A369950 Triangle read by rows: T(n,k) = number of j-covers of [n] with j<=k, k=1..2^n-1.
1, 1, 4, 5, 1, 13, 45, 80, 101, 108, 109, 1, 40, 361, 1586, 4505, 9482, 15913, 22348, 27353, 30356, 31721, 32176, 32281, 32296, 32297, 1, 121, 2681, 27671, 182777, 894103, 3491513, 11348063, 31483113, 75820263, 160485753, 301604003
Offset: 1
Examples
Triangle (with rows n >= 1 and columns k >= 1) begins as follows: 1; 1, 4, 5; 1, 13, 45, 80, 101, 108, 109; 1, 40, 361, 1586, 4505, 9482, 15913, 22348, 27353, 30356, 31721, 32176, 32281, 32296, 32297; ... There are T(3,2) = 13 covers of [3] consisting of up to 2 subsets (brackets and commas omitted): 123 123 1 123 2 123 3 123 12 123 13 123 23 12 13 12 23 13 23 12 3 13 2 23 1
Links
- John Tyler Rascoe, Rows n = 1..9, flattened
Programs
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Mathematica
Flatten[Table[Sum[Sum[StirlingS1[i+1, j+1] (2^j-1)^n, {j, 0, i}]/i!, {i, k}], {n, 6}, {k, 2^n-1}]]
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Python
from math import comb def A369950(n,k): return sum((-1)**j*comb(n, j)*comb(2**(n-j)-1, i) for j in range(n+1) for i in range(1,k+1)) # John Tyler Rascoe, Mar 06 2025
Formula
T(n,k) = Sum_{i=1..k} (1/i!)*Sum_{j=0..i} Stirling1(i+1, j+1)*(2^j-1)^n.
T(n,k) = Sum_{i=1..k} Sum_{j=0..n} (-1)^j*C(n, j)*C(2^(n-j)-1, i).
T(n,2^n-1) = A003465(n).
Comments