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Number of compositions of n into floor(j/3) kinds of j's for all j>=1.
2

%I #30 Oct 14 2024 00:10:53

%S 1,0,0,1,1,1,3,4,5,10,15,21,36,56,83,134,210,320,505,791,1221,1911,

%T 2988,4639,7240,11305,17595,27436,42806,66691,103968,162144,252720,

%U 393965,614285,957581,1492791,2327396,3628273,5656274,8818275,13747425,21431700,33411976,52088551,81204526,126596778,197361904,307682405

%N Number of compositions of n into floor(j/3) kinds of j's for all j>=1.

%C The g.f. for compositions of k_1 kinds of 1's, k_2 kinds of 2's, ..., k_j kinds of j's, ... is 1/(1-sum(j>=1, k_j * x^j )).

%H Jarib R. Acosta, Yadira Caicedo, Juan P. Poveda, José L. Ramírez, Mark Shattuck, <a href="https://www.emis.de/journals/JIS/VOL22/Shattuck/shattuck13.html">Some New Restricted n-Color Composition Functions</a>, J. Int. Seq., Vol. 22 (2019), Article 19.6.4.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 2, -1).

%F G.f.: 1/(1-sum(j>=1, floor(j/3)*x^j )).

%F Conjectural g.f.: (x-1)^2*(x^2+x+1) / (x^4-2*x^3-x+1). - _Colin Barker_, May 15 2013

%F G.f.: 1 + x^3*Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k+1 + 2*x^2 - x^3)/( x*(4*k+3 + 2*x^2 - x^3 ) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Sep 11 2013

%o (PARI) N=66; x='x+O('x^N) /* that many terms */

%o gf= 1/(1-sum(j=1,N, floor(j/3)*x^j ))

%o Vec(gf) /* show terms */

%Y Cf. A121907 (floor((3*j-1)/2)), A055841 (3*j-1), A052156 (2*j-1), A006053 (floor(j/2)), A143787 (floor((3*j)/2)).

%K nonn

%O 0,7

%A _Joerg Arndt_, Jul 06 2011