A note on nested sums. (English) Zbl 1197.05003
Summary: We consider several nested sums, and show how binomial coefficients, Stirling numbers of the second kind and Gaussian binomial coefficients can be written as nested sums. We use this to find the rate of growth for diagonals of Stirling numbers of the second kind, as well as another proof of a known identity for Gaussian binomial coefficients.
MSC:
| 05A10 | Factorials, binomial coefficients, combinatorial functions |
| 11B73 | Bell and Stirling numbers |
| 11B65 | Binomial coefficients; factorials; \(q\)-identities |
Software:
OEISOnline Encyclopedia of Integer Sequences:
Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.4-dimensional pyramidal numbers: a(n) = (3*n+1)*binomial(n+2, 3)/4. Also Stirling2(n+2, n).
Stirling numbers of the second kind S(n+3, n).
a(n) = n*(n+1)*(n+2)/2.
a(n) = n*(n+1)*(n+2)*(n+3)/4.
T(n, k) = [x^k] (-1)^n*Sum_{k=0..n} E2(n, n-k)*(1+x)^(n-k) where E2(n, k) are the second-order Eulerian numbers. Triangle read by rows, T(n, k) for n >= 1 and 0 <= k <= n.
Sixth diagonal of the Stirling2 triangle A048993 and sixth column of triangle A008278.
Number of ways to place 3 points on a triangular grid of side n so that they are not vertices of an equilateral triangle of any orientation.
Triangle read by rows, giving coefficients in an expansion of absolute values of Stirling numbers of the first kind in terms of binomial coefficients.
a(n) = 10*binomial(n+4, 5).
a(n) = 21*binomial(n+6,7).
n*(2*n+1)*binomial(n+2,n)/3.
