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A111578 Triangle T(n, m) = T(n-1, m-1) + (4m-3)*T(n-1, m) read by rows 1<=m<=n.

%I A111578 #19 Sep 06 2024 22:23:06
%S A111578 1,1,1,1,6,1,1,31,15,1,1,156,166,28,1,1,781,1650,530,45,1,1,3906,
%T A111578 15631,8540,1295,66,1,1,19531,144585,126651,30555,2681,91,1,1,97656,
%U A111578 1320796,1791048,646086,86856,4956,120,1,1,488281,11984820,24604420,12774510
%N A111578 Triangle T(n, m) = T(n-1, m-1) + (4m-3)*T(n-1, m) read by rows 1<=m<=n.
%C A111578 From _Peter Bala_, Jan 27 2015: (Start)
%C A111578 Working with an offset of 0, this is the exponential Riordan array [exp(z), (exp(4*z) - 1)/4].
%C A111578 This is the triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*4^n * binomial((x - 1)/4,n) }n>=0. An example is given below.
%C A111578 Call this array M and let P denote Pascal's triangle A007318 then P^2 * M = A225469; P^(-1) * M is a shifted version of A075499.
%C A111578 This triangle is the particular case a = 4, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
%H A111578 P. Bala, <a href="/A143395/a143395.pdf">A 3 parameter family of generalized Stirling numbers</a>
%F A111578 From _Peter Bala_, Jan 27 2015: (Start)
%F A111578 The following formulas assume an offset of 0.
%F A111578 T(n,k) = 1/(4^k*k!)*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*(4*j + 1)^n.
%F A111578 T(n,k) = sum {i = 0..n-1} 4^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
%F A111578 E.g.f.: exp(z)*exp(x/4*(exp(4*z) - 1)) = 1 + (1 + x)*z + (1 + 6*x + x^2)*z^2/2! + ....
%F A111578 O.g.f. for n-th diagonal: exp(-x/4)*sum {k >= 0} (4*k + 1)^(k + n - 1)*((x/4*exp(-x))^k)/k!.
%F A111578 O.g.f. column k: 1/( (1 - x)*(1 - 5*x)*...*(1 - (4*k + 1)*x) ). (End)
%e A111578 The triangle starts in row n=1 as:
%e A111578   1;
%e A111578   1,1;
%e A111578   1,6,1;
%e A111578   1,31,15,1;
%e A111578   1,156,166,28,1;
%e A111578 Connection constants: Row 4: [1, 31, 15, 1] so
%e A111578 x^3 = 1 + 31*(x - 1) + 15*(x - 1)*(x - 5) + (x - 1)*(x - 5)*(x - 9). - _Peter Bala_, Jan 27 2015
%t A111578 T[n_, k_] := 1/(4^(k-1)*(k-1)!) * Sum[ (-1)^(k-j-1) * (4*j+1)^(n-1) * Binomial[k-1, j], {j, 0, k-1}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jan 28 2015, after _Peter Bala_ *)
%o A111578 (Python)
%o A111578 def A096038(n,m):
%o A111578     if n < 1 or m < 1 or m > n:
%o A111578         return 0
%o A111578     elif n <=2:
%o A111578         return 1
%o A111578     else:
%o A111578         return A096038(n-1,m-1)+(4*m-3)*A096038(n-1,m)
%o A111578 print( [A096038(n,m) for n in range(20) for m in range(1,n+1)] )
%o A111578 # _R. J. Mathar_, Oct 11 2009
%Y A111578 Cf. A111577, A008277, A039755, A016234 (3rd column).
%K A111578 nonn,tabl
%O A111578 1,5
%A A111578 _Gary W. Adamson_, Aug 07 2005
%E A111578 Edited and extended by _R. J. Mathar_, Oct 11 2009