A049323 Triangle of coefficients of certain polynomials (exponents in increasing order), equivalent to A033842.
1, 1, 1, 1, 3, 3, 1, 6, 16, 16, 1, 10, 50, 125, 125, 1, 15, 120, 540, 1296, 1296, 1, 21, 245, 1715, 7203, 16807, 16807, 1, 28, 448, 4480, 28672, 114688, 262144, 262144, 1, 36, 756, 10206, 91854, 551124, 2125764, 4782969, 4782969, 1, 45, 1200, 21000, 252000
Offset: 0
Examples
The triangle a(n, m) begins:
n\m 0 1 2 3 4 5 6 7 ...
0: 1
1: 1 1
2: 1 3 3
3: 1 6 16 16
4: 1 10 50 125 125
5: 1 15 120 540 1296 1296
6: 1 21 245 1715 7203 16807 16807
7: 1 28 448 4480 28672 114688 262144 262144
... reformatted. - Wolfdieter Lang, Nov 20 2015
E.g. the third row {1,3,3} corresponds to polynomial p(2,x)= 1 + 3*x + 3*x^2.
Links
- W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Crossrefs
a(n, 0)= A000012 (powers of 1), a(n, 1)= A000217 (triangular numbers), a(n, n)= A000272(n+1), n >= 0 (diagonal), a(n, n-1)= A000272(n+1), n >= 1.
Programs
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Magma
/* As triangle: */ [[Binomial(n+1, k+1)*(n+1)^(k-1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 20 2015
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Maple
seq(seq(binomial(n+1,m+1)*(n+1)^(m-1),m=0..n),n=0..10); # Robert Israel, Oct 19 2015
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Mathematica
Table[Binomial[n + 1, k + 1] (n + 1)^(k - 1), {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 19 2015 *)
Formula
a(n, m) = A033842(n, n-m) = binomial(n+1, m+1)*(n+1)^{m-1}, n >= m >= 0, else 0.
p(k-1, -x)/(1-k*x)^k =(-1+1/(1-k*x)^k)/(x*k^2) is for k=1..5 G.f. for A000012, A001792, A036068, A036070, A036083, respectively.
From Werner Schulte, Oct 19 2015: (Start)
a(2*n,n) = A000108(n)*(2*n+1)^n;
a(3*n,2*n) = A001764(n)*(3*n+1)^(2*n);
a(p*n,(p-1)*n) = binomial(p*n,n)/((p-1)*n+1)*(p*n+1)^((p-1)*n) for p > 0;
Sum_{m=0..n} (m+1)*a(n,m) = (n+2)^n;
Sum_{m=0..n} (-1)^m*(m+1)*a(n,m) = (-n)^n where 0^0 = 1;
p(n,x) = Sum_{m=0..n} a(n,m)*x^m = ((1+(n+1)*x)^(n+1)-1)/((n+1)^2*x).
(End)
Comments