OFFSET
1,1
COMMENTS
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.
FORMULA
t(k, m) = 2^k + k*2^m + 2^(m-1), k >= m >= 1, otherwise 0.
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)).
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
Subt(m, x) = Sum_{k=0..m-1} A288870(m-1, k)*(2*x)^k.
EXAMPLE
The triangle t begins:
k\m 1 2 3 4 5 6 7 8 9 10 ...
1: 5
2: 9 14
3: 15 22 36
4: 25 34 52 88
5: 43 54 76 120 208
6: 77 90 116 168 272 480
7: 143 158 188 248 368 608 108
8: 273 290 324 392 528 800 1344 2432
9: 531 550 588 664 816 1120 1728 2944 5376
10: 1045 1066 1108 1192 1360 1696 2368 3712 6400 11776
...
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.
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).
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Jun 21 2017
STATUS
approved