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%I #33 Mar 14 2024 15:57:03
%S 1,1,4,1,24,16,1,124,240,64,1,624,2656,1792,256,1,3124,26400,33920,
%T 11520,1024,1,15624,250096,546560,331520,67584,4096,1,78124,2313360,
%U 8105664,7822080,2745344,372736,16384,1,390624,21132736,114627072,165398016,88940544,20299776,1966080,65536,1,1953124,191757120,1574682880,3270274560,2529343488,863256576,138215424,10027008,262144
%N Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).
%C For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references.
%C This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0].
%C For the Sheffer triangle (exp(x), (1/4)*(exp(4*x) - 1)) see A111578, also the P. Bala link where this triangle, T(n, m)/4^m, is named S_{(4,0,1)}. The triangle T(n, m)*m! is given in A285066.
%C The a-sequence for this Sheffer triangle has e.g.f. 4*x/log(1+x) and is 4*A006232(n)/A006233(n) (Cauchy numbers of the first kind).
%C The z-sequence has e.g.f. (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)) and is A285062(n)/A285063(n).
%C The main diagonal gives A000302.
%C The row sums give A285064. The alternating row sums give A285065.
%C The first column sequences are A000012, 4*A003463, 4^2*A016234. For the e.g.f.s and o.g.f.s see the formula section.
%C This triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 4*m)^n}_{m>=0}, as G(n, x) = Sum_{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 4*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
%C The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225118 with row reversion.
%C In general the Sheffer triangle S2[d,a] appears in the reordering of the operator (a*1 + d*E_x) = Sum_{m=0..n} S2(d,a;n,m) x^m*(d_x)^m with the derivative d_x and the Euler operator E_x = x*d_x. For [d,a] = [1,0] this becomes the standard Stirling2 property.
%H Michael De Vlieger, <a href="/A285061/b285061.txt">Table of n, a(n) for n = 0..11475</a>, rows n = 0..150, flattened.
%H Paweł Hitczenko, <a href="https://arxiv.org/abs/2403.03422">A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality</a>, arXiv:2403.03422 [math.CO], 2024. See p. 9.
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/1707.04451">On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers</a>, arXiv:1707.04451 [math.NT], 2017.
%F Three term recurrence (from the conversion property mentioned above, with [d,a] =[4,1]): T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m. T(n, m) = 4*T(n-1, m-1) + (1 + 4*m)*T(n-1, m) for n >= 1, m = 0..n.
%F T(n ,m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 4*k)^n/m!, 0 <= m <= n, satisfying the recurrence.
%F E.g.f. of the row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)*exp(x*exp(4*x) -1)). This is the e.g.f. of the triangle.
%F E.g.f. for the sequence of column m: exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
%F O.g.f. for sequence of column m: (4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x) (by Laplace transform of the e.g.f.).
%F T(n, m) = Sum_{k=0..n} binomial(n, k)* 4^k*Stirling2(k, m), 0 <= m <= n, where Stirling2 is given in A048993.
%F A nontrivial recurrence for the column m=0 entries T(n, 0) = 4^n from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
%F Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)* Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
%F Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((1 + 4*x) + 4*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
%F Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*[(n/2)*(2 + 4*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n, p)(-4)^(n-p)*Bernoulli(n-p)*T(p, m)], for n > m >= 0, with input T(m,m) = 4^m. See a comment and references in A282629. An example is given below. - _Wolfdieter Lang_, Aug 11 2017
%e The triangle T(n,m) begins:
%e n\m 0 1 2 3 4 6 7 8 9
%e 0: 1
%e 1: 1 4
%e 2: 1 24 16
%e 3: 1 124 240 64
%e 4: 1 624 2656 1792 256
%e 5: 1 3124 26400 33920 11520 1024
%e 6: 1 15624 250096 546560 331520 67584 4096
%e 7: 1 78124 2313360 8105664 7822080 2745344 372736 16384
%e 8: 1 390624 21132736 114627072 165398016 88940544 20299776 1966080 65536
%e ...
%e Three term recurrence: T(4, 1) = 4*1 + (1 + 4*1)*124 = 624.
%e Recurrence for row polynomial R(3, x) (Meixner type): ((1 + 4*x) + 4*x*d_x)*(1 + 24*x + 16*x^2) = 1 + 124*x + 240*x^2 + 64*x^3.
%e Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) =(1/2)*[2*(2 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2))] = (1/2)*(20*240 + 12*16*(1/6)*16) = 2656. - _Wolfdieter Lang_, Aug 11 2017
%t Table[Sum[Binomial[n, k]*4^k*StirlingS2[k, m], {k, 0, n}], {n, 0, 20}, {m, 0, n}] // Flatten (* _Indranil Ghosh_, May 06 2017 *)
%o (PARI) T(n, m) = sum(k=0, n, binomial(n, k)*4^k*stirling(k, m, 2));
%o for(n=0, 20, for(m=0, n, print1(T(n, m),", ");); print();) \\ _Indranil Ghosh_, May 06 2017
%Y Cf. A000012, A000302, 4*A003463, A111578, A006232/A006233, 16*A016234, A282629, A285062/A285063, A285064, A285065, A285066.
%K nonn,easy,tabl
%O 0,3
%A _Wolfdieter Lang_, Apr 13 2017
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