, ... with exact period 5. (End)

A064571 Binomial transform of (3n)!.

Original entry on oeis.org

1, 7, 733, 365059, 480457465, 1310073012031, 6410226944228437, 51135786265589035963, 620857308610340190325489, 10894455325845833715898309495, 265361776433005578042408841530061, 8686235999260031219427964764144394867, 372097544110411566152076374938114937481193
Offset: 0

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Author

Karol A. Penson, Sep 20 2001

Keywords

Comments

From Peter Bala, Mar 20 2022: (Start)
The congruence a(n+k) == a(n) (mod k) holds for all n and k.
It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with exact period dividing k. For example, taken modulo 5 the sequence becomes [2, 3, 4, 0, 1, 2, 3, 4, 0, 1, ...], a purely periodic sequence with period 5.
More generally, the same property holds for any sequence with an e.g.f. of the form F(x)*exp(x*G(x)), where F(x) and G(x) are power series with integer coefficients and G(0) = 1 (see Bala, Theorem 1). (End)

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] * (3*k)!, {k, 0, n}], {n, 0, 12}] (* Vaclav Kotesovec, Oct 30 2017 *)
  • PARI
    for(n=0,12,print1(round(intnum(x=0,999,exp(-x)*(1+x^3)^n)),", ")) - Gerald McGarvey, Oct 12 2007

Formula

In Maple notation: a(n) = hypergeom([1, 1/3, 2/3, -n], [], -27), n = 0, 1, ....
a(n) = Integral_{x = 0..infinity} (1+x^3)^n*exp(-x) dx. - Gerald McGarvey, Oct 12 2007
From Vaclav Kotesovec, Oct 30 2017: (Start)
a(n) = (27*n^3 - 27*n^2 + 6*n + 1)*a(n-1) - 3*(n-1)*(27*n^2 - 45*n + 20)*a(n-2) + 27*(n-2)*(n-1)*(3*n - 4)*a(n-3) - 27*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ sqrt(2*Pi) * (3*n)^(3*n + 1/2) / exp(3*n). (End)
From Peter Bala, Mar 20 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*(3*k)!.
a(n) = 1 + 6*n*(1 + 60*(n-1)*(1 + 168*(n-2)*(1 + 330*(n-3)*(1 + 546*(n-4)*(1 + ...*(1 + (3*(3*m+1)*(3*m+2))*(n-m)*(1 + ...))))))).
a(n) == 1 (mod 6).
O.g.f: Sum_{k >= 0} (3*k)!*x^k/(1 - x)^(k+1).
E.g.f.: exp(x)*Sum_{k >= 0} (3*k)!/k!*x^k. (End)

Extensions

Corrected and extended by N. J. A. Sloane, Oct 29 2006
Showing 1-5 of 5 results.