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Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.
+10
13
1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
COMMENTS
This is a generalization of the unsigned Stirling1 triangle A132393. In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/ A002209(k+1). This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018
REFERENCES
Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
FORMULA
Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*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) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;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 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393. Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/ A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 ...
O: 1
1: 1 1
2: 4 5 1
3: 28 39 12 1
4: 280 418 159 22 1
5: 3640 5714 2485 445 35 1
6: 58240 95064 45474 9605 1005 51 1
7: 1106560 1864456 959070 227969 28700 1974 70 1
8: 24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
MATHEMATICA
T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[_, -1] = 0; T[0, 0] = 1; T[n_, k_] /; n<k = 0;
CROSSREFS
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], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
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.
+10
11
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
COMMENTS
The definition of the Stirling-Frobenius subset numbers of order m is in A225468. 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. 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)
FORMULA
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, 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 m: T(n, k) = (1/(n - m))*[(n/2)*(4 + 3*m)*T(n-1, k) + m* Sum_{p=m..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
EXAMPLE
[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.
...
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 m = 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)
MAPLE
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);
MATHEMATICA
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 *)
PROG
(Sage)
@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) (PARI) 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 (Python)
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
CROSSREFS
Cf. A000079, A000244, A005057, A016127, A016297, A025999, A006232/ A006233, A225117, A225472, A225468, A282629, A284862/ A284863, A284864, A284865.
Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 4}(x) in descending order.
+10
8
1, 3, 1, 9, 22, 1, 27, 235, 121, 1, 81, 1996, 3446, 620, 1, 243, 15349, 63854, 40314, 3119, 1, 729, 112546, 963327, 1434812, 422087, 15618, 1, 2187, 806047, 12960063, 37898739, 26672209, 4157997, 78117, 1, 6561, 5705752, 162711868, 840642408, 1151050534
COMMENTS
The row sums equal the quadruple factorial numbers A047053 and the alternating row sums, i.e., sum((-1)^k*T(n,k),k=0..n), are up to a sign A079858. - Johannes W. Meijer, May 04 2013
FORMULA
G.f. of the polynomials is gf(n, k) = k^n*n!*(1/x-1)^(n+1)[t^n](x*e^(t*x/k)*(1-x*e(t*x))^(-1)) for k = 4; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
E.g.f. of row polynomials (rising powers of x): (1-x)*exp(3*(1-x)*z)/(1-y*exp(4*(1-x)*z)), i.e. e.g.f. of the triangle.
E.g.f. for the row polynomials with falling powers of x (A_{n, 4}(x) of the name): (1-x)*exp((1-x)*z)/(1 - x*exp(4*(1-x)*z)).
T(n, k) = Sum_{j=0..k} (-1)^(k-j) * binomial(n+1,k-j) * (3+4*j)^n, 0 <= k <= n.
Recurrence: T(n, k) = (4*(n-k) + 1)*T(n-1, k-1) + (3 + 4*k)*T(n-1, k), n >= 1, with T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k. (End)
In terms of Euler's triangle = A123125: T(n, k) = Sum_{m=0..n} (binomial(n, m)*3^(n-m)*4^m*Sum_{p=0..k} (-1)^(k-p)*binomial(n-m, k-p)* A123125(m, p)), 0 <= k <= n. - Wolfdieter Lang, Apr 13 2017
EXAMPLE
[0] 1
[1] 3*x + 1
[2] 9*x^2 + 22*x + 1
[3] 27*x^3 + 235*x^2 + 121*x + 1
[4] 81*x^4 + 1996*x^3 + 3446*x^2 + 620*x + 1
...
The triangle T(n, k) begins:
n\k
0: 1
1: 3 1
2: 9 22 1
3: 27 235 121 1
4: 81 1996 3446 620 1
5: 243 15349 63854 40314 3119 1
6: 729 112546 963327 1434812 422087 15618 1
7: 2187 806047 12960063 37898739 26672209 4157997 78117 1
...
row n=8: 6561 5705752 162711868 840642408 1151050534 442372648 39531132 390616 1,
row n=9: 19683 40156777 1955297356 16677432820 39523450714 29742429982 6818184988 367889284 1953115 1.
MAPLE
gf := proc(n, k) local f; f := (x, t) -> x*exp(t*x/k)/(1-x*exp(t*x));
series(f(x, t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
collect(simplify(%), x) end:
seq(print(seq(coeff(gf(n, 4), x, n-k), k=0..n)), n=0..6);
# Recurrence:
P := proc(n, x) option remember; if n = 0 then 1 else
(n*x+(1/4)*(1-x))*P(n-1, x)+x*(1-x)*diff(P(n-1, x), x);
expand(%) fi end:
A225117 := (n, k) -> 4^n*coeff(P(n, x), x, n-k):
MATHEMATICA
gf[n_, k_] := Module[{f, s}, f[x_, t_] := x*Exp[t*x/k]/(1-x*Exp[t*x]); s = Series[f[x, t], {t, 0, n+2}]; ((1-x)/x)^(n+1)*k^n*n!*SeriesCoefficient[s, {t, 0, n}]]; Table[Table[SeriesCoefficient[gf[n, 4], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
PROG
(Sage)
@CachedFunction
def EB(n, k, x): # Modified cardinal B-splines
if n == 1: return 0 if (x < 0) or (x >= 1) else 1
return k*x*EB(n-1, k, x) + k*(n-x)*EB(n-1, k, x-1)
def EulerianPolynomial(n, k): # Generalized Eulerian polynomials
R.<x> = ZZ[]
if x == 0: return 1
return add(EB(n+1, k, m+1/k)*x^m for m in (0..n))
[EulerianPolynomial(n, 4).coefficients()[::-1] for n in (0..5)]
Triangle read by rows, k!*S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
5
1, 3, 4, 9, 40, 32, 27, 316, 672, 384, 81, 2320, 9920, 13824, 6144, 243, 16564, 127680, 326400, 337920, 122880, 729, 116920, 1536992, 6428160, 11642880, 9584640, 2949120, 2187, 821356, 17842272, 114866304, 324065280, 453304320, 309657600, 82575360, 6561
COMMENTS
The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).
FORMULA
For a recurrence see the Maple program.
T(n, k) = Sum_{m=0..n} binomial(k,m)*(-1)^(k-m)*(3 + 4*m)^n.
Recurrence: T(n, -1) = 0, T(0, 0) = 1, T(n, k) = 0 if n < k and T(n, k) =
4*k*T(n-1, k-1) + (3 + 4*k)*T(n-1, k) for n >= 1, k = 0..n (see the Maple program).
E.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, k)*x^k: exp(3*z)/(1 - x*(exp(4*z) - 1)).
E.g.f. column k: exp(3*x)*(exp(4*x) - 1)^k, k >= 0.
O.g.f. column k: k!*(4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)
EXAMPLE
[n\k][0, 1, 2, 3, 4, 5, 6 ]
[0] 1,
[1] 3, 4,
[2] 9, 40, 32,
[3] 27, 316, 672, 384,
[4] 81, 2320, 9920, 13824, 6144,
[5] 243, 16564, 127680, 326400, 337920, 122880,
[6] 729, 116920, 1536992, 6428160, 11642880, 9584640, 2949120.
MAPLE
SF_SO := 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*k*SF_SO(n-1, k-1, m) + (m*(k+1)-1)*SF_SO(n-1, k, m) end:
seq(print(seq(SF_SO(n, k, 4), k=0..n)), n = 0..5);
MATHEMATICA
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]]); SFSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]; Table[ SFSO[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@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_SO(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))
for n in (0..6): [SF_SO(n, k, 4) for k in (0..n)]
Numerator of Bernoulli(n, 1/4).
+10
4
1, -1, -1, 3, 7, -25, -31, 427, 127, -12465, -2555, 555731, 1414477, -35135945, -57337, 2990414715, 118518239, -329655706465, -5749691557, 45692713833379, 91546277357, -7777794952988025, -1792042792463, 1595024111042171723, 1982765468311237, -387863354088927172625
COMMENTS
The rationals r(n) = Sum_{k=0..n} ((-1)^k / (k+1))* A285061(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))* A225473(n, k) define generalized Bernoulli numbers, named B[4,1](n), in terms of the generalized Stirling2 numbers S2[4,1]. The numerators of r(n) are a(n) and the denominators A141459(n). r(n) = B[4,1](n) = 4^n*B(n, 1/4) with the Bernoulli polynomials B(n, x) = Bernoulli(n, x) from A196838/ A196839 or A053382/ A053383. The generalized Bernoulli numbers B[4,3](n) = Sum_{k=0..n} ((-1)^k/(k+1))* A225467(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))* A225473(n, k) satisfy B[4,3](n) = 4^n*B(n, 3/4) = (-1)^n*B[4,1](n). They have numerators (-1)^n*a(n) and also denominators A141459(n). (End)
FORMULA
a(n) = numerator(Bernoulli(n, 1/4)) with denominator A157818(n) (see the name). a(n) = numerator(4^n*Bernoulli(n, 1/4)) with denominator A141459(n) = A157818(n)/4^n. a(n)*(-1)^n = numerator(4^n*Bernoulli(n, 3/4)) with denominator A141459(n). (End)
MATHEMATICA
Table[Numerator[BernoulliB[n, 1/4]], {n, 0, 50}] (* Vincenzo Librandi, Mar 16 2014 *)
Numerators of the exponential expansion of (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)).
+10
1
1, -3, 9, -363, 6411, -46569, 3615627, -108267435, 2044658079, -27994845375, 5887932942123, -90460390681593, 475997756735954241, -3681053425472669991, 14270353890553782297, -2661381204559253577387, 880641541680797362210263
COMMENTS
This gives one third of the numerators of the z-sequence for the Sheffer triangle (exp(3*x), exp(4*x) - 1) shown in A225467. For the notion and use of a- and z- sequences for Sheffer triangles see the W. Lang link under A006232, also for references. The a-sequence of this Sheffer triangle is given by 4* A006232/ A006233. For the nontrivial recurrence of {3^n} given by the z-sequence for the m = 0 column of the triangle A225467 see the example for n = 3 below.
FORMULA
E.g.f.: (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)) for the rational sequence a(n)/ A285060(n), n >= 0.
EXAMPLE
The rationals a(n)/ A285060(n) start: 1, -3/8, 9/16, -363/256, 6411/1280, -46569/2048, 3615627/28672, -108267435/131072, 2044658079/327680, -27994845375/524288, ... From the z-recurrence for A225467(3, 0) = 3^3 = 27 one finds: 3^3 = 3*3*(1*9 + 40*(-3/8) + 16*(9/16)).
Denominators of the exponential expansion of (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)).
+10
1
1, 8, 16, 256, 1280, 2048, 28672, 131072, 327680, 524288, 11534336, 16777216, 7633633280, 4697620480, 1342177280, 17179869184, 365072220160, 171798691840, 45698452029440, 7696581394432
COMMENTS
For the numerators see A285059, also for the rationals A285059(n)/a(n). The z-sequence for the Sheffer triangle S2[4,3] = A225467 is 3* A285059(n)/a(n).
FORMULA
E.g.f.: (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)) for the rational sequence A285059(n)/a(n), n >= 0.
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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1
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
COMMENTS
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.
FORMULA
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
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 ...
O: 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
...
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. - Wolfdieter Lang, Aug 11 2017
CROSSREFS
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.
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