A059025 Triangle of Stirling numbers of order 6.
1, 1, 1, 1, 1, 1, 1, 462, 1, 1716, 1, 4719, 1, 11440, 1, 25883, 1, 56134, 1, 118456, 2858856, 1, 245480, 23279256, 1, 502588, 124710300, 1, 1020680, 551496660, 1, 2061709, 2181183147, 1, 4149752, 8021782197, 1, 8333153, 28051272535
Offset: 6
Examples
There are 462 ways of partitioning a set N of cardinality 12 into 2 blocks each of cardinality at least 6, so S_6(12,2)=462.
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
- J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.
Links
- Michael De Vlieger, Table of n, a(n) for n = 6..13205 (rows n = 6..400, flattened).
- Bishal Deb and Alan D. Sokal, Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness, arXiv:2507.18959 [math.CO], 2025. See p. 5.
- A. E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.
Programs
-
Mathematica
S6[n_ /; 6 <= n <= 11, 1] = 1; S6[n_, k_] /; 1 <= k <= Floor[n/6] := S6[n, k] = k*S6[n-1, k] + Binomial[n-1, 5]*S6[n-6, k-1]; S6[, ] = 0; Flatten[ Table[ S6[n, k], {n, 6, 24}, {k, 1, Floor[n/6]}]] (* Jean-François Alcover, Feb 21 2012 *)
Formula
S_r(n+1, k)=k S_r(n, k)+binomial(n, r-1)S_r(n-r+1, k-1) for this sequence, r=6.
G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t - Sum_{i=0..r-1} t^i/i!)).
Comments