A264781 -id:A264781 - cp's OEIS frontend

A007696 Quartic (or 4-fold) factorial numbers: a(n) = Product_{k = 0..n-1} (4*k + 1).

1, 1, 5, 45, 585, 9945, 208845, 5221125, 151412625, 4996616625, 184874815125, 7579867420125, 341094033905625, 16713607661375625, 885821206052908125, 50491808745015763125, 3080000333445961550625, 200200021673987500790625, 13813801495505137554553125

Offset: 0

N. J. A. Sloane

a(n), n >= 1, enumerates increasing quintic (5-ary) trees. See David Callan's comment on A007559 (number of increasing quarterny trees).

Hankel transform is A169619. - Paul Barry, Dec 03 2009

			G.f. = 1 + x + 5*x^2 + 45*x^3 + 585*x^4 + 9945*x^5 + 208845*x^6 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq. 3 (2000), Article 00.2.4.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations, (m+1)-ary trees, and m-parking functions, arXiv:1403.5962 [math.CO], 2014.
Maxie D. Schmidt, Generalized j-Factorial Functions, Polynomials, and Applications , J. Integer Seq. 13 (2010), Article 10.6.7; see page 39.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integer Seq. 9 (2006), Article 06.1.1.

Cf. A001147, A001813, A004981, A007559, A008545, A034255, A047053, A051142, A264781.

a(n) = A049029(n, 1) for n >= 1 (first column of triangle).

a:=[1,1];; for n in [3..20] do a[n]:=(4*(n-1)-7)*(a[n-1]+4*a[n-2]); od; a; # G. C. Greubel, Aug 15 2019

[n le 2 select 1 else (4*(n-1)-7)*(Self(n-1) + 4*Self(n-2)): n in [1..20]]; // G. C. Greubel, Aug 15 2019

x:='x'; G(x):=(1-4*x)^(-1/4): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: seq(eval(f[n],x=0),n=0..17);# Zerinvary Lajos, Apr 03 2009
A007696 := n -> mul(k, k = select(k-> k mod 4 = 1, [$ 1 .. 4*n])): seq(A007696(n), n=0..17); # Peter Luschny, Jun 23 2011

a[ n_]:= Pochhammer[ 1/4, n] 4^n; (* Michael Somos, Jan 17 2014 *)
a[ n_]:= If[n < 0, 1 / Product[ -k, {k, 3, -4n-1, 4}], Product[ k, {k, 1, 4n-3, 4}]]; (* Michael Somos, Jan 17 2014 *)
Range[0, 19]! CoefficientList[Series[((1-4x)^(-1/4)), {x, 0, 19}], x] (* Vincenzo Librandi, Oct 08 2015 *)

A007696(n):=prod(4*k+1,k,0,n-1)$
makelist(A007696(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

{a(n) = if( n<0, 1 / prod(k=1, -n, 1 - 4*k), prod(k=1, n, 4*k - 3))}; /* Michael Somos, Jan 17 2014 */

[4^n*rising_factorial(1/4, n) for n in (0..20)] # G. C. Greubel, Aug 15 2019

E.g.f.: (1 - 4*x)^(-1/4).
a(n) ~ 2^(1/2) * Pi^(1/2) * Gamma(1/4)^(-1) * n^(-1/4) * 2^(2*n) * e^(-n) * n^n * (1 - 1/96 * n^(-1) - ...). - Joe Keane (jgk(AT)jgk.org), Nov 23 2001 [corrected by Vaclav Kotesovec, Jul 19 2025]
a(n) = Sum_{k = 0..n} (-4)^(n-k) * A048994(n, k). - Philippe Deléham, Oct 29 2005
G.f.: 1/(1 - x/(1 - 4*x/(1 - 5*x/(1 - 8*x/(1 - 9*x/(1 - 12*x/(1 - 13*x/(1 - .../(1 - A042948(n+1)*x/(1 -... (continued fraction). - Paul Barry, Dec 03 2009
a(n) = (-3)^n * Sum_{k = 0..n} (4/3)^k * s(n+1, n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/T(0), where T(k) =  1 - x * (4*k + 1)/(1 - x * (4*k + 4)/T(k+1)) (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1 + x/Q(0), where Q(k) = 1 + x + 2*(2*k - 1)*x - 4*x*(k+1)/Q(k+1) (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x * (4*k + 1)/(x * (4*k + 1) + 1/G(k+1))) (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013
0 = a(n) * (4*a(n+1) - a(n+2)) + a(n+1) * a(n+1) for all n in Z. - Michael Somos, Jan 17 2014
a(-n) = (-1)^n / A008545(n). - Michael Somos, Jan 17 2014
Let T(x) = 1/(1 - 3*x)^(1/3) be the e.g.f. for the sequence of triple factorial numbers A007559. Then the e.g.f. A(x) for the quartic factorial numbers satisfies T(Integral_{t = 0..x} A(t) dt) = A(x). (Cf. A007559 and A008548.) - Peter Bala, Jan 02 2015
O.g.f.: hypergeom([1, 1/4], [], 4*x). - Peter Luschny, Oct 08 2015
a(n) = A264781(4*n+1, n). - Alois P. Heinz, Nov 24 2015
a(n) = 4^n * Gamma(n + 1/4)/Gamma(1/4). - Artur Jasinski, Aug 23 2016
D-finite with recurrence: a(n) +(-4*n+3)*a(n-1)=0, n>=1. - R. J. Mathar, Feb 14 2020
Sum_{n>=0} 1/a(n) = 1 + exp(1/4)*(Gamma(1/4) - Gamma(1/4, 1/4))/(2*sqrt(2)). - Amiram Eldar, Dec 18 2022

Better description from Wolfdieter Lang, Dec 11 1999

A202213 Number of permutations of [n] avoiding the consecutive pattern 45321.

Original entry on oeis.org

1, 1, 2, 6, 24, 119, 708, 4914, 38976, 347765, 3447712, 37598286, 447294144, 5764747515, 80011430240, 1189835682714, 18873422539776, 318085061976105, 5676223254661760, 106919460527212950, 2119973556022047744, 44136046410218669055, 962630898723772565760

Offset: 0

Ray Chandler, Dec 14 2011

nonn

a(n) is the number of permutations on [n] that avoid the consecutive pattern 45321. It is the same as the number of permutations which avoid 12354, 21345 or 54312.

Alois P. Heinz, Table of n, a(n) for n = 0..200 (terms n = 1..40 from Ray Chandler)
A. Baxter, B. Nakamura, and D. Zeilberger, Automatic generation of theorems and proofs on enumerating consecutive Wilf-classes, 2011.
Sergi Elizalde and Marc Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125; see Theorem 3.2 (p. 116) with m = a = 3 and u = 0.
Eric Weisstein's World of Mathematics, Pochhammer Symbol.
Wikipedia, Falling and rising factorials.

Cf. A177523, A202214-A202236, A329070.

Column k = 0 of A264781 and row m = 2 of A327722.

b:= proc(u, o, t) option remember; `if`(u+o=0, 1,
      add(b(u+j-1, o-j, `if`(u+j-10, -1, `if`(t=-1, -2, 0)))), j=1..u)))
    end:
a:= n-> b(n, 0$2):
seq(a(n), n=0..40);  # Alois P. Heinz, Nov 19 2013

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, Sum[b[u+j-1, o-j, If[u+j-1 < j, 0, j]], {j, 1, o}] + If[t == -2, 0, Sum[b[u-j, o+j-1, If[j0, -1, If[t == -1, -2, 0]]]], {j, 1, u}]]]; a[n_] := b[n, 0, 0]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 12 2015, after Alois P. Heinz *)

From Petros Hadjicostas, Nov 02 2019: (Start)
E.g.f.: 1/W(z), where W(z) = 1 + Sum_{n >= 0} (-1)^(n+1)* z^(4*n+1)/(b(n)*(4*n+1)) with b(n) = A329070(n,4) = (4*n)!/(4^n*(1/4)_n). (Here (x)_n = x*(x + 1)*...*(x + n - 1) is the Pochhammer symbol, or rising factorial, which is denoted by (x)^n in some papers and books.)  The function W(z) satisfies the o.d.e. W^(4)(z) + z*W'(z) = 0 with W(0) = 1, W'(0) = -1, and W^(k)(0) = 0 for k = 2..3. [See Theorem 3.2 (with m = a = 3 and u = 0) in Elizalde and Noy (2003).]
a(n) = Sum_{m = 0..floor((n-1)/4)} (-4)^m * (1/4)_m * binomial(n, 4*m+1) * a(n-4*m-1) for n >= 1 with a(0) = 1. (End)

A197365 T(n,k) gives the number of permutations of the set [n] that contain k occurrences of the subword (132); irregular array read by rows (n >= 0 and 0 <= k <= max(0, floor((n-1)/2))).

Original entry on oeis.org

1, 1, 2, 5, 1, 16, 8, 63, 54, 3, 296, 368, 56, 1623, 2649, 753, 15, 10176, 20544, 9024, 576, 71793, 172596, 104814, 13572, 105, 562848, 1569408, 1228608, 259968, 7968, 4853949, 15398829, 14824314, 4532034, 306729, 945, 45664896, 162412416, 185991936, 75929856

Offset: 0

Peter Bala, Oct 14 2011

A permutation p(1)p(2)...p(n) in the symmetric group S_n contains the subword (132) if there are 3 consecutive elements p(i)p(j)p(k) that have the same order relations as (132), that is, p(i) < p(j) > p(k) and p(i) < p(k). For the enumeration of permutations containing the subword (123) see A162975.
From Petros Hadjicostas, Nov 05 2019: (Start)
The attached Maple program gives a recurrence for the o.g.f. of each row in terms of u for T(n,k), the number of permutations of [n] containing exactly k occurrences of the consecutive pattern 123...(r+1)(r+3)(r+2) for r >= 0. In the program, t = r + 2. Here, n >= 0 and 0 <= k <= max(0, (n-1)/t).
Using that recurrence we may get any row or column from the irregular triangular array T(n, k) for any r >= 0. (Here r = 0, while in array A264781 we have r = 2.)
The recurrence follows from manipulation of the bivariate o.g.f/e.g.f. 1/W(u,z) = Sum_{n, k >= 0} T(n, k)*u^k*z^n/n!, whose reciprocal W(u,z) is the solution of the o.d.e. in Theorem 3.2 in Elizalde and Noy (2003) (with m = a = r + 1). The number t = r + 2 is the order of the o.d.e. in terms of the variable z.
(End)

			Table begins
.n\k.|......0......1.....2......3
= = = = = = = = = = = = = = = = =
..0..|......1
..1..|......1
..2..|......2
..3..|......5......1
..4..|.....16......8
..5..|.....63.....54.....3
..6..|....296....368....56
..7..|...1623...2649...753....15
..8..|..10176..20544..9024...576
...
T(4,0) = 16: The 16 permutations of S_4 not containing the subword (132) are (1234), (2134), (2314), (3124), (3214), (1342), (2341), (3241), (2413), (3412), (3421), (4123), (4213), (4231), (4312), (4321).
T(4,1) = 8: The 8 permutations of S_4 with 1 occurrence of the subword (132) are 1243, 1324, 1423, 1432, 2143, 2431, 3142, 4132.

Alois P. Heinz, Rows n = 0..100, flattened
A. Baxter, B. Nakamura, and D. Zeilberger, Automatic generation of theorems and proofs on enumerating consecutive Wilf-classes, 2011.
S. Elizalde and M. Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125.
Petros Hadjicostas, General Maple program for a recurrence for the number of permutations of [n] containing exactly k times the consecutive pattern 123...(r+1)(r+3)(r+2) for r >= 0. [Here r = 0 and in the program t = r + 2 = 2.]

Columns k=0-10 give: A111004, A263885, A263886, A263887, A263888, A263889, A263890, A263891, A263892, A263893, A263894.
T(2n+1,n) gives A001147.
T(2n+2,n) gives 2*A076729.
Cf. A162975, A264781 (pattern 12354).

b:= proc(u, o, t) option remember; `if`(u+o=0, 1, expand(
      add(b(u-j, o+j-1, 0)*`if`(j (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=0..14);  # Alois P. Heinz, Oct 30 2013

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, Expand[Sum[b[u-j, o+j-1, 0]*If[jJean-François Alcover, Mar 05 2015, after Alois P. Heinz *)

E.g.f.: 1/(1 - int_{t = 0..z} exp((u-1)*t^2/2!)) = sqrt(1 - u)/(sqrt(1 - u) -sqrt(Pi/2) * erf(z/2*sqrt(1 - u))) = 1 + z + 2*z^2/2! + (5 + u)*z^3/3! + (16 + 8*u)*z^4/4! + ....

n-th row sum = n!. First column is A111004.

A062199 Second (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

Original entry on oeis.org

1, 12, 126, 1344, 15120, 181440, 2328480, 31933440, 467026560, 7264857600, 119870150400, 2092278988800, 38532804710400, 746943599001600, 15205637551104000, 324386934423552000, 7237883474325504000, 168600109166641152000, 4093235983656787968000

Offset: 0

Wolfdieter Lang, Jun 19 2001

a(n) is the total number of ascending runs of length 5 over all permutations of {1,2,...,n+5}. a(1) = 12 because we have: [1,2,3,4,6,5], [1,2,3,5,6,4], [1,2,4,5,6,3], [1,3,4,5,6,2], [2,1,3,4,5,6], [2,3,4,5,6,1], [3,1,2,4,5,6], [4,1,2,3,5,6], [5,1,2,3,4,6], [6,1,2,3,4,5], and [1,2,3,4,5,6] which has two runs of length 5. - Geoffrey Critzer, Feb 21 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Milan Janjić, Enumerative Formulas for Some Functions on Finite Sets.
Index entries for sequences related to Laguerre polynomials.

Cf. A001720 (first column of A062140), A264781.

Cf. A001620, A091725, A099285.

[Binomial(n, 5)*Factorial(n-4): n in [5..25]]; // Vincenzo Librandi, Feb 23 2014

Table[Sum[n!/5!, {i, 5, n}], {n, 5, 21}] (* Zerinvary Lajos, Jul 12 2009 *)
With[{nn=20},CoefficientList[Series[(1+5x)/(1-x)^7,{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Nov 10 2016 *)

x='x+O('x^30); Vec(serlaplace((1+5*x)/(1-x)^7)) \\ G. C. Greubel, Feb 07 2018

[binomial(n,5)*factorial (n-4) for n in range(5, 22)] # Zerinvary Lajos, Jul 07 2009

E.g.f.: (1+5*x)/(1-x)^7.
a(n) = A062140(n+1, 1) = (n+1)!*binomial(n+5, 5).
If we define f(n,i,x)= Sum(Sum(binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j),j=i..k),k=i..n) then a(n-1)=(-1)^(n-1)*f(n,1,-6), (n>=1). [Milan Janjic, Mar 01 2009]
a(n) = Sum_{k>0} k * A264781(n+5,k). - Alois P. Heinz, Nov 24 2015
Assuming offset 1: a(n) = -n!*binomial(-n,5). - Peter Luschny, Apr 29 2016
From Amiram Eldar, Sep 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 1565/12 - 50*e - 5*gamma + 5*Ei(1), where gamma is Euler's constant (A001620) and Ei(1) is the exponential integral at 1 (A091725).
Sum_{n>=0} (-1)^n/a(n) = -125/12 + 20/e + 5*gamma - 5*Ei(-1), where -Ei(-1) is the negated exponential integral at -1 (A099285). (End)

More terms from Vincenzo Librandi, Feb 23 2014

A264896 Number of permutations of [n] with exactly one occurrence of the consecutive pattern 45321.

Original entry on oeis.org

1, 12, 126, 1344, 15110, 180736, 2308548, 31481472, 457520055, 7068885600, 115808906178, 2006533573632, 36674815572540, 705453732298240, 14248697953325160, 301564509817810944, 6674811622886359005, 154228999030804811520, 3713903962096887036390

Offset: 5

Alois P. Heinz, Nov 27 2015

nonn

Consecutive patterns 12354, 21345, 54312 give the same sequence.

			a(5) = 1: 45321.
a(6) = 12: 156432, 256431, 356421, 453216, 456321, 463215, 546321, 563214, 564213, 564312, 564321, 645321.

Alois P. Heinz, Table of n, a(n) for n = 5..200

Column k=1 of A264781.

b:= proc(u, o, t) option remember; `if`(u+o=0, 1, add(
       b(u+j-1, o-j, `if`(u+j-30, -1, `if`(t=-1, -2, 0)))), j=1..u),x,2),polynom))
    end:
a:= n-> coeff(b(n, 0$2),x,1):
seq(a(n), n=5..25);

b[u_, o_, t_] := b[u, o, t] = If[u + o == 0, 1, Sum[b[u + j - 1, o - j, If[u + j - 3 < j, 0, j]], {j, 1, o}] + Expand[If[t == -2, x, 1]*Sum[b[u - j, o + j - 1, If[j < t || t == -2, 0, If[t > 0, -1, If[t == -1, -2, 0]]]], {j, 1, u}]]];
a[n_] := Coefficient[b[n, 0, 0], x, 1];
Table[a[n], {n, 5, 25}] (* Jean-François Alcover, Nov 01 2021, after Alois P. Heinz *)

cp's OEIS Frontend

A007696 Quartic (or 4-fold) factorial numbers: a(n) = Product_{k = 0..n-1} (4*k + 1).

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A202213 Number of permutations of [n] avoiding the consecutive pattern 45321.

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A197365 T(n,k) gives the number of permutations of the set [n] that contain k occurrences of the subword (132); irregular array read by rows (n >= 0 and 0 <= k <= max(0, floor((n-1)/2))).

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A062199 Second (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).

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Magma

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A264896 Number of permutations of [n] with exactly one occurrence of the consecutive pattern 45321.

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Maple

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