OFFSET
0,8
COMMENTS
We aim at a q-generalization of the Comtet-Lehmer numbers A354794, which are the case q = 1. Here we consider the case q = 2. The generalization is based on the qStirling numbers, for qStirling1 see A342186 and for qStirling2 see A139382. The general construction is as follows:
Let q <> 1 be a fixed integer and f_q(k) = (q^k - 1)/(q - 1) for k >= 0. Define triangle M(q; n, k) for 0 <= k <= n by M(q; n, 0) = 0^n for n >= 0, and M(q; n, k) = 0 for k > n, and M(q; n, k) = M(q; n-1, k-1) + M(q; n-1, k) * f_q(k) for 0 < k <= n. Then M(q; n, n) = 1 for n >= 0 and the matrix inverse I_q = M_q^(-1) exists. Next define the triangle T(q; n, k) for 0 <= k <= n by T(q; n, 0) = 0^n for n >= 0 and T(q; n, k) = Sum_{i=0..n-k} I(q; n-k, i) * M(q; n-1+i, n-1) for 0 < k <= n. Take account of lim_{q->1} (q^n - 1)/(q - 1) = n for n >= 0.
Conjecture: T(q; n+1, 1) = Sum_{i=0..n} I(q; n, i) * M(q; n+i, n) = (f_q(n))^n = ((q^n - 1)/(q - 1))^n for n >= 0.
Conjecture: T(q; n, k) = (Sum_{i=0..n-k} (-1)^i * q-binomial(n-1-i, k-1) * binomial(n-1, i) * q^((n-k)*(n-k-i))) / (q - 1)^(n-k) for 0 < k <= n.
FORMULA
Conjecture: T(n+1, 1) = (2^n - 1)^n for n >= 0.
Conjecture: T(n, k) = Sum_{i=0..n-k} (-1)^i * binomial(n-1, i) * [n-1-i, k-1]_2 * 2^((n-k)*(n-k-i)) for 0 < k <= n and T(n, 0) = 0^n for n >= 0, where [x, y]_2 = A022166(x, y).
EXAMPLE
Triangle T(n, k) for 0 <= k <= n starts:
n\k : 0 1 2 3 4 5 6 7 8
===============================================================================
0 : 1
1 : 0 1
2 : 0 1 1
3 : 0 9 4 1
4 : 0 343 79 11 1
5 : 0 50625 6028 454 26 1
6 : 0 28629151 1741861 68710 2190 57 1
7 : 0 62523502209 1926124954 38986831 656500 9687 120 1
8 : 0 532875860165503 8264638742599 84816722571 734873171 5760757 40929 247 1
etc.
MAPLE
# using qStirling2 from A333143.
A355282 := proc(n, k) if k = 0 then 0^n elif n = k then 1 else
add(A342186(n - k, i)*qStirling2(n + i - 2, n - 2, 2), i = 1..n-k) fi end:
seq(print(seq(A355282(n, k), k = 0..n)), n = 0..8); # Peter Luschny, Jun 28 2022
PROG
(PARI) mat(nn) = my(m = matrix(nn, nn)); for (n=1, nn, for(k=1, nn, m[n, k] = if (n==1, if (k==1, 1, 0), if (k==1, 1, (2^k-1)*m[n-1, k] + m[n-1, k-1])); ); ); m; \\ A139382
tabl(nn) = my(m=mat(3*nn), im=1/m); matrix(nn, nn, n, k, n--; k--; if (k==0, 0^n, k<n, sum(i=1, n-k, im[n-k, i]*m[n-1+i, n-1]), if (k==n, 1, 0))); \\ Michel Marcus, Jun 27 2022
CROSSREFS
KEYWORD
AUTHOR
Werner Schulte, Jun 26 2022
STATUS
approved