A022917 Multinomial coefficient n!/ ([n/4]!, [(n+1)/4]!, [(n+2)/4]!, [(n+3)/4]!).

1, 1, 2, 6, 24, 60, 180, 630, 2520, 7560, 25200, 92400, 369600, 1201200, 4204200, 15765750, 63063000, 214414200, 771891120, 2933186256, 11732745024, 41064607584, 150570227808, 577185873264, 2308743493056, 8245512475200

Offset: 0

Author

Clark Kimberling, Jun 14 1998

Keywords

• nonn
• easy
• nice

Comments

Multinomial coefficients(TOP, BOTTOM), where TOP = n, BOTTOM = ( a b c d ) where a = [ n/4 ], b = [ n/4 ], c = [ n/4 ], d = [ n/4 ] if n==0 mod 4; a = [ n/4 ], b = [ n/4 ], c = [ n/4 ], d = [ n/4 ]+1 if n==1 mod 4; a = [ n/4 ], b = [ n/4 ], c = [ n/4 ]+1, d = [ n/4 ]+1 if n==2 mod 4; a = [ n/4 ], b = [ n/4 ]+1, c = [ n/4 ]+1, d = [ n/4 ]+1 if n==3 mod 4.

Number of permutation patterns modulo 4. This matches the multinomial formula. - Olivier Gérard, Feb 25 2011

Also the number of permutations of n elements where p(k-4) < p(k) for all k. - Joerg Arndt, Jul 23 2011

Examples

			Starting from n=5, several permutations have the same pattern mod 4. Both (4,1,5,2,3) and (4,5,1,2,3) have pattern (0,1,1,2,3) modulo 4.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A001405 (permutation patterns mod 2).

Cf. A022916 (permutation patterns mod 3).

Programs

Maple

A022917 := proc(n)
    n!/floor(n/4)!/floor((n+1)/4)!/floor((n+2)/4)!/floor((n+3)/4)! ;
end proc: # R. J. Mathar, Aug 06 2015

Mathematica

Table[ n!/(Quotient[n, 4]!*Quotient[n + 1, 4]!*Quotient[n + 2, 4]!*
    Quotient[n + 3, 4]!), {n, 0, 30}]
Table[n!/Times@@(Floor[Range[n,n+3]/4]!),{n,0,30}] (* Harvey P. Dale, May 30 2018 *)

PARI

{a(n)= if(n<0, 0, n!/(n\4)!/((n+1)\4)!/((n+2)\4)!/((n+3)\4)!)} /* Michael Somos, Jun 20 2007 */

Formula

Conjecture: -(126*n-1)*(n+3)*(n+2)*(n+1)*a(n) +4*(-208*n^4-51*n^3+793*n^2+376*n-6)*a(n-1) +16*(334*n^3-114*n^2-5*n-369)*a(n-2) +64*(n-2)*(334*n^2+264*n-213)*a(n-3) +768*(n-2)*(n-3)*(42*n^2+153*n+113)*a(n-4) +1024*(208*n+177)*(n-2)*(n-3)*(n-4)*a(n-5)=0. - R. J. Mathar, Aug 06 2015

From Vaclav Kotesovec, Mar 15 2019: (Start)

Recurrence: (n+1)*(n+2)*(n+3)*(4*n^3 + 8*n^2 + 4*n - 1)*a(n) = 8*(2*n^2 + 2*n - 1)*(n^3 + 4*n^2 + 7*n + 3)*a(n-1) + 16*(n-1)*(4*n^4 + 16*n^3 + 16*n^2 + 18*n + 15)*a(n-2) + 128*(n-2)*(n-1)*(2*n^3 + 14*n^2 + 31*n + 18)*a(n-3) + 256*(n-3)*(n-2)*(n-1)*(4*n^3 + 20*n^2 + 32*n + 15)*a(n-4).

a(n) ~ 2^(2*n + 5/2) / (Pi*n)^(3/2). (End)

Extensions

Corrected by Michael Somos, Jun 20 2007