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Number of 4-gonal compositions of n into positive parts.
2

%I #29 Sep 26 2019 09:41:26

%S 0,0,0,0,1,4,10,16,31,40,68,80,125,140,206,224,315,336,456,480,633,

%T 660,850,880,1111,1144,1420,1456,1781,1820,2198,2240,2675,2720,3216,

%U 3264,3825,3876,4506,4560,5263,5320,6100,6160,7021,7084,8030

%N Number of 4-gonal compositions of n into positive parts.

%H Colin Barker, <a href="/A069982/b069982.txt">Table of n, a(n) for n = 0..1000</a>

%H G. E. Andrews, P. Paule and A. Riese, <a href="https://www.researchgate.net/publication/277299165_MacMahon&#39;s_Partition_Analysis_III_The_Omega_Package">MacMahon's partition analysis III. The Omega package</a>, p. 17.

%H G. E. Andrews, P. Paule and A. Riese, <a href="https://doi.org/10.1006/eujc.2001.0527">MacMahon's Partition Analysis: The Omega Package</a>, European Journal of Combinatorics, Vol. 22, No. 7 (2001), 887-904.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).

%F G.f.: q^4/(1-q)^4-4*q^7/(1-q)^4/(1+q)^3.

%F a(n) = (2*n^3-3*n^2-23*n+3*(13+(n^2-7*n+11)*(-1)^n))/24. - _Luce ETIENNE_, Jul 02 2015; edited by _Mo Li_, Sep 18 2019

%F a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n>7. - _Colin Barker_, Sep 18 2019

%t Table[Piecewise[{

%t {Binomial[k - 1, k - 4] - 4*Binomial[(k - 1)/2, (k - 7)/2], Mod[k, 2] == 1},

%t {Binomial[k - 1, k - 4] - 4*Binomial[(k - 2)/2, (k - 8)/2], Mod[k, 2] == 0}}], {k, 1, 20}] (* _Mo Li_, Sep 18 2019 *)

%o (PARI) concat([0,0,0,0], Vec(x^4*(1 + 3*x + 3*x^2 - 3*x^3) / ((1 - x)^4*(1 + x)^3) + O(x^40))) \\ _Colin Barker_, Sep 18 2019

%Y Cf. A069981, A069983, A005044.

%K nonn,easy

%O 0,6

%A _N. J. A. Sloane_, May 06 2002