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A049323 Triangle of coefficients of certain polynomials (exponents in increasing order), equivalent to A033842.

%I A049323 #25 Sep 08 2022 08:44:58
%S A049323 1,1,1,1,3,3,1,6,16,16,1,10,50,125,125,1,15,120,540,1296,1296,1,21,
%T A049323 245,1715,7203,16807,16807,1,28,448,4480,28672,114688,262144,262144,1,
%U A049323 36,756,10206,91854,551124,2125764,4782969,4782969,1,45,1200,21000,252000
%N A049323 Triangle of coefficients of certain polynomials (exponents in increasing order), equivalent to A033842.
%C A049323 These polynomials p(n, x) appear in the W. Lang reference as c1(-(n+1);x), n >= 0 on p.12. The coefficients are given there in eq.(44) on p. 6. - _Wolfdieter Lang_, Nov 20 2015
%H A049323 W. Lang, <a href="https://www.cs.uwaterloo.ca/journals/JIS/VOL3/LANG/lang.html">On generalizations of Stirling number triangles</a>, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
%F A049323 a(n, m) = A033842(n, n-m) = binomial(n+1, m+1)*(n+1)^{m-1}, n >= m >= 0, else 0.
%F A049323 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.
%F A049323 From _Werner Schulte_, Oct 19 2015: (Start)
%F A049323 a(2*n,n) = A000108(n)*(2*n+1)^n;
%F A049323 a(3*n,2*n) = A001764(n)*(3*n+1)^(2*n);
%F A049323 a(p*n,(p-1)*n) = binomial(p*n,n)/((p-1)*n+1)*(p*n+1)^((p-1)*n) for p > 0;
%F A049323 Sum_{m=0..n} (m+1)*a(n,m) = (n+2)^n;
%F A049323 Sum_{m=0..n} (-1)^m*(m+1)*a(n,m) = (-n)^n where 0^0 = 1;
%F A049323 p(n,x) = Sum_{m=0..n} a(n,m)*x^m = ((1+(n+1)*x)^(n+1)-1)/((n+1)^2*x).
%F A049323 (End)
%e A049323 The triangle a(n, m) begins:
%e A049323 n\m 0  1   2    3     4      5      6      7 ...
%e A049323 0:  1
%e A049323 1:  1  1
%e A049323 2:  1  3   3
%e A049323 3:  1  6  16   16
%e A049323 4:  1 10  50  125  125
%e A049323 5:  1 15 120  540  1296  1296
%e A049323 6:  1 21 245 1715  7203  16807  16807
%e A049323 7:  1 28 448 4480 28672 114688 262144 262144
%e A049323 ... reformatted. - Wolfdieter Lang, Nov 20 2015
%e A049323 E.g. the third row {1,3,3} corresponds to polynomial p(2,x)= 1 + 3*x + 3*x^2.
%p A049323 seq(seq(binomial(n+1,m+1)*(n+1)^(m-1),m=0..n),n=0..10); # _Robert Israel_, Oct 19 2015
%t A049323 Table[Binomial[n + 1, k + 1] (n + 1)^(k - 1), {n, 0, 9}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Nov 19 2015 *)
%o A049323 (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
%Y A049323 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.
%Y A049323 For n = 0..5 the row sequences a(n, m), m >= 0, are the first columns of the triangles A023531 (unit matrix), A030528, A049324, A049325, A049326, A049327, respectively.
%Y A049323 Cf. A033842, A046757.
%K A049323 nonn,tabl,easy
%O A049323 0,5
%A A049323 _Wolfdieter Lang_