A039755 -id:A039755 - cp's OEIS frontend

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

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

A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

Original entry on oeis.org

1, 2, 1, 8, 8, 1, 48, 72, 18, 1, 384, 768, 288, 32, 1, 3840, 9600, 4800, 800, 50, 1, 46080, 138240, 86400, 19200, 1800, 72, 1, 645120, 2257920, 1693440, 470400, 58800, 3528, 98, 1, 10321920, 41287680, 36126720, 12042240, 1881600, 150528, 6272, 128, 1, 185794560, 836075520, 836075520, 325140480, 60963840, 6096384, 338688, 10368, 162, 1, 3715891200, 18579456000, 20901888000, 9289728000, 2032128000, 243855360, 16934400, 691200, 16200, 200, 1

Offset: 0

Wolfdieter Lang, Jun 16 2017

These generalized unsigned Lah numbers are the instance L[2,1] of the Sheffer triangles called L[d,a], with integers d >= 1 and integers 0 <= a < d with gcd(d,a) = 1. The standard unsigned Lah numbers are L[1,0] = A271703.
The Sheffer structure of L[d,a] is ((1 - d*t)^(-2*a/d), t/(1 - d*t)). This follows from the defining property
  risefac[d,a](x, n) = Sum_{m=0..n} L[d,a](n, m)*fallfac[d,a](x, m), where risefac[d,a](x, n):= Product_{0..n-1} (x  + (a+d*j)) for n >= 1 and risefac[d,a](x, 0) := 1, and fallfac[d,a](x, n):= Product_{0..n-1} (x - (a+d*j)) = for n >= 1 and fallfac[d,a](x, 0) := 1. Such rising and falling factorials arise in the generalization of Stirling numbers of both kinds S2[d,a] and S1[d,a]. See the Peter Bala link under A143395 for these falling factorials called there [t;a,b,c]_n with t=x, a=d, b=0, c=a.
In matrix notation: L[d,a] = S1phat[d,a].S2hat[d,a] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations with Sheffer structures S1phat[d,a] = ((1 - d*t)^(-a/d), -(1/d)*(log(1 - d*t))) and S2hat[d,a] = (exp(a*t), (1/d)*(exp(d*t) - 1). See, e.g., S1phat[2,1] = A028338 and S2hat[2,1] = A039755.
The a- and z-sequences for these Sheffer matrices have e.g.f.s 1 + d*t and ((1 + d*t)/t)*(1 - (1 + d*t)^(-2*a/d)), respectively. See a W. Lang link under A006232 for these types of sequences.
E.g.f. of row polynomials R[d,a](n, x) := Sum_{m=0..n} L[d,a](n, m)*x^m
  (1 - d*x)^(-2*a/d)*exp(t*x/(1 - d*x)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - d*t)^(-2*a/d)*(t/(1 - d*t))^m/m, m >= 0.
Meixner type identity for (monic) row polynomials: (D_x/(1 + d*D_x)) * R[d,a](n, x) = n*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1. The series in the differentiations D_x = d/dx terminates.
General Sheffer recurrence for row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
  R[d,a](n, x) = [(2*a+x)*1 + 2*d*(a + x)*D_x + d^2*x*(D_x)^2]*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1.
The inverse matrix L^(-1)[d,a] is Sheffer (g[d,a](-t), -f[d,a](-t)) with L[d,a] Sheffer (g[d,a](t), f[d,a](t)) from above. This means (see the column e.g.f. of Sheffer matrices) that L^(-1)[d,a](n, m) = (-1)^(n-m)*L[d,a](n, m). Therefore, the recurrence relations can easily be rewritten for L^(-1)[d,a] by replacing a -> -a and d -> -d.
fallfac[d,a](x, n) = Sum_{m=0..n} L^(-1)[d,a](n, m)*risefac[d,a](x, m), n >= 0.
From Wolfdieter Lang, Aug 12 2017: (Start)
The Sheffer row polynomials R[d,a](n, x) belong to the Boas-Buck class and satisfy therefore the Boas-Buck identity (see the reference, and 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*(2*a*1 + d*E_x) * Sum_{k=0..n-1} d^k*R(d,a;n-1-k,x)/(n-1-k)!, with E_x = x*d/dx (Euler operator).
This implies a recurrence for the sequence of column m: L[d,a](n, m) = (n!*(2*a + d*m)/(n-m))*Sum_{p=0..n-1-m} d^p*L[d,a](n-1-p, m)/(n-1-p)!, for n > m>=0, and input L[d,a](m, m) = 1. For the present [d,a] = [2,1] instance see the formula and example sections. (End)
From Wolfdieter Lang, Sep 14 2017: (Start)
The diagonal sequences are 2^D*D!*(binomial(m+D, m))^2, m >= 0, for D >= 0 (main diagonal D = 0). From the o.g.f.s obtained via Lagrange's theorem. See the second W. Lang link below for the general Sheffer case.
The o.g.f. of the diagonal D sequence is 2^D*D!*Sum_{m=0..D} A008459(D, m)*x^m /(1- x)^(2*D + 1),  D >= 0. (End)
It appears that this is also the matrix square of unsigned triangle of coefficients of Laguerre polynomials n!*L_n(x), abs(A021009(n, k)). - Ali Pourzand, Mar 10 2025 [This observation is correct. - Peter Luschny, Mar 10 2025]

			The triangle T(n, m) begins:
  n\m        0         1         2         3        4       5      6     7   8 9
  0:         1
  1:         2         1
  2:         8         8         1
  3:        48        72        18         1
  4:       384       768       288        32        1
  5:      3840      9600      4800       800       50       1
  6:     46080    138240     86400     19200     1800      72      1
  7:    645120   2257920   1693440    470400    58800    3528     98     1
  8:  10321920  41287680  36126720  12042240  1881600  150528   6272   128   1
  9: 185794560 836075520 836075520 325140480 60963840 6096384 338688 10368 162 1
  ...
From _Wolfdieter Lang_, Aug 12 2017: (Start)
Recurrence for column elements with m >= 1, and input column m = 0: T(3, 2) = (3/2)*T(2, 1) + 2*3*T(2, 2) = (3/2)*8 + 6 = 18.
Four term recurrence: T(3, 2) = T(2, 1) + 2*5*T(2, 2) - 4*2^2*T(1, 2) = 8 + 10 + 0 = 18.
Meixner type identity, n=2: 2*R(1, x) = (D_x - 2*(D_x)^2)*R(2, x), 2*(2 + x) = (8 + 2*x) - 2*2.
Sheffer recurrence: R(2, x) = (2 + x)*(2 + x) + 4*(1 + x)*1 + 0 = 8 + 8*x + x^2.
Boas-Buck recurrence for column m = 2 and n = 4: T(4, 2) = (2*4!*3/2)*(1*T(3, 2)/3! + 2*T(2, 2)/2!) = 4!*3*(18/3! + 1) = 288. (End)
Diagonal sequence D = 1: o.g.f. 2*1!*(1 + 1*x)/(1- x)^3 generating
{2*(binomial(m+1, m))^2}_{m >= 0} = {2, 8, 18, 32, ...}. - _Wolfdieter Lang_, Sep 14 2017

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.
Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Peter Bala, The white diamond product of power series
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
Wolfdieter Lang, On Generating functions of Diagonal Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Emanuele Munarini, Combinatorial identities involving the central coefficients of a Sheffer matrix, Applicable Analysis and Discrete Mathematics (2019) Vol. 13, 495-517.

Cf. A006232, A025167, A028338, A039755, A143395, A271703, A021009.
Column sequences (no leading zeros): A000165, A014479, A286725.
Diagonal sequences: A000012, 2*A000290(m+1), 8*A000537(n+1), 48*A001249, 384*A288876. - Wolfdieter Lang, Sep 14 2017
Row sums are A025167. - Michael Somos, Sep 27 2017

T := (n, k) -> ifelse(n < k, 0, ifelse(k = 0, n!*2^n, (n/k)*T(n-1, k-1) + 2*n*T(n-1, k))): seq(seq(T(n, k), k = 0..n), n = 0..10);  # Peter Luschny, Mar 10 2025

T[ n_, k_] := Coefficient[ Integrate[ Exp[-x^2 - y x] HermiteH[n, x]^2, {x, -Infinity, Infinity}] / (Sqrt[Pi] Exp[y^2 / 4]), y, 2 k]; (* Michael Somos, Sep 27 2017 *)

# Using the function A021009_triangle, displays as a matrix. Following the observation of Ali Pourzand.
print(A021009_triangle(9)^2)  # Peter Luschny, Mar 10 2025

T(n, m) = L[2,1](n, m) = Sum_{k=m..n} A028338(n, k)*A039755(k, m).
Three term recurrence for column elements with m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 2*n*T(n-1, m) with T(n, m) = 0 for n < m and the column m = 0 is T(n, 0) = (2*n)!! = n*2^n = A000165(n). (From the a- and z-sequences {1, 2, repeat(0)} and {2, repeat(0)}, respectively.)
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(2*n-1)*T(n-1, m) - 4*(n-1)^2*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
E.g.f. of row polynomials R(n, x) = R[2,1](n, x) (i.e., e.g.f. of the triangle): (1/(1-2*t))*exp(x*t/(1-2*t)).
E.g.f. of column m sequences: (t^m/(1-2*t)^(m+1))/m!, m >= 0.
Meixner type identity: Sum_{k=0..n-1} (-1)^k*2^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx.
Sheffer recurrence: R(n, x) = [(2 + x)*1 + 4*(1 + x)*D_x + 4*x*(D_x)^2]*R(n-1, x), n >= 1, and R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment above): T(n, m) = (2*n!*(1 + m)/(n-1))*Sum_{p=0..n-1-m} 2^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, and input T(m, m) = 1. - Wolfdieter Lang, Aug 12 2017
Explicit form (from the diagonal sequences with the o.g.f.s given as a comment above): T(n, m) = 2^(n-m)*(n-m)!*(binomial(n, n-m))^2 for n >= m >= 0. - Wolfdieter Lang, Sep 23 2017
Let R(n,x) denote the n-th row polynomial. Then x^n*R(n,x) = x^n o x^n, where o denotes the deformed Hadamard product of power series defined in Bala, Section 3.1. - Peter Bala, Jan 18 2018

A186695 A Galton triangle: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)): a type B analog of the ordered Bell numbers A019538.

Original entry on oeis.org

1, 1, 3, 1, 12, 15, 1, 39, 135, 105, 1, 120, 870, 1680, 945, 1, 363, 4950, 17850, 23625, 10395, 1, 1092, 26565, 159600, 373275, 374220, 135135, 1, 3279, 138285, 1303155, 4795875, 8222445, 6621615, 2027025

Offset: 1

Peter Bala, Mar 26 2011

The row polynomials R(n,x) of A019538 satisfy the recurrence relation R(n+1,x) = x*d/dx((1+x)*R(n,x)), and have the expansion R(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
Here we consider a sequence of polynomials P(n,x) (n>=1) defined by means of the similar recursion P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with starting value P(1,x) = x.
The first few polynomials are P(1,x) = x, P(2,x) = x + 3*x^3, P(3,x) = x + 12*x^3 + 15*x^5, and P(4,x) = x + 39*x^3 + 135*x^5 + 105*x^7.
Clearly, the P(n,x) are odd polynomials of the form P(n,x) = Sum_{k = 1..n} T(n,k)*x^(2*k-1).
This triangle lists the coefficients T(n,k). They are related to A039755, the type B Stirling numbers of Suter, by T(n,k) = (2*k-1)!!*A039755(n-1,k-1).

			Triangle begins
  n\k.|..1.....2.....3......4......5......6
  =========================================
  ..1.|..1
  ..2.|..1.....3
  ..3.|..1....12....15
  ..4.|..1....39...135....105
  ..5.|..1...120...870...1680....945
  ..6.|..1...363..4950..17850..23625..10395
  ..
Examples of recurrence relation
  T(4,3) = 5*(T(3,3)+T(3,2)) = 5*(15+12) = 135;
  T(6,4) = 7*(T(5,4)+T(5,3)) = 7*(1680+870) = 17850.

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. 8.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Tech Report TR 99-05, 1999.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).

Cf. A001147, A019538, A039755, A124212, A145901, A156919, A187075.

A186695 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 then 1; else (2*k-1)*(procname(n-1, k) + procname(n-1, k-1)) ; end if; end proc: seq(seq(A186695(n,k),k = 1..n),n = 1..10);

T[n_, k_] := (2k-1)! Sum[(-1)^(k-j-1) (2j+1)^(n-1) Binomial[k-1, j], {j, 0, k-1}] / (2^(k-1) (k-1)!)^2;
Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

T(n+1,k+1) = ((2*k+1)!/(2^k*k!)^2)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
Recurrence relation: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)) with boundary conditions T(n,1) = 1, T(1,k) = 0 for k >= 2.
E.g.f.: F(x,t) = (x/(1+x))*(exp(t)/sqrt((1+x) - x*exp(2*t)) - 1) = Sum_{n>=1} R(n,x)*t^n/n! = x*t + (x + 3*x^2)*t^2/2! + (x + 12*x^2 + 15*x^3)*t^3/3! + ....
Compare with the e.g.f. for A019538, which is (x/(1+x))*(exp(t)/((1+x) - x*exp(t))-1).
The row polynomials R(n,x) are related to the polynomials P(n,x) of the comments section by P(n,x) = 1/x*R(n,x^2).
The generating function F(x,t) satisfies the partial differential equation d/dt(F) = 2*x*(1+x)*d/dx(F) + (x-1)*F + x.
It follows that the polynomials P(n,x) := Sum_{k = 1..n} T(n,k)*x^(2*k-1) satisfy the recurrence P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with P(1,x) = x. (Cf. the recurrence relation for row polynomials of A185896.)
The recurrence relation for T(n,k) given above now follows.
The row polynomials R(n,x) = Sum_{k = 1..n} T(n,k)*x^k satisfy R(n,-x-1) = (-1)^n*((1+x)/x)*S(n,x), where S(n,x) is the n-th row polynomial of A187075.
In addition, R(n,1/(x-1)) = (1/(x-1)^n)*Q(n-1,x), where Q(n,x) is the n-th row polynomial of A156919.
Row sums are [1,4,28,280,3616...] = 1/2*A124212(n) for n >= 1.
Main diagonal is [1,3,15,105,...] = A001147(k) for k >= 1.
Put S(n) = sum {k = 1..n} (-1)^k*T(n,k)/(k+1). Then for m>=2, S(2*m-1) = S(2*m) = (4^m-1)*Bernoulli(2*m)/m.
From Peter Bala, Aug 30 2016: (Start)
n-th row polynomial R(n,x) = 1/(1 + x)^(3/2) * Sum_{k >= 0} (1/4)^k*(x/(1 + x))^k*binomial(2*k,k)*(2*k + 1)^n.
R(n,x) = (1/(1 + x))*Sum_{k = 0..n} binomial(2*k,k)*A145901(n,k)*(x/4)^k. (End)

A111669 Triangle read by rows, based on a simple Fibonacci recursion rule.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 11, 7, 1, 1, 5, 26, 32, 12, 1, 1, 6, 57, 122, 92, 20, 1, 1, 7, 120, 423, 582, 252, 33, 1, 1, 8, 247, 1389, 3333, 2598, 681, 54, 1, 1, 9, 502, 4414, 18054, 24117, 11451, 1815, 88, 1, 1, 10, 1013, 13744, 94684, 210990, 172980, 49566, 4807, 143, 1

Offset: 0

Gary W. Adamson, Aug 14 2005

Subdiagonal is A000071(n+3). Row sums of inverse are 0^n.

Row sums are given by A135934. - Emanuele Munarini, Dec 05 2017

			Triangle begins
  1....1....2....3....5....8...13....F(k+1)
  1
  1....1
  1....2....1
  1....3....4....1
  1....4...11....7....1
  1....5...26...32...12....1
  1....6...57..122...92...20....1
For example, T(6,3) = 122 = 26 + 3*32 = T(5,2) + F(4)*T(5,3).

Michel Marcus, Rows n = 0..20, flattened

Cf. A111577, A111578, A111579, A008277, A039755, A135934.

(* To generate the triangle *)
Grid[RecurrenceTable[{F[n,k] == F[n-1,k-1] + Fibonacci[k+1] F[n-1,k], F[0,k] == KroneckerDelta[k]}, F, {n,0,10}, {k,0,10}]] (* Emanuele Munarini, Dec 05 2017 *)

T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1, k-1) + fibonacci(k+1)*T(n-1, k))); \\ Michel Marcus, May 25 2024

T(n, k) = T(n-1, k-1) + F(k+1)*T(n-1, k) where F(n)=A000045(n).

Column k has g.f. x^k/Product_{j=0..k} (1 - F(j+1)*x).

Edited by Paul Barry, Nov 14 2005

A182825 E.g.f. 1/(cos(sqrt(3)x) - sin(sqrt(3)x)/sqrt(3)).

Original entry on oeis.org

1, 1, 5, 21, 153, 1209, 12285, 140589, 1871217, 27773361, 460041525, 8363802501, 166064229513, 3570030632169, 82674532955565, 2051044762727709, 54279654050034657, 1526205561241263201, 45438086217150617445, 1427921718081647393781, 47235337785416646609273

Offset: 0

Paul Barry, Dec 05 2010

nonn

First column of A182824. Hankel transform is 4^C(n+1,2)*(A000178(n))^2.

Moments of orthogonal polynomials whose coefficient array is A182826.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Shi-Mei Ma, Jun Ma, Yeong-Nan Yeh, On certain combinatorial expansions of descent polynomials and the change of grammars, arXiv:1802.02861 [math.CO], 2018.

nn = 20; Table[n!, {n, 0, nn}] CoefficientList[Series[1/(Cos[Sqrt[3]*x] - Sin[Sqrt[3]*x]/Sqrt[3]), {x, 0, nn}], x] (* T. D. Noe, Jun 28 2011 *)

From Peter Bala, Jan 21 2011: (Start)
By comparing the e.g.f. for this sequence with the e.g.f for the type B Eulerian numbers A060187 we can show that
(1)... a(n) = B(n,w)/(1+w)^(n+1), where w = exp(2*Pi*I/3) and {B(n,x)}n>=1 = [x,x+x^2,x+6*x^2+x^3,x+23*x^2+23*x^3+x^4,...] are the type B Eulerian polynomials.
Equivalently,
(2)... a(n) = (-I*sqrt(3))^n*Sum_{k = 0..n} 2^k*k!*A039755(n,k)*(-1/2+sqrt(3)*I/6)^k,
where A039755(n,k) are the type B analogs of the Stirling numbers of the second kind. We can rewrite this as
(3)... a(n) = (-I*sqrt(3))^n*sum {k = 0..n} (-1/2+sqrt(3)*I/6)^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(4a)... a(p) = 1 (mod p) for prime p = 6*n+1,
(4b)... a(p) = -1 (mod p) for prime p = 6*n+5.
For similar results see A000111. Let u = exp(2*Pi*I/6) = 1/2+sqrt(3)/2*I be a primitive sixth root of unity.
(5)... a(n) = Sum_{k = 0..n+1} u^(n+2-2*k)*Sum_{j = 1..n+1} (-1)^(k-j)*binomial(n+1,k-j)*(2*j-1)^n. Cf. A002439. (End)
G.f.: 1/Q(0), where Q(k) = 1 - x*(2*k+1) - x^2*(2*k+2)^2/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
E.g.f.: 1/E(0), where E(k) = 1 - x/( 2*k+1 - 3*x*(2*k+1)/(3*x + 2*(k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 28 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 4*x^2*(k+1)^2/(4*x^2*(k+1)^2 - (1 -x -2*x*k)*(1 -3*x -2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 10 2013

A021424 Expansion of 1/((1-x)(1-3x)(1-5x)(1-7x)).

Original entry on oeis.org

1, 16, 170, 1520, 12411, 96096, 719860, 5278240, 38153621, 273134576, 1942326750, 13748476560, 97001079631, 682818667456, 4798793396840, 33686888924480, 236284962774441, 1656378634646736, 11606570499786130

Offset: 0

N. J. A. Sloane

a(n) is h^{(4)}n, the complete homogeneous symmetric function of the four symbols s_j = 1 + 2*j, j = 0..3, of degree n >= 1, with h^{(4)}_0 = 1. See an example below. Thus it is the (dimensionless) volume of all multichoose(4, n) = binomial(n+3, 3) polytopes of dimension n with side lengths from the set {1, 3, 5, 7}. - _Wolfdieter Lang, May 26 2017

			a(2) = h^{(4)}_2 = (1^2 + 3^2 + 5^2 + 7^2) +  (1^1*(3^1 + 5^1 + 7^1) + 3^1*(5^1 + 7^1) + 5^1*7^1)  = 84 + 86  = 120. - _Wolfdieter Lang_, May 26 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (16,-86,176,-105).

Cf. A039755 (column k=3), A016209.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-x)*(1-3*x)*(1-5*x)*(1-7*x)))); // Vincenzo Librandi, Jul 09 2013

I:=[1, 16, 170, 1520]; [n le 4 select I[n] else 16*Self(n-1)-86*Self(n-2)+176*Self(n-3)-105*Self(n-4): n in [1..25]]; // Vincenzo Librandi, Jul 09 2013

Table[(7^n - 3*5^n + 3^(n + 1) - 1)/48, {n, 3, 60}]
CoefficientList[Series[1 / ((1 - x) (1 - 3 x) (1 - 5 x) (1 - 7 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Jul 09 2013 *)
LinearRecurrence[{16,-86,176,-105},{1,16,170,1520},30] (* Harvey P. Dale, May 26 2014 *)

x='x+O('x^99); Vec(1/((1-x)*(1-3*x)*(1-5*x)*(1-7*x))) \\ Altug Alkan, Oct 11 2017

a(n) = (7^n- 3*5^n+ 3^(n+1)-1)/48. - Victor Adamchik (adamchik(AT)cs.cmu.edu), Jul 21 2001
a(n) = 12*a(n-1) - 35*a(n-2) + (3^n-1)/2 with a(0)=1, a(1)=16. - Vincenzo Librandi, Jul 09 2013
a(n) = 16*a(n-1) - 86*a(n-2) + 176*a(n-3) - 105*a(n-4), with a(0)=1, a(1)=16, a(2)=170, a(3)=1520. - Vincenzo Librandi, Jul 09 2013
G.f.: 1/((1-x)*(1-3*x)*(1-5*x)*(1-7*x)). See the name.
E.g.f.: (343*exp(7*x) - 375*exp(5*x) + 81*exp(3*x) - exp(x))/48, from the e.g.f. of the fourth column (k=3) of A039755. - Wolfdieter Lang, May 26 2017

A039756 Triangle of B-analogs of Stirling numbers of 2nd kind.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 13, 1, 1, 16, 58, 40, 1, 1, 25, 170, 330, 121, 1, 1, 36, 395, 1520, 1771, 364, 1, 1, 49, 791, 5075, 12411, 9219, 1093, 1, 1, 64, 1428, 13776, 58086, 96096, 47188, 3280, 1, 1, 81, 2388, 32340, 209622, 618870, 719860, 239220, 9841, 1

Offset: 0

Ruedi Suter (suter(AT)math.ethz.ch)

This is a variant of A039755 with reflected rows. - Tilman Piesk, Oct 27 2019

			1;
1,  1;
1,  4,   1;
1,  9,  13,    1;
1, 16,  58,   40,     1;
1, 25, 170,  330,   121,    1;
1, 36, 395, 1520,  1771,  364,    1;
1, 49, 791, 5075, 12411, 9219, 1093,  1;

Alois P. Heinz, Rows n = 0..100, 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. 8.
Ruedi Suter, Two analogues of a classical sequence, J. Integer Sequences, Vol. 3 (2000), #P00.1.8.

Cf. A039755.

T(n,k)=if(k<0||k>n,0,n!*polcoeff(polcoeff(exp(x*y+(exp(2*x*y+x*O(x^n))-1)/(2*y)),n),k))

Sum a(n,n-k) x^n*y^k/n! = exp(x + y/2*(exp(2*x) - 1)).

T(n, k) = A039755(n, n-k). - Tilman Piesk, Oct 27 2019

A111579 Triangle A(r,c) read by rows, which contains the row sums of the triangle T(n,k)= T(n-1,k-1)+((c-1)k+1)T(n-1,k) in column c.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 5, 2, 1, 1, 16, 15, 6, 2, 1, 1, 32, 52, 24, 7, 2, 1, 1, 64, 203, 116, 35, 8, 2, 1, 1, 128, 877, 648, 214, 48, 9, 2, 1, 1, 256, 4140, 4088, 1523, 352, 63, 10, 2, 1, 1, 512, 21147, 28640, 12349, 3008, 536, 80, 11, 2, 1

Offset: 0

Gary W. Adamson, Aug 07 2005

Triangles of generalized Stirling numbers of the second kind may be defined by recurrences T(n,k) = T(n-1,k-1) + Q*T(n-1,k) initialized by T(0,0)=T(1,0)=T(1,1)=1. Q=1 generates Pascal's triangle A007318,
Q=k+1 generates A008277, Q=2k+1 generates A039755, Q=3k+1 generates A111577, Q=4k+1 generates A111578, Q=5k+1 generates A166973.
(These definitions assume row and column enumeration 0<=n, 0<=k<=n.)
Each of these triangles characterized by Q=(c-1)*k+1 has row sums sum_{k=0..n} T(n,k), which define the column A(.,c).

Cf. A008277, A000110, A039755, A004211, A111577, A111578.

T := proc(n,k,c) if k < 0 or k > n then 0 ; elif n <= 1 then 1; else procname(n-1,k-1,c)+((c-1)*k+1)*procname(n-1,k,c) ; fi; end:
A111579 := proc(r,c) local n; if c = 0 then 1 ; else n := r-c ; add( T(n,k,c),k=0..n) ; end if; end:
seq(seq(A111579(r,c),c=0..r),r=0..10) ; # R. J. Mathar, Oct 30 2009

T[n_, k_, c_] := T[n, k, c] = If[k < 0 || k > n, 0, If[n <= 1, 1, T[n-1, k-1, c] + ((c-1)*k+1)*T[n-1, k, c]]];
A111579[r_, c_] := Module[{n}, If[c == 0, 1, n = r - c; Sum[T[n, k, c], {k, 0, n}]]];
Table[A111579[r, c], {r, 0, 10}, {c, 0, r}] // Flatten (* Jean-François Alcover, Aug 01 2023, after R. J. Mathar *)

A(r=n+c,c) = sum_{k=0..n} T(n,k,c), 0<=c<=r where T(n,k,c) = T(n-1,k-1,c) + ((c-1)*k+1)*T(n-1,k,c).
A(r,0) = 1.
A(r,1) = 2^(r-1).
A(r,2) = A000110(r-1).
A(r,3) = A007405(r-3).

Edited by R. J. Mathar, Oct 30 2009

A112351 Triangle read by rows, generated from (..., 5, 3, 1).

Original entry on oeis.org

1, 1, 3, 1, 6, 5, 1, 9, 19, 7, 1, 12, 42, 44, 9, 1, 15, 74, 138, 85, 11, 1, 18, 115, 316, 363, 146, 13, 1, 21, 165, 605, 1059, 819, 231, 15, 1, 24, 224, 1032, 2470, 2984, 1652, 344, 17, 1, 27, 292, 1624, 4974, 8378, 7380, 3060

Offset: 0

Gary W. Adamson, Sep 05 2005

A039755 (Analogs of a Stirling number of the second kind triangle) is generated through an analogous set of operations (but using the matrix M = [1 / 1 3 / 1 3 5 /...]). First few rows of the array are 1, 3, 5, 7, 9, 11, ...; 1, 6, 19, 44, 85, ...; 1, 9, 42, 138, 363, ...; 1, 12, 74, 316, 1059, ....
A112351 is jointly generated with A209414 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x) = x*u(n-1,x) + v(n-1,x) and v(n,x) = 2x*u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica and Example sections. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle T(n,k) given by (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 3, -4/3, 1/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 12 2012

			The antidiagonal 1 9 19 7 of the array becomes row 3 of the triangle.
From _Clark Kimberling_, Mar 09 2012: (Start)
When jointly generated with A209414, the format as a triangle has the following first five rows:
  1;
  1,  3;
  1,  6,  5;
  1,  9, 19,   7;
  1, 12, 42,  44,  9;
  1, 15, 74, 138, 85, 11;
The corresponding first five polynomials are
  1,
  1 + 3x,
  1 + 6x + 5x^2,
  1 + 9x + 19x^2 + 7x^3,
  1 + 12x + 42x^2 + 44x^3 + 9x^4. (End)
(1, 0, 0, 0, 0, ...) DELTA (0, 3, -4/3, 1/3, 0, 0, 0, ...) begins:
  1;
  1,  0;
  1,  3,   0;
  1,  6,   5,   0;
  1,  9,  19,   7,   0;
  1, 12,  42,  44,   9,   0;
  1, 15,  74, 138,  85,  11,  0;
  1, 18, 115, 316, 363, 146, 13, 0;
- _Philippe Deléham_, Mar 12 2012

Cf. A039755, A005900 (array row 2), A061927 (array row 3), A209414.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + (x + 1)*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209414 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A112351 *)
(* Clark Kimberling, Mar 09 2012 *)

Let M = an infinite lower triangular matrix of the form [1 / 3 1 / 5 3 1 / ...] (with the rest of the terms zeros). Perform M^n * [1 0 0 0 ...] forming an array. Antidiagonals of the array become rows of the triangle A112351.
From Philippe Deléham, Mar 12 2012: (Start)
As DELTA-triangle T(n,k) with 0 <= k <= n:
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = 1, T(1,1) = T(2,1) = 0, T(2,1) = 3 and T(n,k) = 0 if k < 0 or if k > n.
G.f.: (1-y*x)^2/(1-x-2*y*x-y*x^2+y^2*x^2). (End)

cp's OEIS Frontend

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A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

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A186695 A Galton triangle: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)): a type B analog of the ordered Bell numbers A019538.

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A111669 Triangle read by rows, based on a simple Fibonacci recursion rule.

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A182825 E.g.f. 1/(cos(sqrt(3)*x) - sin(sqrt(3)*x)/sqrt(3)).

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A021424 Expansion of 1/((1-x)(1-3x)(1-5x)(1-7x)).

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A182825 E.g.f. 1/(cos(sqrt(3)x) - sin(sqrt(3)x)/sqrt(3)).

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A348085 a(n) = [x^n] Product_{k=1..2n} 1/(1 - (2k-1) * x).