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
Examples
G.f. = 1 + x + 5*x^2 + 45*x^3 + 585*x^4 + 9945*x^5 + 208845*x^6 + ...
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- 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.
Crossrefs
Programs
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GAP
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
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Magma
[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
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Maple
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
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Mathematica
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 *)
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Maxima
A007696(n):=prod(4*k+1,k,0,n-1)$ makelist(A007696(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
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PARI
{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 */
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Sage
[4^n*rising_factorial(1/4, n) for n in (0..20)] # G. C. Greubel, Aug 15 2019
Formula
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
Extensions
Better description from Wolfdieter Lang, Dec 11 1999
Comments