A040075 5-fold convolution of A000302 (powers of 4); expansion of 1/(1-4*x)^5.

1, 20, 240, 2240, 17920, 129024, 860160, 5406720, 32440320, 187432960, 1049624576, 5725224960, 30534533120, 159719096320, 821412495360, 4161823309824, 20809116549120, 102821517066240, 502682972323840, 2434043865989120, 11683410556747776, 55635288365465600

Offset: 0

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Also convolution of A020920 with A000984 (central binomial coefficients).

With a different offset, number of n-permutations (n=5) of 5 objects u, v, w, z, x with repetition allowed, containing exactly four (4)u's. Example: a(1)=20 because we have uuuuv, uuuvu, uuvuu, uvuuu, vuuuu, uuuuw, uuuwu, uuwuu, uwuuu, wuuuu, uuuuz, uuuzu, uuzuu, uzuuu, zuuuu, uuuux, uuuxu, uuxuu, uxuuu and xuuuu. - Zerinvary Lajos, May 19 2008

Also convolution of A000302 with A038846, also convolution of A002457 with A020918, also convolution of A002697 with A038845, also convolution of A002802 with A002802. [Rui Duarte, Oct 08 2011]

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..400
• Eric Weisstein's World of Mathematics, Idempotent Number.
• Index entries for linear recurrences with constant coefficients, signature (20, -160, 640, -1280, 1024).

Crossrefs

Cf. A000302, A020920, A000984.

Cf. A038231.

Programs

GAP

List([0..30], n-> 4^n*Binomial(n+4, 4)); # G. C. Greubel, Jul 20 2019

Magma

[4^n*Binomial(n+4, 4): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

Maple

seq(seq(binomial(i, j)*4^(i-4), j =i-4), i=4..22); # Zerinvary Lajos, Dec 03 2007
seq(binomial(n+4,4)*4^n,n=0..30); # Zerinvary Lajos, May 19 2008
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, Z, Z, Z, B, B, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/24, n=4..34); # Zerinvary Lajos, Apr 05 2009

Mathematica

Table[Binomial[n+4,4]*4^n, {n,0,30}] (* Michael De Vlieger, Aug 21 2015 *)

PARI

vector(30, n, n--; 4^n*binomial(n+4, 4)) \\ G. C. Greubel, Jul 20 2019

Sage

[lucas_number2(n, 4, 0)*binomial(n,4)/2^8 for n in range(4, 34)] # Zerinvary Lajos, Mar 11 2009

Formula

a(n) = binomial(n+4, 4)*4^n.

G.f.: 1/(1-4*x)^5.

a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10 = n } f(i_1)*f(i_2) *f(i_3)*f(i_4)*f(i_5)*f(i_6)*f(i_7)*f(i_8)*f(i_9)*f(i_10) with f(k)=A000984(k). - Rui Duarte, Oct 08 2011

E.g.f.: (3 + 48*x + 144*x^2 + 128*x^3 + 32*x^4)*exp(4*x)/3. - G. C. Greubel, Jul 20 2019

From Amiram Eldar, Mar 25 2022: (Start)

Sum_{n>=0} 1/a(n) = 376/3 - 432*log(4/3).

Sum_{n>=0} (-1)^n/a(n) = 2000*log(5/4) - 1336/3. (End)