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

%I #30 Oct 11 2019 21:12:47

%S 1,1,2,6,24,60,180,630,2520,7560,25200,92400,369600,1201200,4204200,

%T 15765750,63063000,214414200,771891120,2933186256,11732745024,

%U 41064607584,150570227808,577185873264,2308743493056,8245512475200

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

%C 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.

%C Number of permutation patterns modulo 4. This matches the multinomial formula. - _Olivier GĂ©rard_, Feb 25 2011

%C Also the number of permutations of n elements where p(k-4) < p(k) for all k. - _Joerg Arndt_, Jul 23 2011

%H Alois P. Heinz, <a href="/A022917/b022917.txt">Table of n, a(n) for n = 0..1000</a>

%F 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

%F From _Vaclav Kotesovec_, Mar 15 2019: (Start)

%F 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).

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

%e 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.

%p A022917 := proc(n)

%p n!/floor(n/4)!/floor((n+1)/4)!/floor((n+2)/4)!/floor((n+3)/4)! ;

%p end proc: # _R. J. Mathar_, Aug 06 2015

%t Table[ n!/(Quotient[n, 4]!*Quotient[n + 1, 4]!*Quotient[n + 2, 4]!*

%t Quotient[n + 3, 4]!), {n, 0, 30}]

%t Table[n!/Times@@(Floor[Range[n,n+3]/4]!),{n,0,30}] (* _Harvey P. Dale_, May 30 2018 *)

%o (PARI) {a(n)= if(n<0, 0, n!/(n\4)!/((n+1)\4)!/((n+2)\4)!/((n+3)\4)!)} /* _Michael Somos_, Jun 20 2007 */

%Y Cf. A001405 (permutation patterns mod 2).

%Y Cf. A022916 (permutation patterns mod 3).

%K nonn,easy,nice

%O 0,3

%A _Clark Kimberling_, Jun 14 1998

%E Corrected by _Michael Somos_, Jun 20 2007