cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A286723 Column k = 1 of the triangle A225471; Sheffer ((1 - 3*x)^(-3/4), (-1/4)*log(1 - 4*x)).

Original entry on oeis.org

1, 10, 131, 2196, 45189, 1105182, 31354119, 1012861224, 36717532425, 1476342400050, 65212709985675, 3139386801018300, 163605030141437325, 9176588125543758150, 551225830134140520975, 35305848011347321438800, 2401944921672748059294225, 172980447467901106243829850
Offset: 0

Views

Author

Wolfdieter Lang, May 29 2017

Keywords

Comments

a(n) is, for n >= 1, the total volume of the binomial(n+1, n) rectangular polytopes (hyper-cuboids) built from n orthogonal vectors with lengths of the sides from the set {3 + 4*j | j=0..n}. See the formula a(n) = sigma[4,3]^{(n+1)}_n and an example below.

Examples

			a(2) = 131 because sigma[4,3]^{(3)}_2 = 3*(7 + 11) + 7*11 = 131. There are three rectangles (2D rectangular polytopes) built from two orthogonal vectors of different lengths from the set of {3,7,11} of total area 131.
		

Crossrefs

Cf. A008545 (k=0), A225471.

Formula

a(n) = A225471(n+1, 1), n >= 1.
E.g.f.: (d/dx) ((1 - 4*x)^(-3/4)*((-1/4)*log(1 - 4*x))) = (4 - 3*log(1-4*x)) / (4*(1-4*x)^(7/4)).
a(n) = sigma[4,3]^{(n+1)}_n, n >= 0, with the elementary symmetric function sigma[4,3]^{(n+1)}_n of degree n of the n+1 numbers 3, 7, 11, ..., (1 + 4*n), and sigma[4,3]^{(n+1)}_0 := 1.

A008545 Quadruple factorial numbers: Product_{k=0..n-1} (4*k + 3).

Original entry on oeis.org

1, 3, 21, 231, 3465, 65835, 1514205, 40883535, 1267389585, 44358635475, 1729986783525, 74389431691575, 3496303289504025, 178311467764705275, 9807130727058790125, 578620712896468617375, 36453104912477522894625, 2442358029135994033939875
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

a(n-1), n >= 1, enumerates increasing plane (a.k.a. ordered) trees with n vertices (one of them a root labeled 1) with one version of a vertex with out-degree r = 0 (a leaf or a root) and each vertex with out-degree r >= 1 comes in binomial(r + 2, 2) types (like a binomial(r + 2, 2)-ary vertex). See the increasing tree comments under A001498. For example, a(1) = 3 from the three trees with n = 2 vertices (a root (out-degree r = 1, label 1) and a leaf (r = 0), label 2). There are three such trees because of the three types of out-degree r = 1 vertices. - Wolfdieter Lang, Oct 05 2007 [corrected by Karen A. Yeats, Jun 17 2013]
a(n) is the product of the positive integers less than or equal to 4n that have modulo 4 = 3. - Peter Luschny, Jun 23 2011

Examples

			G.f. = 1 + 3*x + 21*x^2 + 231*x^3 + 3465*x^4 + 65835*x^5 + 1514205*x^6 + ...
a(3) = sigma[4,3]^{3}_3 = 3*7*11 = 231. See the name. - _Wolfdieter Lang_, May 29 2017
		

Crossrefs

a(n)= A000369(n+1, 1) (first column of triangle).
Partial products of A004767.
Cf. A007696, A014601, A225471 (first column).

Programs

  • GAP
    List([0..20], n-> Product([0..n-1], k-> 4*k+3) ); # G. C. Greubel, Aug 18 2019
  • Haskell
    a008545 n = a008545_list !! n
    a008545_list = scanl (*) 1 a004767_list
    -- Reinhard Zumkeller, Oct 25 2013
    
  • Magma
    [1] cat [(&*[4*k+3: k in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Aug 18 2019
    
  • Maple
    f := n->product( (4*k-1),k=0..n);
    A008545 := n -> mul(k, k = select(k-> k mod 4 = 3, [$1 .. 4*n])): seq(A008545(n), n=0..15); # Peter Luschny, Jun 23 2011
  • Mathematica
    FoldList[Times, 1, 4 Range[0, 20] + 3] (* Harvey P. Dale, Jan 19 2013 *)
    a[n_]:= Pochhammer[3/4, n] 4^n; (* Michael Somos, Jan 17 2014 *)
    a[n_]:= If[n < 0, 1 / Product[ -k, {k, 1, -4 n - 3, 4}], Product[k, {k, 3, 4 n - 1, 4}]]; (* Michael Somos, Jan 17 2014 *)
  • PARI
    a(n)=prod(k=0,n-1,4*k+3) \\ Charles R Greathouse IV, Jun 23 2011
    
  • PARI
    {a(n) = if( n<0, 1 / prod(k=1, -n, 3 - 4*k), prod(k=1, n, 4*k - 1))}; /* Michael Somos, Jan 17 2014 */
    
  • Sage
    [product(4*k+3 for k in (0..n-1)) for n in (0..20)] # G. C. Greubel, Aug 18 2019
    

Formula

a(n) = 3*A034176(n) = (4*n-1)(!^4), n >= 1, a(0) := 1.
E.g.f.: (1-4*x)^(-3/4).
a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(3/4)^(-1)*n^(1/4)*2^(2*n)*e^(-n)*n^n*{1 - 1/96*n^(-1) + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
G.f.: 1/(1 - 3x/(1 - 4x/(1 - 7x/(1 - 8x/(1 - 11x/(1 - 12x/(1 - 15x/(1 - 16x/(1 - 19x/(1 - 20x/(1 - 23x/(1 - 24x/(1 - ...))))))))))))) (continued fraction). - Paul Barry, Dec 03 2009
a(n) = (-1)^n*Sum_{k = 0..n} 4^k*s(n + 1, n + 1 - k), where s(n, k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
D-finite with recurrence: a(n) + (-4*n + 1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
G.f.: 1/x - G(0)/(2*x), where G(k)= 1 + 1/(1 - x*(4*k-1)/(x*(4*k-1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013
a(-n) = (-1)^n / A007696(n). - Michael Somos, Jan 17 2014
G.f.: 1/(1 - b(1)*x / (1 - b(2)*x / ...)) where b = A014601. - Michael Somos, Jan 17 2014
a(n) = 4^n * Gamma(n+3/4) / Gamma(3/4). - Vaclav Kotesovec, Jan 28 2015
G.f.: 1/(1 - 3*x - 12*x^2/(1 - 11*x - 56*x^2/(1 - 19*x - 132*x^2/(1 - 27*x - 240*x^2/(1 - ...))))) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 28 2020
Sum_{n>=0} 1/a(n) = 1 + exp(1/4)*(Gamma(3/4) - Gamma(3/4, 1/4))/sqrt(2). - Amiram Eldar, Dec 18 2022

A286718 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.

Original entry on oeis.org

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
Offset: 0

Views

Author

Wolfdieter Lang, May 18 2017

Keywords

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.
From Wolfdieter Lang, Aug 09 2017: (Start)
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

Examples

			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
...
From _Wolfdieter Lang_, Aug 09 2017: (Start)
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)
		

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.

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.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, 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.
Column sequences for k = 0..4: A007559, A024216(n-1), A286721(n-2), A382984, A382985.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).

Programs

  • 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_] /; nJean-François Alcover, Jun 20 2018 *)

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

A225470 Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 1, 10, 7, 1, 80, 66, 15, 1, 880, 806, 231, 26, 1, 12320, 12164, 4040, 595, 40, 1, 209440, 219108, 80844, 14155, 1275, 57, 1, 4188800, 4591600, 1835988, 363944, 39655, 2415, 77, 1, 96342400, 109795600, 46819324, 10206700, 1276009, 95200, 4186, 100, 1
Offset: 0

Views

Author

Peter Luschny, May 16 2013

Keywords

Comments

The Stirling-Frobenius subset numbers S_{m}(n,k), for m >= 1 fixed, regarded as an infinite lower triangular matrix, can be inverted by Sum_{k} S_{m}(n,k)*s_{m}(k,j)*(-1)^(n-k) = [j = n]. The inverse numbers s_{m}(k,j), which are unsigned, are the Stirling-Frobenius cycle numbers. For m = 1 this gives the classical Stirling cycle numbers A132393. The Stirling-Frobenius subset numbers are defined in A225468.
Triangle T(n,k), read by rows, given by (2, 3, 5, 6, 8, 9, 11, 12, 14, 15, ... (A007494)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015

Examples

			Triangle starts:
  [n\k][    0,      1,     2,     3,    4,  5,  6]
  [0]       1,
  [1]       2,      1,
  [2]      10,      7,     1,
  [3]      80,     66,    15,     1,
  [4]     880,    806,   231,    26,    1,
  [5]   12320,  12164,  4040,   595,   40,  1,
  [6]  209440, 219108, 80844, 14155, 1275, 57,  1.
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see Maple program): T(4, 2) = T(3, 1) + (3*4 - 1)*T(3, 2) = 66 + 11*15 = 231.
Boas-Buck type recurrence for column k = 2 and n = 4: T(4, 2) = (4!/2)*(3*(2 + 6*(5/12))*T(2, 2)/2! + 1*(2 + 6*(1/2))*T(3,2)/3!) = (4!/2)*(3*9/4 + 5*15/3!) = 231. (End)
		

Crossrefs

Cf. A225468; A132393 (m=1), A028338 (m=2), A225471(m=4).
Column k=0..4 give A008544, A024395(n-1), A286722(n-2), A383221, A383222.
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024391; T(n, n-3) ~ A024392.
row sums ~ A032031; alternating row sums ~ A007559.
Cf. A132393.

Programs

  • Maple
    SF_C := 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;
    SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m) end:
    seq(print(seq(SF_C(n, k, 3), k = 0..n)), n = 0..8);
  • Mathematica
    SFC[0, 0, ] = 1; SFC[n, k_, ] /; (k > n || k < 0) = 0; SFC[n, k_, m_] := SFC[n, k, m] = SFC[n-1, k-1, m] + (m*n-1)*SFC[n-1, k, m]; Table[SFC[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2013, after Maple *)

Formula

For a recurrence see the Maple program.
From Wolfdieter Lang, May 18 2017: (Start)
This is the Sheffer triangle (1/(1 - 3*x)^{-2/3}, -(1/3)*log(1-3*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(3,0,2)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 3*z)^{-(2+x)/3}.
E.g.f. of column k: (1-3*x)^(-2/3)*(-(1/3)*log(1-3*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+2)*R(n-1,x+3), with R(0, x) = 1.
R(n, x) = risefac(3,2;x,n) := Product_{j=0..(n-1)} (x + (2 + 3*j)). (See the P. Bala link, eq. (16) for the signed s_{3,0,2} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*2^(n-k-j)*3^j, with S1p(n, m) = A132393(n, m). (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 3^(n-1-p)*(2 + 3*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 {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

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.

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

Views

Author

Wolfdieter Lang, Aug 08 2017

Keywords

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.

Examples

			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)
		

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.
Columns k=0..3 give A007696, A024382(n-1), A383700, A383701.
Row sums: A001813. Alternating row sums: A000007.

Programs

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

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

A292219 Triangle read by rows. A generalization of unsigned Lah numbers, called L[4,3].

Original entry on oeis.org

1, 6, 1, 60, 20, 1, 840, 420, 42, 1, 15120, 10080, 1512, 72, 1, 332640, 277200, 55440, 3960, 110, 1, 8648640, 8648640, 2162160, 205920, 8580, 156, 1, 259459200, 302702400, 90810720, 10810800, 600600, 16380, 210, 1, 8821612800, 11762150400, 4116752640, 588107520, 40840800, 1485120, 28560, 272, 1
Offset: 0

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Author

Wolfdieter Lang, Sep 23 2017

Keywords

Comments

For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[4,3], the Sheffer triangle ((1 - 4*t)^(-3/2), t/(1 - 4*t)). It is defined as transition matrix
risefac[4,3](x, n) = Sum_{m=0..n} L[4,3](n, m)*fallfac[4,3](x, m), where risefac[4,3](x, n) := Product_{0..n-1} (x + (3 + 4*j)) for n >= 1 and risefac[4,3](x, 0) := 1, and fallfac[4,3](x, n):= Product_{0..n-1} (x - (3 + 4*j)) for n >= 1 and fallfac[4,3](x, 0) := 1.
In matrix notation: L[4,3] = S1phat[4,3]*S2hat[4,3] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A225471 and A225469, respectively.
The a- and z-sequences for this Sheffer matrix have e.g.f.s Ea(t) = 1 + 4*t and Ez(t) = (1 + 4*t)*(1 - (1 + 4*t)^(-3/2))/t, respectively. That is, a = {1, 4, repeat(0)} and z(n) = 2*A292221(n). See the W. Lang link on a- and z-sequences there.
The inverse matrix T^(-1) = L^(-1)[4,3] is Sheffer ((1 + 4*t)^(-3/2), t/(1 + 4*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[4,3](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[4,3](x, m), n >= 0.
Diagonal sequences have o.g.f. G(d, x) = A001813(d)*Sum_{m=0..d} A103327(d, m)*x^m/(1 - x)^(2*d + 1), for d >= 0 (d=0 main diagonal). G(d, x) generates {A001813(d)*binomial(2*(m + d) + 1, 2*d)}{m >= 0}. See the second W. Lang link on how to compute o.g.f.s of diagonal sequences of general Sheffer triangles. - _Wolfdieter Lang, Oct 12 2017

Examples

			The triangle T(n, m) begins:
  n\m          0           1          2         3        4       5     6   7  8
  0:           1
  1:           6           1
  2:          60          20          1
  3:         840         420         42         1
  4:       15120       10080       1512        72        1
  5:      332640      277200      55440      3960      110       1
  6:     8648640     8648640    2162160    205920     8580     156     1
  7:   259459200   302702400   90810720  10810800   600600   16380   210   1
  8:  8821612800 11762150400 4116752640 588107520 40840800 1485120 28560 272  1
  ...
Recurrence from a-sequence: T(4, 2) = (4/2)*T(3, 1) + 4*4*T(3, 2) = 2*420 + 16*42 = 1512.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2)+ z(3)*T(3, 3)) = 4*(6*840 - 6*420 + 40*42 -420*1) = 15120.
Meixner type identity for n = 2: (D_x - 4*(D_x)^2)*(60 + 20*x + 1*x^2 ) = (20 + 2*x) - 4*2 = 2*(6 + x).
Sheffer recurrence for R(3, x): [(6 + x) + 8*(3 + x)*D_x + 16*x*(D_x)^2] (60 + 20*x + 1*x^2) = (6 + x)*(60 + 20*x + x^2) + 8*(3 + x)*(20 + 2*x) + 16*2*x = 840 + 420*x + 42*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (2*4!/2)*(3 + 2*2)*(1*42/3! + 4*1/2!) = 1512.
Diagonal sequence d = 2: {60, 420, 1512, ...} has o.g.f. 12*(5 + 10*x + x^2)/(1 - x)^5 (see A001813(2) and row n=2 of A103327) generating {12*binomial(2*(m + 2) + 1, 4)}_{m >= 0}. - _Wolfdieter Lang_, Oct 12 2017
		

References

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

Crossrefs

Cf. A225469, A225471, A271703 L[1,0], A286724 L[2,1], A290596 L[3,1], A290597 L[3,2], A048854 L[4,1], A292221, A103327,
Diagonal sequences: A000012, 2*A014105(m+1), 12*A053126(m+4), 120*A053128(m+6), A053130(n+8), ... - Wolfdieter Lang, Oct 12 2017

Formula

T(n, m) = L[4,3](n,m) = Sum_{k=m..n} A225471(n, k)*A225469(k, m), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 4*t)^(-3/2)*exp(x*t/(1 - 4*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 4*t)^(-3/2)*(t/(1 - 4*t))^m/m!, m >= 0.
Three term recurrence for column entries k >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 4*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), n >= 1, T(0, 0) = 0, from the a-sequence {1, 4 repeat(0)} and z(j) = 2*A292221(j) (see above).
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(4*n - 1)*T(n-1, m) - 8*(n-1)*(2*n - 1)*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.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 4*D_x)) * R(n, x) = n * R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-4)^k*{D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(6 + x)*1 + 8*(3 + x)*D_x + 16*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (2*n!/(n-m))*(3 + 2*m)*Sum_{p=0..n-1-m} 4^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.
Explicit form (from the o.g.f.s of diagonal sequences): ((2*(n-m))!/(n-m)!)*binomial(2*n + 1, 2*(n-m)), n >= m >= 0, and vanishing for n < m. - Wolfdieter Lang, Oct 12 2017

A383702 Coefficient of x^2 in expansion of (x+3) * (x+7) * ... * (x+4*n-1).

Original entry on oeis.org

0, 0, 1, 21, 446, 10670, 290599, 8951355, 308846124, 11822475564, 497794079421, 22881487815153, 1140642637297866, 61312161303209466, 3535773901817957955, 217787248332803277495, 14271822475100747003160, 991517953843097370650520, 72799719644532661375481145
Offset: 0

Views

Author

Seiichi Manyama, May 06 2025

Keywords

Crossrefs

Column k=2 of A225471.

Programs

  • Magma
    [&+[(4*n-1)^(k-2) * 4^(n-k) * Binomial(k,2) * StirlingFirst(n,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, May 07 2025
  • Mathematica
    Array[Residue[Product[(x + 4*k - 1), {k, 1, #}]/x^3, {x, 0}] &, 19, 0] (* Shenghui Yang, May 06 2025 *)
    Table[Sum[(4*n-1)^(k-2)*4^(n-k)*Binomial[k,2]*StirlingS1[n,k],{k,0,n}],{n,0,18}] (* Vincenzo Librandi, May 07 2025 *)
  • PARI
    a(n) = polcoef(prod(k=0, n-1, x+4*k+3), 2);
    

Formula

a(n) = Sum_{k=2..n} 3^(k-2) * 4^(n-k) * binomial(k,2) * |Stirling1(n,k)|.
a(n) = Sum_{k=2..n} (4*n-1)^(k-2) * 4^(n-k) * binomial(k,2) * Stirling1(n,k).
E.g.f.: f(x)^3 * log(f(x))^2 / 2, where f(x) = 1/(1 - 4*x)^(1/4).
Conjecture D-finite with recurrence a(n) +3*(-4*n+5)*a(n-1) +(48*n^2-168*n+151)*a(n-2) -(4*n-9)^3*a(n-3)=0. - R. J. Mathar, May 07 2025

A383703 Coefficient of x^3 in expansion of (x+3) * (x+7) * ... * (x+4*n-1).

Original entry on oeis.org

0, 0, 0, 1, 36, 1130, 36660, 1280419, 48644344, 2011398164, 90267003960, 4379275249701, 228707424551100, 12804721289403966, 765571832220427596, 48704512002823186119, 3286171504510664002992, 234445313277315235203624, 17637135196532479070107824, 1395584859384468591633567945
Offset: 0

Views

Author

Seiichi Manyama, May 06 2025

Keywords

Crossrefs

Column k=3 of A225471.

Programs

  • Magma
    [&+[(4*n-1)^(k-3) * 4^(n-k) * Binomial(k,3) * StirlingFirst(n,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, May 07 2025
  • Mathematica
    Table[Sum[(4*n-1)^(k-3)*4^(n-k)*Binomial[k,3]*StirlingS1[n,k],{k,3,n}],{n,0,25}] (* Vincenzo Librandi, May 07 2025 *)
  • PARI
    a(n) = polcoef(prod(k=0, n-1, x+4*k+3), 3);
    

Formula

a(n) = Sum_{k=3..n} 3^(k-3) * 4^(n-k) * binomial(k,3) * |Stirling1(n,k)|.
a(n) = Sum_{k=3..n} (4*n-1)^(k-3) * 4^(n-k) * binomial(k,3) * Stirling1(n,k).
E.g.f.: f(x)^3 * log(f(x))^3 / 6, where f(x) = 1/(1 - 4*x)^(1/4).

A225478 Triangle read by rows, 4^k*s_4(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 3, 4, 21, 40, 16, 231, 524, 336, 64, 3465, 8784, 7136, 2304, 256, 65835, 180756, 170720, 72320, 14080, 1024, 1514205, 4420728, 4649584, 2346240, 613120, 79872, 4096, 40883535, 125416476, 143221680, 81946816, 25939200, 4609024, 430080, 16384, 1267389585, 4051444896, 4941537984, 3113238016, 1131902464, 246636544, 31768576, 2228224, 65536
Offset: 0

Views

Author

Peter Luschny, May 17 2013

Keywords

Comments

Triangle T(n,k), read by rows, given by (3, 4, 7, 8, 11, 12, 15, 16, ... (A014601)) DELTA (4, 0, 4, 0, 4, 0, 4, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015.

Examples

			[n\k][    0,       1,       2,       3,      4,     5,    6 ]
[0]       1,
[1]       3,       4,
[2]      21,      40,      16,
[3]     231,     524,     336,      64,
[4]    3465,    8784,    7136,    2304,    256,
[5]   65835,  180756,  170720,   72320,  14080,  1024,
[6] 1514205, 4420728, 4649584, 2346240, 613120, 79872, 4096.
		

Crossrefs

T(n, 0) ~ A008545; T(n, n) ~ A000302; T(n, n-1) ~ A002700.
row sums ~ A034176; alternating row sums ~ A008545.
Cf. A225471, A132393 (m=1), A028338 (m=2), A225477 (m=3).

Programs

  • Mathematica
    s[][0, 0] = 1; s[m][n_, k_] /; (k > n || k < 0) = 0; s[m_][n_, k_] := s[m][n, k] = s[m][n - 1, k - 1] + (m*n - 1)*s[m][n - 1, k];
    T[n_, k_] := 4^k*s[4][n, k];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 02 2019 *)
  • Sage
    @CachedFunction
    def SF_CS(n, k, m):
        if k > n or k < 0 : return 0
        if n == 0 and k == 0: return 1
        return m*SF_CS(n-1, k-1, m) + (m*n-1)*SF_CS(n-1, k, m)
    for n in (0..8): [SF_CS(n, k, 4) for k in (0..n)]

Formula

For a recurrence see the Sage program.
T(n,k) = 4^k * A225471(n,k). - Philippe Deléham, May 14 2015.
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