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A288871 Triangle t needed for the e.g.f.s of the column sequences of A288870 with leading zeros.

%I A288871 #7 Jul 10 2017 04:18:30
%S A288871 5,9,14,15,22,36,25,34,52,88,43,54,76,120,208,77,90,116,168,272,480,
%T A288871 143,158,188,248,368,608,1088,273,290,324,392,528,800,1344,2432,531,
%U A288871 550,588,664,816,1120,1728,2944,5376,1045,1066,1108,1192,1360,1696,2368,3712,6400,11776
%N A288871 Triangle t needed for the e.g.f.s of the column sequences of A288870 with leading zeros.
%C A288871 See the triangle T = A288870. The e.g.f. of the sequence of column k (k >= 0) without the leading k zeros is E(k, x) = (2*k+1)*exp(2*x) + exp(x). In order to get the e.g.f. for the column k sequence with the leading k zeros one has to integrate k times for k >=1; but this will first generate unwanted fractional numbers for the first k entries (when no integration constants are taken into account). These rational polynomials of degree k to be subtracted are S(k, x) = 2^(-k)* Sum_{m=1..k} t(k,m)*x^(m-1)/(m-1)! if k >=1.
%F A288871 t(k, m) = 2^k + k*2^m + 2^(m-1), k >= m >= 1, otherwise 0.
%F A288871 O.g.f. column m: G(m, x) =x*(2*x)^(m-1)*(3 - 5*x + 2*(1 - 3*x + 2*x^2)*m)/((1-x)^2*(1-2*x)).
%F A288871 O.g.f. G(m, x) = 1/(1-2*x) + 2^m*x/(1-x)^2 + 2^(m-1)/(1-x) - Subt(m ,x), with
%F A288871   Subt(m, x) = Sum_{k=0..m-1} A288870(m-1, k)*(2*x)^k.
%e A288871 The triangle t begins:
%e A288871 k\m     1    2    3    4    5    6    7    8    9    10 ...
%e A288871 1:      5
%e A288871 2:      9   14
%e A288871 3:     15   22   36
%e A288871 4:     25   34   52   88
%e A288871 5:     43   54   76  120  208
%e A288871 6:     77   90  116  168  272  480
%e A288871 7:    143  158  188  248  368  608  108
%e A288871 8:    273  290  324  392  528  800 1344 2432
%e A288871 9:    531  550  588  664  816 1120 1728 2944 5376
%e A288871 10:  1045 1066 1108 1192 1360 1696 2368 3712 6400 11776
%e A288871 ...
%e A288871 k = 1: E(1, x) = 3*exp(2*x) + exp(x) generates exponentially: 4, 7, 13, 25, 49, ..., the column k = 1 of T = A288870 without leading zero. Integration gives (without integration constant) (3/2)*exp(2*x) + exp(x), generating 5/2, 4, 7, 13, 25, 49, ..., therefore 5/2 = 2^(-1)* t(1,1)*x^(1-1)/(1-1)!= 2^(-1)*5*x^0 = 5/2.
%e A288871 Column o.g.f. for m=2: G(2, x) = 1/(1-2*x) + 4*x/(1-x)^2 + 2/(1-x) - (3 + 2^1*4*x) = 2*x^2*(7-17*x+8*x^2)/((1 - 2*x)*( 1 - x)^2).
%Y A288871 Cf. A288870.
%K A288871 nonn,tabl,easy
%O A288871 1,1
%A A288871 _Wolfdieter Lang_, Jun 21 2017