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.

Previous Showing 41-50 of 73 results. Next

A052503 Number of permutations sigma of [2n] without fixed points such that sigma^4 = Id.

Original entry on oeis.org

1, 1, 9, 105, 2625, 76545, 3440745, 176080905, 12034447425, 922995698625, 87505195602825, 9203114782686825, 1141501848477415425, 155540530213013570625, 24232951756530007115625, 4112826185329479728735625, 781060320618828163499210625
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Crossrefs

Bisection of column k=4 of A261430.

Programs

  • Magma
    m:=40; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( Exp(x^2/2 + x^4/4) )); [Factorial(2*n-2)*b[2*n-1]: n in [1..Floor((m-2)/2)]]; // G. C. Greubel, May 14 2019
    
  • Maple
    spec := [S,{S=Set(Union(Cycle(Z,card=2),Cycle(Z,card=4)))},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    With[{nmax = 40}, CoefficientList[Series[Exp[x^2*(2 + x^2)/4], {x, 0, nmax}], x]*(Range[0, nmax])!][[1 ;; -1 ;; 2]] (* G. C. Greubel, May 14 2019 *)
  • PARI
    x='x+O('x^40); v=Vec(serlaplace( exp(x^2/2 + x^4/4) )); vector(#v\2, n, v[2*n-1]) \\ G. C. Greubel, May 14 2019
    
  • Sage
    m = 40; T = taylor(exp(x^2/2 + x^4/4), x, 0, 2*m+2); [factorial(2*n)*T.coefficient(x, 2*n) for n in (0..m)] # G. C. Greubel, May 14 2019

Formula

a(n) = (2n)! * [x^(2n)] exp(x^2/2 + x^4/4).
D-finite with recurrence a(n) +(-2*n+1)*a(n-1) -2*(n-1)*(2*n-1)*(2*n-3)*a(n-2)=0, with a(0)=1, a(1)=1, a(2)=9. - Corrected by R. J. Mathar, Feb 20 2020 to skip zeros.
a(n) = 2^n*Gamma(n+1/2)*A047974(n)/Pi^(1/2). - Mark van Hoeij, Oct 30 2011

A293528 E.g.f.: exp(x * Product_{k>0} (1 + x^k)).

Original entry on oeis.org

1, 1, 3, 13, 97, 741, 7291, 81313, 1027713, 14231017, 220911571, 3730744821, 68096325793, 1339705629133, 28225576881867, 634123159354441, 15127595174135041, 381586517104288593, 10147599723510322723, 283846981316172613597, 8324822922497497733601
Offset: 0

Views

Author

Seiichi Manyama, Oct 11 2017

Keywords

Comments

From Peter Bala, Mar 28 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, the sequence taken modulo 10 becomes [1, 1, 3, 3, 7, 1, 1, 3, 3, 7, ...], a purely periodic sequence with period 5.
3 divides a(3*n+2); 13 divides a(13*n+3) and a(13*n+5); 19 divides a(19*n+5), a(19*n+12) and a(19*n+14). (End)

Crossrefs

Programs

  • Mathematica
    nmax = 25; CoefficientList[Series[E^(x*QPochhammer[-1, x]/2), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 11 2017 *)
  • PARI
    N=66; x='x+O('x^N); Vec(serlaplace(exp(x*prod(k=1, N, (1+x^k)))))

Formula

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} k*A000009(k-1)*a(n-k)/(n-k)! for n > 0.

A293841 E.g.f.: exp(Sum_{n>=1} n*A000009(n)*x^n).

Original entry on oeis.org

1, 1, 5, 49, 409, 4841, 66541, 1006825, 17349809, 333948529, 6997459861, 159199648961, 3918175462345, 103227624161689, 2901807752857469, 86684932131301561, 2738566218754961761, 91236821580866560865, 3196113263245038385189
Offset: 0

Views

Author

Seiichi Manyama, Oct 17 2017

Keywords

Comments

From Peter Bala, Mar 28 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, the sequence taken modulo 10 becomes [1, 1, 5, 9, 9, 1, 1, 5, 9, 9, ...], a purely periodic sequence with exact period 5.
5 divides a(5*n+2), 7 divides a(7*n+3); 17 divides a(17*n+7), a(17*n+8) and a(17*n+11). (End)

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[E^Sum[k*PartitionsQ[k]*x^k, {k, 1, nmax}], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 18 2017 *)

Formula

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} k^2*A000009(k)*a(n-k)/(n-k)! for n > 0.

A361277 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..n} binomial(k*j,n-j)/j!.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 7, 1, 1, 1, 7, 19, 25, 1, 1, 1, 9, 37, 97, 81, 1, 1, 1, 11, 61, 241, 581, 331, 1, 1, 1, 13, 91, 481, 1981, 3661, 1303, 1, 1, 1, 15, 127, 841, 4881, 17551, 26335, 5937, 1, 1, 1, 17, 169, 1345, 10001, 55321, 171697, 202049, 26785, 1
Offset: 0

Views

Author

Seiichi Manyama, Mar 06 2023

Keywords

Examples

			Square array begins:
  1,  1,   1,    1,    1,     1, ...
  1,  1,   1,    1,    1,     1, ...
  1,  3,   5,    7,    9,    11, ...
  1,  7,  19,   37,   61,    91, ...
  1, 25,  97,  241,  481,   841, ...
  1, 81, 581, 1981, 4881, 10001, ...
		

Crossrefs

Columns k=0..4 give A000012, A047974, A361278, A361279, A361280.
Main diagonal gives A361281.
Cf. A293012.

Programs

  • PARI
    T(n, k) = n!*sum(j=0, n, binomial(k*j, n-j)/j!);

Formula

E.g.f. of column k: exp(x * (1+x)^k).
T(0,k) = 1; T(n,k) = (n-1)! * Sum_{j=1..n} j * binomial(k,j-1) * T(n-j,k)/(n-j)!.

A361568 Expansion of e.g.f. exp(x^3/6 * (1+x)^3).

Original entry on oeis.org

1, 0, 0, 1, 12, 60, 130, 420, 8400, 101080, 781200, 4435200, 37714600, 607807200, 8660652000, 94007313400, 914497584000, 11566931376000, 198256136478400, 3275456501116800, 46558791351072000, 636647461257808000, 10238792220969312000, 194852563745775936000
Offset: 0

Views

Author

Seiichi Manyama, Mar 16 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(x^3/6*(1+x)^3)))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!/6*sum(j=3, i, j*binomial(3, j-3)*v[i-j+1]/(i-j)!)); v;

Formula

a(n) = n! * Sum_{k=0..floor(n/3)} binomial(3*k,n-3*k)/(6^k * k!).
a(0) = 1; a(n) = ((n-1)!/6) * Sum_{k=3..n} k * binomial(3,k-3) * a(n-k)/(n-k)!.
a(n) = (n-1)*(n-2)/6 * (3*a(n-3) + 12*(n-3)*a(n-4) + 15*(n-3)*(n-4)*a(n-5) + 6*(n-3)*(n-4)*(n-5)*a(n-6)). -Seiichi Manyama, Jun 16 2024

A083886 Expansion of e.g.f. exp(3*x)*exp(x^2).

Original entry on oeis.org

1, 3, 11, 45, 201, 963, 4899, 26253, 147345, 862083, 5238459, 32957037, 214117209, 1433320515, 9867008979, 69734001357, 505212273441, 3747124863747, 28418591888235, 220152270759597, 1740363304031721, 14027180742479043, 115176800996769411, 962726355659386125, 8186311912829551281, 70769800810139187843
Offset: 0

Views

Author

Paul Barry, May 09 2003

Keywords

Comments

Binomial transform of A000898.
Hankel transform is A108400. - Paul Barry, Jun 13 2009
a(n) is the number of self-inverse signed permutations of length 2n that are equal to their reverse-complements and avoid the pattern (-2,-1). As a result, a(n) also gives the same thing but for avoiding any one of (-1,-2), (+2,+1) or (+1,+2) instead of (-2,-1) (See the Hardt and Troyka reference). - Justin M. Troyka, Aug 05 2011
a(n) is also the number of skew-symmetric (n,n)-clans, or the number of B-orbits in the symmetric space of type CI, Sp_{2n}(C)/GL_n(C) where B is a Borel subgroup of Sp_{2n}(C). - Aram Bingham, Oct 08 2019

Examples

			Since a(2) = 11, there are 11 self-inverse signed permutations of 4 that are equal to their reverse-complements and avoid (-2,-1). Some of these are: (+3,+4,+1,+2), (+4,-2,-3,+1), (-1,+3,+2,-4), (-1,-2,-3,-4). - _Justin M. Troyka_, Aug 05 2011
		

Crossrefs

Programs

  • Mathematica
    a = {1, 3}; For[n = 2, n < 26, n++, a = Append[a, 3 a[[n]] + 2 (n - 1) a[[n - 1]]]]; a (* Justin M. Troyka, Aug 05 2011 *)
    CoefficientList[Series[Exp[3*x+x^2], {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 25 2013 *)
    Table[Abs[HermiteH[n, 3 I/2]], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 11 2016 *)
  • PARI
    my(x='x+O('x^26)); Vec(serlaplace(exp(3*x)*exp(x^2))) /* Joerg Arndt, Jul 12 2012 */

Formula

E.g.f.: exp(3*x+x^2).
From Paul Barry, Jun 13 2009: (Start)
G.f.: 1/(1-3x-2x^2/(1-3x-4x^2/(1-3x-6x^2/(1-3x-8x^2/(1-... (continued fraction);
a(n) = Sum_{k=0..floor(n/2)} C(n,2k) * (2k)! * 3^(n-2k) / k!. (End)
a(n) = i^n*Hermite_H(n, -3i/2), i=sqrt(-1). - Paul Barry, Jun 15 2009
a(0) = 1; a(1) = 3; a(n) = 3*a(n-1) + 2*(n-1)*a(n-2) for n >= 2. - Justin M. Troyka, Aug 05 2011
E.g.f. 1 + (x+3)*x/(G(0)-x^2-3*x) where G(k)= x^2 + 3*x + k + 1 - (x+3)*x*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jul 12 2012
G.f.: 1/Q(0) where Q(k) = 1 + 2*x*k - 2*x - x/(1 - 2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 07 2013
a(n) ~ n^(n/2)*2^(n/2-1/2)*exp(3*sqrt(n/2)-n/2-9/8) * (1+21*sqrt(2)/(32*sqrt(n))). - Vaclav Kotesovec, Jun 25 2013

A193930 E.g.f.: exp(x+x^2+x^3+x^4).

Original entry on oeis.org

1, 1, 3, 13, 73, 381, 2611, 19993, 165873, 1436473, 14004451, 145099461, 1584090553, 18196817653, 223416271443, 2865429498961, 38330181602401, 535448870264433, 7823019065848003, 118402856414023933, 1856454825152993961, 30160691907215561581
Offset: 0

Views

Author

Vladimir Kruchinin, Aug 09 2011

Keywords

Crossrefs

Programs

  • Magma
    I:=[1,1,3,13]; [n le 4 select I[n] else Self(n-1)+2*(n-2)*Self(n-2)+3*(n-2)*(n-3)*Self(n-3)+4*(n-2)*(n-3)*(n-4)*Self(n-4): n in [1..25]]; // Vincenzo Librandi, Oct 12 2014
  • Maple
    A193930:=n->`if`(n=0,1,n!*add(add( (-1)^i*binomial(k,k-i)*binomial(n-4*i-1,k-1)/k!, i=0..(n-k)/4), k=1..n)): seq(A193930(n), n=0..20); # Wesley Ivan Hurt, Oct 11 2014
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add(
          a(n-j)*binomial(n-1, j-1)*j!, j=1..min(n, 4)))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 29 2017
  • Mathematica
    RecurrenceTable[{a[n] == a[n - 1] + 2 (n - 1) a[n - 2] + (n - 1) (n - 2) 3 a[n - 3] + (n - 1) (n - 2) (n - 3) 4 a[n - 4], a[0] == 1,
    a[1] == 1, a[2] == 3, a[3] == 13}, a, {n, 0, 20}] (* Geoffrey Critzer, Oct 11 2014 *)
    With[{nn=30},CoefficientList[Series[Exp[x+x^2+x^3+x^4],{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Aug 20 2021 *)
  • Maxima
    a(n):=if n=0 then 1 else n!*sum(sum((-1)^i*binomial(k,k-i)*binomial(n-4*i-1,k-1),i,0,(n-k)/4)/k!,k,1,n);
    
  • PARI
    x = y + O(y^50); concat(0, Vec(serlaplace(exp(x+x^2+x^3+x^4)))) \\ Michel Marcus, Oct 12 2014
    

Formula

a(n) = n!*sum(k=1..n, sum(i=0..(n-k)/4, (-1)^i*binomial(k,k-i)*binomial(n-4*i-1,k-1))/k!), n>0, a(0)=1.
a(n) = a(n-1) + 2*(n-1)*a(n-2) + 3*(n-1)(n-2)*a(n-3) + 4*(n-1)*(n-2)*(n-3)*a(n-4). - Geoffrey Critzer, Oct 11 2014

A293527 E.g.f.: exp(x/Product_{k>0} (1 - x^k)).

Original entry on oeis.org

1, 1, 3, 19, 145, 1401, 15331, 198283, 2840769, 45744625, 807769891, 15590922051, 325339538833, 7316871562729, 175934564213955, 4508362093795771, 122558873094082561, 3522465207528093153, 106681726559176156099, 3395601487535927589235, 113287948824653903674641
Offset: 0

Views

Author

Seiichi Manyama, Oct 11 2017

Keywords

Comments

From Peter Bala, Mar 25 2022: (Start)
The sequence terms are odd. 3 divides a(3*n+2), 5 divides a(5*n+4), 9 divides a(9*n+8), 15 divides a(15*n+14) and 19 divides a(19*n+3).
More generally, 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 7 the sequence becomes [1, 1, 3, 5, 5, 1, 1, 1, 3, 3, 5, 5, 1, 1, ...], a purely periodic sequence with period 7. (End)

Crossrefs

Programs

  • Mathematica
    nmax = 25; CoefficientList[Series[E^(x/QPochhammer[x, x]), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 11 2017 *)
  • PARI
    N=66; x='x+O('x^N); Vec(serlaplace(exp(x/prod(k=1, N, (1-x^k)))))

Formula

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} k*A000041(k-1)*a(n-k)/(n-k)! for n > 0.

A349640 a(n) = Sum_{k=0..n} binomial(n,k) * A000108(k) * k!.

Original entry on oeis.org

1, 2, 7, 46, 485, 7066, 130987, 2946182, 77923561, 2369742130, 81467904431, 3124302688222, 132237820201357, 6123150708289226, 307903794151741075, 16709463201832993846, 973385368533058021457, 60583668821975488285282, 4012342371757905842648791, 281735471040327667890013070
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 23 2021

Keywords

Comments

For each positive integer k, the sequence obtained by reducing a(n) modulo k is a periodic sequence with period dividing k. For example, modulo 5 the sequence becomes [1, 2, 2, 1, 0, 1, 2, 2, 1, 0, ...] with period 5. In particular, a(5*n+4) == 0 (mod 5). Cf. A047974. - Peter Bala, Mar 13 2025

Crossrefs

Programs

  • Maple
    gf := exp(x)*(1 - sqrt(1 - 4*x))/(2*x): ser := series(gf, x, 24):
    seq(n!*coeff(ser, x, n), n = 0..19);
    # Alternative:
    a := n -> `if`(n < 4, [1, 2, 7, 46][n + 1], ((4*n^2 - 12*n + 8)*a(n - 3) - (8*n^2 - 13*n + 5)*a(n - 2) + 4*n^2*a(n - 1))/(n + 1)):
    seq(a(n), n = 0..19);  # Peter Luschny, Nov 23 2021
    # Alternative
    seq(simplify(hypergeom([-n, 1/2, 1], [2], -4)), n = 0..19); # Peter Bala, Mar 13 2025
  • Mathematica
    Table[Sum[Binomial[n, j]*CatalanNumber[j]*j!, {j, 0, n}], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k) * (binomial(2*k,k)/(k+1)) * k!); \\ Michel Marcus, Nov 23 2021

Formula

a(n) ~ 2^(2*n + 1/2) * n^(n-1) / exp(n - 1/4).
From Peter Luschny, Nov 23 2021: (Start)
a(n) = n! * [x^n](exp(x)*(1 - sqrt(1 - 4*x))/(2*x)).
a(n) = (4*(n-1)*(n-2)*a(n - 3) - (n-1)*(8*n-5)*a(n - 2) + 4*n^2*a(n - 1))/(n + 1) for n >= 4.
a(n-1) = A224500(n) / n for n >= 1. (End)
a(n) = hypergeom([-n, 1/2, 1], [2], -4). - Peter Bala, Mar 13 2025

A359760 Triangle read by rows. The Kummer triangle, the coefficients of the Kummer polynomials. K(n, k) = binomial(n, k) * oddfactorial(k/2) if k is even, otherwise 0, where oddfactorial(z) := (2*z)!/(2^z*z!).

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 3, 0, 1, 0, 6, 0, 3, 1, 0, 10, 0, 15, 0, 1, 0, 15, 0, 45, 0, 15, 1, 0, 21, 0, 105, 0, 105, 0, 1, 0, 28, 0, 210, 0, 420, 0, 105, 1, 0, 36, 0, 378, 0, 1260, 0, 945, 0, 1, 0, 45, 0, 630, 0, 3150, 0, 4725, 0, 945, 1, 0, 55, 0, 990, 0, 6930, 0, 17325, 0, 10395, 0
Offset: 0

Views

Author

Peter Luschny, Jan 13 2023

Keywords

Comments

The Kummer numbers K(n, k) are a refinement of the oddfactorial numbers (A001147) in the sense that they are the coefficients of polynomials K(n, x) = Sum_{n..k} K(n, k) * x^k that take the value oddfactorial(n) at x = 1. The coefficients of x^n are the aerated oddfactorial numbers A123023.
These numbers appear in many different versions (see the crossrefs). They are the coefficients of the Chebyshev-Hermite polynomials in signed form when ordered in decreasing powers. Our exposition is based on the seminal paper by Kummer, which preceded the work of Chebyshev and Hermite for more than 20 years. They are also referred to as Bessel numbers of the second kind (Mansour et al.) when the odd powers are omitted.

Examples

			Triangle K(n, k) starts:
 [0] 1;
 [1] 1, 0;
 [2] 1, 0,  1;
 [3] 1, 0,  3, 0;
 [4] 1, 0,  6, 0,   3;
 [5] 1, 0, 10, 0,  15, 0;
 [6] 1, 0, 15, 0,  45, 0,   15;
 [7] 1, 0, 21, 0, 105, 0,  105, 0;
 [8] 1, 0, 28, 0, 210, 0,  420, 0, 105;
 [9] 1, 0, 36, 0, 378, 0, 1260, 0, 945, 0;
		

References

  • John Riordan, Introduction to Combinatorial Analysis, Dover (2002), pp. 85-86.

Crossrefs

Variants: Signed version: A073278. Other variants are the irregular triangle A100861 with zeros deleted, A066325 and A099174 with reversed rows, A111924, A144299, A104556.

Programs

  • Maple
    oddfactorial := proc(z) (2*z)! / (2^z*z!) end:
    K := (n, k) -> ifelse(irem(k, 2) = 1, 0, binomial(n, k) * oddfactorial(k/2)):
    seq(seq(K(n, k), k = 0..n), n = 0..11);
    # Alternative, as coefficients of polynomials:
    p := (n, x) -> 2^(n/2)*(-1/x^2)^(-n/2)*KummerU(-n/2, 1/2, -1/(2*x^2)):
    seq(print(seq(coeff(simplify(p(n, x)), x, k), k = 0..n)), n = 0 ..9);
    # Using the exponential generating function:
    egf := exp(x + (t*x)^2 / 2): ser := series(egf, x, 12):
    seq(print(seq(coeff(n! * coeff(ser, x, n), t, k), k = 0..n)), n = 0..9);
  • Mathematica
    K[n_, k_] := K[n, k] = Which[OddQ[k], 0, k == 0, 1, n == k, K[n - 1, n - 2], True, K[n - 1, k] n/(n - k)];
    Table[K[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 25 2023 *)
  • Python
    from functools import cache
    @cache
    def K(n: int, k: int) -> int:
        if k %  2: return 0
        if n <  3: return 1
        if n == k: return K(n - 1, n - 2)
        return (K(n - 1, k) * n) // (n - k)
    for n in range(10): print([K(n, k) for k in range(n + 1)])

Formula

Let p(n, x) = 2^(n/2)*(-1/x^2)^(-n/2)*KummerU(-n/2, 1/2, -1/(2*x^2)).
p(n, 1) = A000085(n); p(n, sqrt(2)) = A047974(n); p(n, 2) = A115329(n);
p(2, n) = A002522(n) (n >= 1); p(3, n) = A056107(n) (n >= 1);
p(n, n) = A359739(n) (n >= 1); 2^n*p(n, 1/2) = A005425(n).
K(n, k) = [x^k] p(n, x).
K(n, k) = [t^k] (n! * [x^n] exp(x + (t*x)^2 / 2)).
K(n, n) = A123023(n).
K(n, n-1) = A123023(n + 1).
K(2*n, 2*n) = A001147(n).
K(4*n, 2*n) = A359761, the central terms without zeros.
K(2*n+2, 2*n) = A001879.
Sum_{k=0..n} (-1)^n * i^k * K(n, k) = A001464(n), ((the number of even involutions) - (the number of odd involutions) in the symmetric group S_n (Robert Israel)).
Sum_{k=0..n} Sum_{j=0..k} K(n, j) = A000085(n + 1).
For a recursion see the Python program.
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