A232394 The number of compositions of n with no more than 3 consecutive identical parts (summands).

1, 1, 2, 4, 7, 15, 29, 57, 111, 218, 429, 841, 1651, 3239, 6355, 12473, 24475, 48029, 94249, 184946, 362932, 712194, 1397569, 2742507, 5381729, 10560797, 20723884, 40667338, 79803197, 156601100, 307304821, 603036937, 1183364302, 2322164658, 4556879623

Offset: 0

Author

Geoffrey Critzer, Nov 23 2013

Keywords

• nonn

Examples

			a(6) = 29 because there are 32 compositions of 6 but we exclude: 1+1+1+1+1+1, 1+1+1+1+2, 2+1+1+1+1.
		

Links

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

Crossrefs

Cf. A003242, A128695, A232464.

Programs

Maple

b:= proc(n, t, c) option remember;
       `if`(n=0, 1, add(`if`(t<>j, b(n-j, j, 1),
       `if`(c<3, b(n-j, j, c+1), 0)), j=1..n))
    end:
a:= n-> b(n, 0, 1):
seq(a(n), n=0..50);  # Alois P. Heinz, Nov 24 2013

Mathematica

nn=34; CoefficientList[Series[1/(1-Sum[(z^j+z^(2j)+z^(3j))/(1+z^j+z^(2j)+z^(3j)),{j,1,nn}]),{z,0,nn}],z]

Formula

The g.f. for the number of compositions of n with no more than m consecutive identical parts is 1/( 1 - sum_{j>=1} x^j*(1 - x^(j*m))/(1 - x^j)/ (1 + x^j*(1 - x^(j*m))/(1 - x^j)) ); set m = 3 for this sequence.

a(n) ~ c * d^n, where d=1.962341312018097075518216734398388302205091029921968626465436021267458..., c=0.506212613637348069558928622560083229757824786467201325660889396545904... - Vaclav Kotesovec, May 01 2014