A054340 10-fold convolution of A000302 (powers of 4).

1, 40, 880, 14080, 183040, 2050048, 20500480, 187432960, 1593180160, 12745441280, 96865353728, 704475299840, 4931327098880, 33381291130880, 219362770288640, 1403921729847296, 8774510811545600, 53679360258867200, 322076161553203200, 1898554215471513600

Offset

0,2

Comments

With a different offset, number of n-permutations (n>=9) of 5 objects: u, v, z, x, y with repetition allowed, containing exactly nine (9) u's. - Zerinvary Lajos, Jul 02 2008

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..400

Index entries for linear recurrences with constant coefficients, signature (40,-720,7680,-53760,258048,-860160,1966080,-2949120,2621440,-1048576).

Formula

a(n) = binomial(n+9, 9)*4^n.

G.f.: 1/(1 - 4*x)^10.

a(n) = A054335(n+19, 19).

E.g.f.: (2^7/9!)*(2835 + 102060*x + 816480*x^2 + 2540160*x^3 + 3810240*x^4 + 3048192*x^5 + 1354752*x^6 + 331776*x^7 + 41472*x^8 + 2048*x^9)*exp(4*x).

From Amiram Eldar, Mar 27 2022: (Start)

Sum_{n>=0} 1/a(n) = 236196*log(4/3) - 4756383/70.

Sum_{n>=0} (-1)^n/a(n) = 14062500*log(5/4) - 43931373/14. (End)

Maple

seq(binomial(n+9, 9)*4^n, n=0..20); # Zerinvary Lajos, Jul 02 2008

Mathematica

Table[4^n*Binomial[n+9, 9], {n, 0, 20}] (* G. C. Greubel, Jul 21 2019 *)

Prog

(Magma) [4^n*Binomial(n+9, 9): n in [0..20]]; // Vincenzo Librandi, Oct 15 2011
(PARI) vector(20, n, n--; 4^n*binomial(n+9, 9)) \\ G. C. Greubel, Jul 21 2019
(GAP) List([0..20], n-> 4^n*Binomial(n+9, 9)); # G. C. Greubel, Jul 21 2019

Crossrefs

Cf. A000302, A054335.

Sequence in context: A247411 A250986 A190825 * A271030 A013350 A013346

Adjacent sequences: A054337 A054338 A054339 * A054341 A054342 A054343

Keyword

easy,nonn

Author

Wolfdieter Lang, Mar 13 2000

Status

approved