A367198 - OEIS

%I #19 Nov 10 2023 18:01:55

%S 1,1,2,4,6,3,15,30,18,4,76,165,125,40,5,455,1075,930,380,75,6,3186,

%T 8015,7679,3675,945,126,7,25487,67536,70042,37688,11550,2044,196,8,

%U 229384,634935,702372,414078,144417,30870,3990,288,9,2293839,6591943,7696245,4886390,1885065,463092,73080,7200,405,10

%N T(n, k) = Sum_{m = 0..n-1} Stirling1(m+1, k)*binomial(n, m)*(-1)^(n + k), where "Stirling1" are the signed Stirling numbers of the first kind.

%C To use the unsigned Stirling numbers rewrite the formula as: T(n, k) = Sum_{m = 0..n-1} abs(Stirling1(m+1, k))*binomial(n, m)*(-1)^(1+m+n). Replacing in this formula Stirling1 (A008275) by Stirling2 (A048993) one obtains a shifted version of A321331.

%F T(n+1, n) = n^2*(n+1)/2 = A002411(n).

%F T(n, n-2) = 6*T(n-1, n-3) - 15*T(n-2, n-4) + 20*T(n-3, n-5) - 15*T(n-4, n-6) + 6*T(n-5, n-7) - T(n-6, n-8), for n > 8.

%F T(n, n-k) = (-1)^k*Sum_{m=0..n-1} Stirling1(m+1, n-k)*binomial(n, m).

%e Triangle begins:

%e     1;

%e     1,    2;

%e     4,    6,   3;

%e    15,   30,  18,   4;

%e    76,  165, 125,  40,  5;

%e   455, 1075, 930, 380, 75, 6;

%p T := (n, k) -> local m; add(Stirling1(m+1, k)*binomial(n, m)*(-1)^(n + k), m = 0..n-1): seq(seq(T(n, k), k = 1..n), n = 1..9);  # _Peter Luschny_, Nov 10 2023

%o (PARI) T(n,k) = sum(m=0, n-1, stirling(m+1, k)*binomial(n, m)*(-1)^(n+k))

%Y Cf. A002411, A002467 (first column), A000027 (main diagonal), A008275.

%Y Cf. A180191(n+1) (row sums), A321331 (variant with Stirling2).

%K nonn,tabl

%O 1,3

%A _Thomas Scheuerle_, Nov 10 2023