A016234 -id:A016234 - cp's OEIS frontend

A285061 Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).

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

Wolfdieter Lang, Apr 13 2017

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.

			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

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.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.

Cf. A000012, A000302, 4*A003463, A111578, A006232/A006233, 16*A016234, A282629, A285062/A285063, A285064, A285065, A285066.

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 *)

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

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*z) - 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

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

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 31, 15, 1, 1, 156, 166, 28, 1, 1, 781, 1650, 530, 45, 1, 1, 3906, 15631, 8540, 1295, 66, 1, 1, 19531, 144585, 126651, 30555, 2681, 91, 1, 1, 97656, 1320796, 1791048, 646086, 86856, 4956, 120, 1, 1, 488281, 11984820, 24604420, 12774510

Offset: 1

Gary W. Adamson, Aug 07 2005

From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the exponential Riordan array [exp(z), (exp(4*z) - 1)/4].
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.
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.
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)

			The triangle starts in row n=1 as:
  1;
  1,1;
  1,6,1;
  1,31,15,1;
  1,156,166,28,1;
Connection constants: Row 4: [1, 31, 15, 1] so
x^3 = 1 + 31*(x - 1) + 15*(x - 1)*(x - 5) + (x - 1)*(x - 5)*(x - 9). - _Peter Bala_, Jan 27 2015

P. Bala, A 3 parameter family of generalized Stirling numbers

Cf. A111577, A008277, A039755, A016234 (3rd column).

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 *)

def A096038(n,m):
    if n < 1 or m < 1 or m > n:
        return 0
    elif n <=2:
        return 1
    else:
        return A096038(n-1,m-1)+(4*m-3)*A096038(n-1,m)
print( [A096038(n,m) for n in range(20) for m in range(1,n+1)] )
# R. J. Mathar, Oct 11 2009

From Peter Bala, Jan 27 2015: (Start)
The following formulas assume an offset of 0.
T(n,k) = 1/(4^k*k!)*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*(4*j + 1)^n.
T(n,k) = sum {i = 0..n-1} 4^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
E.g.f.: exp(z)*exp(x/4*(exp(4*z) - 1)) = 1 + (1 + x)*z + (1 + 6*x + x^2)*z^2/2! + ....
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!.
O.g.f. column k: 1/( (1 - x)*(1 - 5*x)*...*(1 - (4*k + 1)*x) ). (End)

Edited and extended by R. J. Mathar, Oct 11 2009

A285066 Triangle read by rows: T(n, m) = A285061(n, m)*m!, 0 <= m <= n.

Original entry on oeis.org

1, 1, 4, 1, 24, 32, 1, 124, 480, 384, 1, 624, 5312, 10752, 6144, 1, 3124, 52800, 203520, 276480, 122880, 1, 15624, 500192, 3279360, 7956480, 8110080, 2949120, 1, 78124, 4626720, 48633984, 187729920, 329441280, 268369920, 82575360, 1, 390624, 42265472, 687762432, 3969552384, 10672865280, 14615838720, 9909043200, 2642411520, 1, 1953124, 383514240, 9448097280, 78486589440, 303521218560, 621544734720, 696605736960, 404288962560, 95126814720

Offset: 0

Wolfdieter Lang, Apr 19 2017

This is the Sheffer triangle S2[4,1] = A285061 with column m scaled by m!. This is the fourth member of the triangle family A131689, A145901 and A284861.
This triangle appears in the o.g.f. G(n, x) = Sum_{m=0..n} T(n, m)*x^m/(1-x)^(m+1), n >= 0, of the power sequence {(1+4*m)^n}_{m >= 0}.
The diagonal sequence is A047053. The row sums give A285067. The alternating sum of row n is A141413(n+2), n >= 0.
The first column sequences are: A000012, 4*A003463, 2!*4^2*A016234.

			The triangle T(n, m) begins:
  n\m 0     1       2        3         4         5         6        7
  0:  1
  1:  1     4
  2:  1    24      32
  3:  1   124     480      384
  4:  1   624    5312    10752      6144
  5:  1  3124   52800   203520    276480    122880
  6:  1 15624  500192  3279360   7956480   8110080   2949120
  7:  1 78124 4626720 48633984 187729920 329441280 268369920 82575360
  ...
row 8: 1 390624 42265472 687762432 3969552384 10672865280 14615838720 9909043200 2642411520
row 9: 1 1953124 383514240 9448097280 78486589440 303521218560 621544734720 696605736960 404288962560 95126814720
...

Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.

Cf. A000012, 4*A003463, 32*A016234, A111578 (T(n, m)/4^m), A131689, A141413, A145901, A225118, A225472, A225473, A284861, A285061.

T[n_, m_]:=Sum[Binomial[m, k]*(-1)^(k - m)*(1 + 4k)^n, {k, 0, n}]; Table[T[n, m], {n, 0, 10},{m, 0, n}] // Flatten (* Indranil Ghosh, May 02 2017 *)

from sympy import binomial
def T(n, m):
    return sum([binomial(m, k)*(-1)**(k - m)*(1 + 4*k)**n for k in range(n + 1)])
for n in range(21):
    print([T(n, m) for m in range(n + 1)])
# Indranil Ghosh, May 02 2017

T(n, m) = A285061(n, m)*m! = A111578(n, m)*(4^m*m!), 0 <= m <= n.
T(n, m) = Sum_{k=0..n} binomial(m,k)*(-1)^(k-m)*(1+4*k)^n.
T(n, m) = Sum_{j=0..n} binomial(n-j,m-j)*A225118(n,n-j).
Recurrence: T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m and T(n, m) =
4*m*T(n-1, m-1) + (1+4*m)*T(n-1, m) for n >= 1, m=0..n.
E.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)/(1 - x*(exp(4*z) - 1)).
E.g.f. column m: exp(x)*(exp(4*x) - 1)^m, m >= 0.
O.g.f. column m: m!*(4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x), m >= 0.

A016247 Expansion of 1/((1-x) * (1-6x) (1-11*x)).

Original entry on oeis.org

1, 18, 241, 2910, 33565, 378546, 4219993, 46755846, 516329845, 5691721530, 62681496241, 689931815118, 7591862105101, 83526155988930, 918881752875145, 10108263503608086, 111194283871577893, 1223157434578690506, 13454853652313597665, 148004121407137586910, 1628049722868641531581

Offset: 0

N. J. A. Sloane

nonn

Index entries for linear recurrences with constant coefficients, signature (18, -83, 66).

Cf. A016209, A016223, A016234.

CoefficientList[Series[1/((1-x)(1-6x)(1-11x)),{x,0,30}],x] (* or *) LinearRecurrence[{18,-83,66},{1,18,241},30] (* Harvey P. Dale, Sep 23 2012 *)

a(n) = (11^(n+2) - 2*6^(n+2) + 1)/50; \\ Joerg Arndt, Aug 13 2013

a(0)=1, a(1)=18, a(n)=17*a(n-1)-66*a(n-2)+1. - Vincenzo Librandi, Feb 10 2011
a(0)=1, a(1)=18, a(2)=241, a(n)=18*a(n-1)-83*a(n-2)+66*a(n-3). - Harvey P. Dale, Sep 23 2012
a(n) = (11^(n+2) - 2*6^(n+2) + 1)/50. [Yahia Kahloune, Aug 13 2013]
From Seiichi Manyama, May 05 2025: (Start)
a(n) = Sum_{k=0..n} 5^k * binomial(n+2,k+2) * Stirling2(k+2,2).
a(n) = Sum_{k=0..n} (-5)^k * 11^(n-k) * binomial(n+2,k+2) * Stirling2(k+2,2). (End)

cp's OEIS Frontend

A285061 Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).

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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.

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A285066 Triangle read by rows: T(n, m) = A285061(n, m)*m!, 0 <= m <= n.

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A016247 Expansion of 1/((1-x) * (1-6*x) * (1-11*x)).

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A016247 Expansion of 1/((1-x) * (1-6x) (1-11*x)).