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A285061
Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).
10
1, 1, 4, 1, 24, 16, 1, 124, 240, 64, 1, 624, 2656, 1792, 256, 1, 3124, 26400, 33920, 11520, 1024, 1, 15624, 250096, 546560, 331520, 67584, 4096, 1, 78124, 2313360, 8105664, 7822080, 2745344, 372736, 16384, 1, 390624, 21132736, 114627072, 165398016, 88940544, 20299776, 1966080, 65536, 1, 1953124, 191757120, 1574682880, 3270274560, 2529343488, 863256576, 138215424, 10027008, 262144
OFFSET
0,3
COMMENTS
For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references.
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].
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.
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).
The z-sequence has e.g.f. (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)) and is A285062(n)/A285063(n).
The main diagonal gives A000302.
The row sums give A285064. The alternating row sums give A285065.
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.
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.
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.
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.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..11475, rows n = 0..150, flattened.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.
FORMULA
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.
T(n ,m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 4*k)^n/m!, 0 <= m <= n, satisfying the recurrence.
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.
E.g.f. for the sequence of column m: exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
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.).
T(n, m) = Sum_{k=0..n} binomial(n, k)* 4^k*Stirling2(k, m), 0 <= m <= n, where Stirling2 is given in A048993.
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.
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.
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.
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
EXAMPLE
The triangle T(n,m) begins:
n\m 0 1 2 3 4 6 7 8 9
0: 1
1: 1 4
2: 1 24 16
3: 1 124 240 64
4: 1 624 2656 1792 256
5: 1 3124 26400 33920 11520 1024
6: 1 15624 250096 546560 331520 67584 4096
7: 1 78124 2313360 8105664 7822080 2745344 372736 16384
8: 1 390624 21132736 114627072 165398016 88940544 20299776 1966080 65536
...
Three term recurrence: T(4, 1) = 4*1 + (1 + 4*1)*124 = 624.
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.
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
MATHEMATICA
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 *)
PROG
(PARI) T(n, m) = sum(k=0, n, binomial(n, k)*4^k*stirling(k, m, 2));
for(n=0, 20, for(m=0, n, print1(T(n, m), ", "); ); print(); ) \\ Indranil Ghosh, May 06 2017
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Apr 13 2017
STATUS
approved