A154537 -id:A154537 - cp's OEIS frontend

A225466 Triangle read by rows, 3^k*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

1, 2, 3, 4, 21, 9, 8, 117, 135, 27, 16, 609, 1431, 702, 81, 32, 3093, 13275, 12015, 3240, 243, 64, 15561, 115479, 171990, 81405, 13851, 729, 128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187, 256, 390369, 7998111, 27533142, 29893941, 13121514, 2561706

Offset: 0

Peter Luschny, May 08 2013

The definition of the Stirling-Frobenius subset numbers of order m is in A225468.
From Wolfdieter Lang, Apr 09 2017: (Start)
This is the Sheffer triangle (exp(2*x), exp(3*x) - 1), denoted by S2[3,2]. See also A282629 for S2[3,1]. The stirling2 triangle A048993 is in this notation denoted by S2[1,0].
The a-sequence for this Sheffer triangle has e.g.f. 3*x/log(1+x) and is 3*A006232(n)/A006233(n) (Cauchy numbers of the first kind). For a- and z-sequences for Sheffer triangles see the W. Lang link under A006232, also with references).
The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(2/3)) and gives 2*A284862/A284863.
The first column k sequences divided by 3^k are A000079, A016127, A016297, A025999. For the e.g.f.s and o.g.f.s see below.
The row sums give A284864. The alternating row sums give A284865.
This triangle appears in the o.g.f. G(n, x) of the sequence {(2 + 3*m)^n}{m>=0}, as G(n, x) = Sum{k=0..n} T(n, k)*k!*x^k/(1-x)^(k+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0} (2 + 3*m)^n t^m/m! = exp(t)*Sum_{k=0..n} T(n, k)*t^k.
The corresponding Eulerian number triangle is A225117(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, m)*m!, 0 <= k <= n. (End)

			[n\k][ 0,     1,      2,       3,       4,      5,     6,    7]
[0]    1,
[1]    2,     3,
[2]    4,    21,      9,
[3]    8,   117,    135,      27,
[4]   16,   609,   1431,     702,      81,
[5]   32,  3093,  13275,   12015,    3240,    243,
[6]   64, 15561, 115479,  171990,   81405,  13851,   729,
[7]  128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see the Maple program): T(4, 2) = 3*T(3, 1) + (3*2+2)*T(3, 2) = 3*117 + 8*135 = 1431.
Boas-Buck recurrence for column k = 2, and n = 4: T(4,2) = (1/2)*(2*(4 + 3*2)*T(3, 2) + 2*6*(-3)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(20*135 + 12*9*(1/6)*9) = 1431. (End)

Vincenzo Librandi, Rows n = 0..50, 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.
Peter Luschny, Eulerian polynomials.
Peter Luschny, The Stirling-Frobenius numbers.
Shi-Mei Ma, Toufik Mansour, and Matthias Schork, Normal ordering problem and the extensions of the Stirling grammar, Russian Journal of Mathematical Physics, 2014, 21(2), arXiv:1308.0169 [math.CO], 2013, p. 12.

Cf. A048993 (m=1), A154537 (m=2), A225467 (m=4), A225468.

Cf. A000079, A000244, A005057, A016127, A016297, A025999, A006232/A006233, A225117, A225472, A225468, A282629, A284862/A284863, A284864, A284865.

SF_SS := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or  k < 0 then return(0) fi;
m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:
seq(print(seq(SF_SS(n, k, 3), k=0..n)), n=0..5);

EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)

T(n, k) = sum(j=0, k, binomial(k, j)*(-1)^(j - k)*(2 + 3*j)^n/k!);
for(n=0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 10 2017

from sympy import binomial, factorial
def T(n, k): return sum(binomial(k, j)*(-1)**(j - k)*(2 + 3*j)**n//factorial(k) for j in range(k + 1))
for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017

@CachedFunction
def EulerianNumber(n, k, m) :
    if n == 0: return 1 if k == 0 else 0
    return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m) + (m*k+1)*EulerianNumber(n-1,k,m)
def SF_SS(n, k, m):
    return add(EulerianNumber(n,j,m)*binomial(j,n-k) for j in (0..n))/ factorial(k)
def A225466(n): return SF_SS(n, k, 3)

T(n, k) = (1/k!)*Sum_{j=0..n} binomial(j, n-k)*A_3(n, j) where A_m(n, j) are the generalized Eulerian numbers A225117.
For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A005057; T(n, n) ~ A000244.
From Wolfdieter Lang, Apr 09 2017: (Start)
T(n, k) = Sum_{j=0..k} binomial(k,j)*(-1)^(j-k)*(2 + 3*j)^n/k!, 0 <= k <= n.
E.g.f. of triangle: exp(2*z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column k is exp(2*x)*((exp(3*x) - 1)^k)/k! (Sheffer property).
O.g.f. for sequence of column k is 3^k*x^k/Product_{j=0..k} (1 - (2+3*j)*x).
A nontrivial recurrence for the column m=0 entries T(n, 0) = 2^n from the z-sequence given above: T(n,0) = n*Sum_{k=0..n-1} z(k)*T(n-1,k), n >= 1, T(0, 0) = 1.
The corresponding recurrence for columns k >= 1 from the a-sequence is T(n, k) = (n/k)* Sum_{j=0..n-k} binomial(k-1+j, k-1)*a(j)*T(n-1, k-1+j).
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+2) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
(End)
Boas-Buck recurrence for column sequence k: T(n, k) = (1/(n - k))*((n/2)*(4 + 3*k)*T(n-1, k) + k*Sum_{p=k..n-2} binomial(n, p)*(-3)^(n-p)*Bernoulli(n-p)*T(p, k)), for n > k >= 0, with input T(k, k) = 3^k. See a comment and references in A282629, An example is given below. - Wolfdieter Lang, Aug 11 2017

A016756 a(n) = (2*n+1)^4.

Original entry on oeis.org

1, 81, 625, 2401, 6561, 14641, 28561, 50625, 83521, 130321, 194481, 279841, 390625, 531441, 707281, 923521, 1185921, 1500625, 1874161, 2313441, 2825761, 3418801, 4100625, 4879681, 5764801, 6765201, 7890481, 9150625, 10556001, 12117361, 13845841, 15752961, 17850625

Offset: 0

N. J. A. Sloane

a(n) is the number of ordered pairs of lattice points (vectors in R^2 with integer coordinates) that are in or on a square centered at the origin with side length 2*n. - Geoffrey Critzer, Apr 20 2013

			a(1) = 81 because there are 9 lattice points in or on the 2 x 2 square centered at the origin, so there are 9*9 =81 ordered pairs. - _Geoffrey Critzer_, Apr 20 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1)

Cf. A016755, A060187, A154537, A300707.

[(2*n+1)^4: n in [0..40]]; // Vincenzo Librandi, Sep 07 2011

Table[(2n+1)^4,{n,0,25}]  (* Geoffrey Critzer, Apr 20 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,81,625,2401,6561},30] (* Harvey P. Dale, Mar 24 2020 *)

vector(40, n, n--; (2*n+1)^4) \\ G. C. Greubel, Sep 15 2018

From Wolfdieter Lang, Mar 12 2017: (Start)
G.f.: (1+76*x+230*x^2+76*x^3+x^4)/(1-x)^5; see row n=5 of A060187.
E.g.f.: (1 + 80*x + 232*x^2 + 128*x^3 + 16*x^4)*exp(x); see row n=4 of A154537. (End)
Sum_{n>=0} 1/a(n) = Pi^4/96 (A300707). - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = (cos(Pi/sqrt(2)) + cosh(Pi/sqrt(2)))/2.
Product_{n>=1} (1 - 1/a(n)) = Pi*cosh(Pi/2)/8. (End)

A157779 Numerator of Bernoulli(n, 1/2).

Original entry on oeis.org

1, 0, -1, 0, 7, 0, -31, 0, 127, 0, -2555, 0, 1414477, 0, -57337, 0, 118518239, 0, -5749691557, 0, 91546277357, 0, -1792042792463, 0, 1982765468311237, 0, -286994504449393, 0, 3187598676787461083, 0, -4625594554880206790555, 0, 16555640865486520478399, 0

Offset: 0

N. J. A. Sloane, Nov 08 2009

Included for completeness, normally alternating zeros like this are omitted. A001896 is the official version of this sequence.

The sequence {a(n)/A141459(n)} gives the generalized Bernoulli numbers B[2,1] obtained from the generalized Stirling2 triangle S3[2,1] = A154537. See the formula section. - Wolfdieter Lang, Apr 27 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..250
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.

For denominators see A157780 and A141459.

Numerator[BernoulliB[Range[0,40],1/2]] (* Harvey P. Dale, May 04 2013 *)

a(n) = numerator(subst(bernpol(n, x), x, 1/2)); \\ Altug Alkan, Jul 05 2016

def A157779_list(size):
    f = x / sum(x^(n*2+1)/factorial(n*2+1) for n in (0..2*size))
    t = taylor(f, x, 0, size)
    return [(factorial(n)*s).numerator() for n,s in enumerate(t.list())]
print(A157779_list(33)) # Peter Luschny, Jul 05 2016

Let P(x) = Sum_{n>=0} x^(2*n+1)/(2*n+1)!; then a(n) = numerator( n! [x^n] x/P(x) ). - Peter Luschny, Jul 05 2016
a(n) = numerator(r(n)) with the rationals r(n) = Sum_{k=0..n} ((-1)^k / (k+1))*A154537(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))*A145901(n, k). The denominators are in A141459. r(n) = B[2,1](n) = 2^n*B(n, 1/2) with the Bernoulli polynomials A196838/A196839 or A053382/A053383. - Wolfdieter Lang, Apr 27 2017
a(n) = numerator(-(1-2^(1-n))*Bernoulli(n)). - Fabián Pereyra, Dec 31 2022

A225467 Triangle read by rows, T(n, k) = 4^k*S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 3, 4, 9, 40, 16, 27, 316, 336, 64, 81, 2320, 4960, 2304, 256, 243, 16564, 63840, 54400, 14080, 1024, 729, 116920, 768496, 1071360, 485120, 79872, 4096, 2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384, 6561, 5758240, 101417920, 322850304

Offset: 0

Peter Luschny, May 08 2013

The definition of the Stirling-Frobenius subset numbers of order m is in A225468.

This is the Sheffer triangle (exp(3*x), exp(4*x) - 1). See also the P. Bala link under A225469, the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x) - 1)), which is named there exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017

			[n\k][  0,      1,       2,        3,        4,       5,      6,     7]
[0]     1,
[1]     3,      4,
[2]     9,     40,      16,
[3]    27,    316,     336,       64,
[4]    81,   2320,    4960,     2304,      256,
[5]   243,  16564,   63840,    54400,    14080,    1024,
[6]   729, 116920,  768496,  1071360,   485120,   79872,   4096,
[7]  2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see the Maple program): T(4, 2) = 4*T(3, 1) + (4*2+3)*T(3, 2) = 4*316 + 11*336 = 4960.
Boas-Buck recurrence for column k = 2, and n = 4: T(4, 2) = (1/2)*(2*(6 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(28*336 + 12*16*(1/6)*16) = 4960. (End)

Vincenzo Librandi, Rows n = 0..50, 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.
Peter Luschny, Eulerian polynomials.
Peter Luschny, The Stirling-Frobenius numbers.

Cf. A048993 (m=1), A154537 (m=2), A225466 (m=3). A225469 (scaled).

Cf. Columns: A000244, 4*A016138, 16*A018054. A225118.

SF_SS := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or  k < 0 then return(0) fi;
m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:
seq(print(seq(SF_SS(n, k, 4), k=0..n)), n=0..5);

EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)

T(n, k) = sum(m=0, k, binomial(k, m)*(-1)^(m - k)*((3 + 4*m)^n)/k!);
for(n = 0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 13 2017

from sympy import binomial, factorial
def T(n, k): return sum(binomial(k, m)*(-1)**(m - k)*((3 + 4*m)**n)//factorial(k) for m in range(k + 1))
for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 13 2017

@CachedFunction
def EulerianNumber(n, k, m) :
    if n == 0: return 1 if k == 0 else 0
    return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m)+(m*k+1)*EulerianNumber(n-1,k,m)
def SF_SS(n, k, m):
    return add(EulerianNumber(n,j,m)*binomial(j,n-k) for j in (0..n))/factorial(k)
def A225467(n): return SF_SS(n, k, 4)

T(n, k) = (1/k!)*sum_{j=0..n} binomial(j, n-k)*A_4(n, j) where A_m(n, j) are the generalized Eulerian numbers A225118.
For a recurrence see the Maple program.
T(n, 0) ~ A000244; T(n, 1) ~ A190541.
T(n, n) ~ A000302; T(n, n-1) ~ A002700.
From Wolfdieter Lang, Apr 13 2017: (Start)
T(n, k) = Sum_{m=0..k} binomial(k,m)*(-1)^(m-k)*((3+4*m)^n)/k!, 0 <= k <= n.
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^(n-k)*4^k*Stirling2(k, m), 0 <= m <= n.
E.g.f. exp(3*z)*exp(x*(exp(4*z) - 1)) (Sheffer property).
E.g.f. column k: exp(3*x)*((exp(4*x) - 1)^k)/k!, k >= 0.
O.g.f. column k: (4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)
Boas-Buck recurrence for column sequence k: T(n, k) = (1/(n - k))*((n/2)*(6 + 4*k)*T(n-1, k) + k*Sum_{p=k..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, k)), for n > k >= 0, with input T(k, k) = 4^k. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017

A290319 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 4x)^(-1/4), (-1/4)log(1 - 4*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 45, 59, 15, 1, 585, 812, 254, 28, 1, 9945, 14389, 5130, 730, 45, 1, 208845, 312114, 122119, 20460, 1675, 66, 1, 5221125, 8011695, 3365089, 633619, 62335, 3325, 91, 1, 151412625, 237560280, 105599276, 21740040, 2441334, 158760, 5964, 120, 1, 4996616625, 7990901865, 3722336388, 823020596, 102304062, 7680414, 355572, 9924, 153, 1, 184874815125, 300659985630, 145717348221, 34174098440, 4608270890, 386479380, 20836578, 722760, 15585, 190, 1

Offset: 0

Wolfdieter Lang, Aug 08 2017

This generalization of the unsigned Stirling1 triangle A132393 is called here |S1hat[4,1]|.
The signed matrix S1hat[4,1] with elements (-1)^(n-k)*|S1hat[4,1]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[4,1] with elements S2[4,1](n, k)/d^k, where S2[4,1] is Sheffer (exp(x), exp(4*x) - 1), given in A285061. See also the P. Bala link below for the scaled and signed version s_{(4,0,1)}.
For the general |S1hat[d,a]| case see a comment in A286718.

			The triangle T(n, k) begins:
  n\k         0         1         2        3       4      5    6   7  8 ...
  0:          1
  1:          1         1
  2:          5         6         1
  3:         45        59        15        1
  4:        585       812       254       28       1
  5:       9945     14389      5130      730      45      1
  6:     208845    312114    122119    20460    1675     66    1
  7:    5221125   8011695   3365089   633619   62335   3325   91   1
  8:  151412625 237560280 105599276 21740040 2441334 158760 5964 120  1
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (16 - 3)*T(3, 2) = 59 + 13*15 = 254.
Boas-Buck recurrence for column k=2 and n=4:
T(4, 2) = (4!/2)*(4*(1 + 8*(5/12))*T(2, 2)/2! + 1*(1 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(2*13/3 + 5*15/3!) = 254. (End)

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
Peter Bala, A 3 parameter family of generalized Stirling numbers.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,1], [3,2] and [4,3] is A132393, A028338, A286718, A225470 and A225471, respectively.
Columns k=0..3 give A007696, A024382(n-1), A383700, A383701.
Row sums: A001813. Alternating row sums: A000007.

FoldList[Join[Table[If[i == 1, 0, #[[i-1]]] + (4*#2 - 3)*#[[i]], {i, Length[#]}], {1}] &, {1}, Range[10]] (* Paolo Xausa, Aug 18 2025 *)

Recurrence: T(n, k) = T(n-1, k-1) + (4*n - 3)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle): (1 - 4*z)^{-(x + 1)/4}.
E.g.f. of column k is (1 - 4*x)^(-1/4)*((-1/4)*log(1 - 4*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+4), with R(0, x) = 1. Row polynomial R(n, x) = risefac(4,1;x,n) with the rising factorial risefac(d,a;x,n) :=Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0, a_1, ..., a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 4*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*4^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(1 + 4*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

cp's OEIS Frontend

A225466 Triangle read by rows, 3^k*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

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A016756 a(n) = (2*n+1)^4.

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A157779 Numerator of Bernoulli(n, 1/2).

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A225467 Triangle read by rows, T(n, k) = 4^k*S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

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A290319 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 4*x)^(-1/4), (-1/4)*log(1 - 4*x)). A generalized Stirling1 triangle.

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A290319 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 4x)^(-1/4), (-1/4)log(1 - 4*x)). A generalized Stirling1 triangle.