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 1-10 of 15 results. Next

A286721 Column k=2 of the triangle A286718; Sheffer ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)).

Original entry on oeis.org

1, 12, 159, 2485, 45474, 959070, 22963996, 616224492, 18331744896, 599061555136, 21339235262784, 823098817664448, 34183157124707200, 1520908498941532800, 72182781516370886400, 3640264913563748243200, 194408478299496756556800, 10961007293837647131724800
Offset: 0

Views

Author

Wolfdieter Lang, May 29 2017

Keywords

Comments

a(n) is, for n >= 1, the total volume of the binomial(n+2, n) rectangular polytopes (hyper-cuboids) built from n orthogonal vectors with lengths of the sides from the set {1 + 3*j | j=0..n+1}. See the formula a(n) = sigma[3,1]^{(n+2)}_n and an example below.

Examples

			a(2) = 159 because sigma[3,1]^{(4)}_2 = 1*(4 + 7 + 10) + 4*(7 + 10) + 7*10 = 159. There are six rectangles (2D rectangular polytopes) built from two orthogonal vectors of different lengths from the set of {1,4,7,10} of total area 159.
		

Crossrefs

Cf. A007559 (k=0), A024216 (k=1), A286718.

Formula

a(n) = A286718(n+2, 2), n >= 0.
E.g.f.: (d^2/dx^2)((1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^2/2!) = (2*(log(1-3*x))^2 - 15*log(1-3*x) + 9)/(3^2*(1-3*x)^(7/3)).
a(n) = sigma[3,1]^{(n+2)}_n, n >= 0, with the elementary symmetric function sigma[3,1]^{n+2}_n of degree n of the n+2 numbers 1, 4, 7, ..., (1 + 3*(n+1)).

A290595 Triangle T(n, k) read by rows: row n gives the coefficients of the numerator polynomials of the o.g.f. of the (n+1)-th diagonal of the Sheffer triangle A286718 (|S1hat[3,1]| generalized Stirling 1), for n >= 0.

Original entry on oeis.org

1, 1, 2, 4, 19, 4, 28, 222, 147, 8, 280, 3194, 4128, 887, 16, 3640, 55024, 113566, 52538, 4835, 32, 58240, 1107336, 3268788, 2562676, 555684, 25167, 64, 1106560, 25526192, 100544412, 117517960, 45415640, 5301150, 128203, 128, 24344320, 663605680, 3325767376, 5352311764, 3189383200, 695714590, 47537320, 646519, 256, 608608000, 19213911360, 118361719296, 248493947496, 208996478388, 72479948400, 9696965250, 410038434, 3245139, 512
Offset: 0

Views

Author

Wolfdieter Lang, Aug 08 2017

Keywords

Comments

The ordinary generating function (o.g.f.) of the (n+1)-th diagonal sequence of the Sheffer triangle A286718 = ((1 - 3*x)^(-1/3), -log(1 - 3*x)/3), called |S1hat[3,1]|, is GD(3,1;n,x) = P(n, x)/(1 - x)^(2*n+1), with the row polynomials P(n, x) = Sum_{k=0..n} T(n, k)*x^k, n >= 0.
For the two parameter Sheffer case |S1hat[d,a]| = ((1 - d*x)^{-a/d}, -log(1 - d*x)/d) (with gcd(d,a) = 1, d >=0, a >= 0, and for d = 1 one takes a = 0) the e.g.f. ED(t, x) of the o.g.f.s {GD(d,a;n,x)}_{n>=0} of the diagonal sequences with elements D(d,a;n,m) = |S1hat[d,a]|(n+m, m) (n=0 for the main diagonal) is of interest. It can be computed via Lagrange's theorem. For the special Sheffer case (1, f(x)) this has been done by P. Bala (see the link). This method can be generalized for Sheffer (g(x), f(x)), as shown in the W. Lang link.

Examples

			The triangle T(n, k) begins:
n\k        0        1         2         3        4       5      6   7 ...
0:         1
1:         1        2
2:         4       19         4
3:        28      222       147         8
4:       280     3194      4128       887       16
5:      3640    55024    113566     52538     4835      32
6:     58240  1107336   3268788   2562676   555684   25167      6
7:   1106560 25526192 100544412 117517960 45415640 5301150 128203 128
...
n = 8: 24344320 663605680 3325767376 5352311764 3189383200 695714590 47537320 646519 256,
n = 9: 608608000 19213911360 118361719296 248493947496 208996478388 72479948400 9696965250 410038434 3245139 512.
n = 3: The o.g.f. of the 4th diagonal sequence of A286718, [28, 418, 2485, ...] = A024213(n+1), n >= 0, is P(3, x) = (28 + 222*x + 147*x^2 + 8*x^3)/(1 - 3*x)^7.
		

Crossrefs

Cf. A024213, A286718, A288875 ([2,1] case).

Formula

T(n, k) = [x^k] P(n, x) with the numerator polynomials of the o.g.f. GD(n, x) = P(n, x)/(1-x)^(2*n+1) of the (n+1)-th diagonal sequence of the triangle A286718. See a comment above.

A007559 Triple factorial numbers (3*n-2)!!! with leading 1 added.

Original entry on oeis.org

1, 1, 4, 28, 280, 3640, 58240, 1106560, 24344320, 608608000, 17041024000, 528271744000, 17961239296000, 664565853952000, 26582634158080000, 1143053268797440000, 52580450364682240000, 2576442067869429760000, 133974987529210347520000, 7368624314106569113600000
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of increasing quaternary trees on n vertices. (See A001147 for ternary and A000142 for binary trees.) - David Callan, Mar 30 2007
a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 1. - Peter Luschny, Jun 23 2011
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
Partial products of A016777. - Reinhard Zumkeller, Sep 20 2013
For n > 2, a(n) is a Zumkeller number. - Ivan N. Ianakiev, Jan 28 2020
a(n) is the number of generalized permutations of length n related to the degenerate Eulerian numbers (see arXiv:2007.13205), cf. A336633. - Orli Herscovici, Jul 28 2020

Examples

			G.f. = 1 + x + 4*x^2 + 28*x^3 + 280*x^4 + 3640*x^5 + 58240*x^6 + ...
a(3) = 28 and a(4) = 280; with top row of M^3 = (28, 117, 108, 27), sum = 280.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n)= A035469(n, 1), n >= 1, (first column of triangle A035469(n, m)).
Cf. A107716. - Gary W. Adamson, Oct 22 2009
Cf. A095660. - Gary W. Adamson, Jul 19 2011
Subsequence of A007661. A007696, A008548.
a(n) = A286718(n,0), n >= 0.
Row sums of A336633.

Programs

  • GAP
    List([0..20], n-> Product([0..n-1], k-> 3*k+1 )); # G. C. Greubel, Aug 20 2019
  • Haskell
    a007559 n = a007559_list !! n
    a007559_list = scanl (*) 1 a016777_list
    -- Reinhard Zumkeller, Sep 20 2013
    
  • Magma
    b:= func< n | (n lt 2) select n else (3*n-2)*Self(n-1) >;
    [1] cat [b(n): n in [1..20]]; // G. C. Greubel, Aug 20 2019
    
  • Maple
    A007559 := n -> mul(k, k = select(k-> k mod 3 = 1, [$1 .. 3*n])): seq(A007559(n), n = 0 .. 17); # Peter Luschny, Jun 23 2011
    # second Maple program:
    b:= proc(n) option remember; `if`(n<1, 1, n*b(n-3)) end:
    a:= n-> b(3*n-2):
    seq(a(n), n=0..20);  # Alois P. Heinz, Dec 18 2024
  • Mathematica
    a[ n_] := If[ n < 0, 1 / Product[ k, {k, - 2, 3 n - 1, -3}],
      Product[ k, {k, 1, 3 n - 2, 3}]]; (* Michael Somos, Oct 14 2011 *)
    FoldList[Times,1,Range[1,100,3]] (* Harvey P. Dale, Jul 05 2013 *)
    Range[0, 19]! CoefficientList[Series[((1 - 3 x)^(-1/3)), {x, 0, 19}], x] (* Vincenzo Librandi, Oct 08 2015 *)
  • Maxima
    a(n):=if n=1 then 1 else (n)!*(sum(m/n*sum(binomial(k,n-m-k)*(-1/3)^(n-m-k)* binomial (k+n-1,n-1),k,1,n-m),m,1,n)+1); /* Vladimir Kruchinin, Aug 09 2010 */
    
  • PARI
    {a(n) = if( n<0, (-1)^n / prod(k=0,-1-n, 3*k + 2), prod(k=0, n-1, 3*k + 1))}; /* Michael Somos, Oct 14 2011 */
    
  • PARI
    my(x='x+O('x^33)); Vec(serlaplace((1-3*x)^(-1/3))) /* Joerg Arndt, Apr 24 2011 */
    
  • Sage
    def A007559(n) : return mul(j for j in range(1,3*n,3))
    [A007559(n) for n in (0..17)]  # Peter Luschny, May 20 2013
    

Formula

a(n) = Product_{k=0..n-1} (3*k + 1).
a(n) = (3*n - 2)!!!.
a(n) = A007661(3*n-2).
E.g.f.: (1-3*x)^(-1/3).
a(n) ~ sqrt(2*Pi)/Gamma(1/3)*n^(-1/6)*(3*n/e)^n*(1 - (1/36)/n - ...). - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = 3^n*Pochhammer(1/3, n).
a(n) = Sum_{k=0..n} (-3)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
a(n) = n!*(1+Sum_{m=1..n} (m/n)*Sum_{k=1..n-m} binomial(k, n-m-k)*(-1/3)^(n-m-k)*binomial(k+n-1, n-1)), n>1. - Vladimir Kruchinin, Aug 09 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term in M^n, M = a variant of Pascal (1,3) triangle (Cf. A095660); as an infinite square production matrix:
1, 3, 0, 0, 0,...
1, 4, 3, 0, 0,...
1, 5, 7, 3, 0,...
...
a(n+1) = sum of top row terms of M^n. (End)
a(n) = (-2)^n*Sum_{k=0..n} (3/2)^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
G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+1)/( 1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 21 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+1) + (k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
Let D(x) = 1/sqrt(1 - 2*x) be the e.g.f. for the sequence of double factorial numbers A001147. Then the e.g.f. A(x) for the triple factorial numbers satisfies D( Integral_{t=0..x} A(t) dt ) = A(x). Cf. A007696 and A008548. - Peter Bala, Jan 02 2015
O.g.f.: hypergeom([1, 1/3], [], 3*x). - Peter Luschny, Oct 08 2015
a(n) = 3^n * Gamma(n + 1/3)/Gamma(1/3). - Artur Jasinski, Aug 23 2016
a(n) = (-1)^n / A008544(n), 0 = a(n)*(+3*a(n+1) -a(n+2)) +a(n+1)*a(n+1) for all n in Z. - Michael Somos, Sep 30 2018
D-finite with recurrence: a(n) +(-3*n+2)*a(n-1)=0, n>=1. - R. J. Mathar, Feb 14 2020
Sum_{n>=1} 1/a(n) = (e/9)^(1/3) * (Gamma(1/3) - Gamma(1/3, 1/3)). - Amiram Eldar, Jun 29 2020

Extensions

Better description from Wolfdieter Lang

A132393 Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

Views

Author

Philippe Deléham, Nov 10 2007, Oct 15 2008, Oct 17 2008

Keywords

Comments

Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.
A094645*A007318 as infinite lower triangular matrices.
Row sums are the factorial numbers. - Roger L. Bagula, Apr 18 2008
Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014
Also the Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015
This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017
T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017
From Wolfdieter Lang, Jul 20 2017: (Start)
The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End)
For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019

Examples

			Triangle T(n,k) begins:
  1;
  0,    1;
  0,    1,     1;
  0,    2,     3,     1;
  0,    6,    11,     6,    1;
  0,   24,    50,    35,   10,    1;
  0,  120,   274,   225,   85,   15,   1;
  0,  720,  1764,  1624,  735,  175,  21,  1;
  0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
  ...
---------------------------------------------------
Production matrix is
  0, 1
  0, 1, 1
  0, 1, 2,  1
  0, 1, 3,  3,  1
  0, 1, 4,  6,  4,  1
  0, 1, 5, 10, 10,  5,  1
  0, 1, 6, 15, 20, 15,  6, 1
  0, 1, 7, 21, 35, 35, 21, 7, 1
  ...
From _Wolfdieter Lang_, May 09 2017: (Start)
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - _Wolfdieter Lang_, May 28 2017
O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - _Wolfdieter Lang_, Jul 20 2017
Boas-Buck recurrence for column k = 2 and n = 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*(3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - _Wolfdieter Lang_, Aug 11 2017
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150

Crossrefs

Essentially a duplicate of A048994. Cf. A008275, A008277, A112007, A130534, A288874, A354795.

Programs

  • Haskell
    a132393 n k = a132393_tabl !! n !! k
    a132393_row n = a132393_tabl !! n
    a132393_tabl = map (map abs) a048994_tabl
    -- Reinhard Zumkeller, Nov 06 2013
    
  • Maple
    a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x,n)),x,k),k=0..n) end: # Peter Luschny, Nov 28 2010
  • Mathematica
    p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *)
    Flatten[Table[Abs[StirlingS1[n,i]],{n,0,10},{i,0,n}]] (* Harvey P. Dale, Feb 04 2014 *)
  • Maxima
    create_list(abs(stirling1(n,k)),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    column(n,k) = my(v1, v2); v1 = vector(n-1, i, 0); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, v1[i] = (i+k)*(i+k-1)/2*v2[i]; for(j=1, i-1, v1[j] *= (i-j)*(i+k)/(i-j+2)); v2[i+1] = vecsum(v1)/i); v2 \\ generates n first elements of the k-th column starting from the first nonzero element. - Mikhail Kurkov, Mar 05 2025

Formula

T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007
Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008
Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008
a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008
Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008
Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008
From Wolfdieter Lang, Feb 21 2017: (Start)
E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/A027642. The e.g.f. for the z-sequence is 0. (End)
From Wolfdieter Lang, May 28 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1)}_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} 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). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)*A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023
n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1.
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link). (End)

A028338 Triangle of coefficients in expansion of (x+1)*(x+3)*...*(x + 2n - 1) in rising powers of x.

Original entry on oeis.org

1, 1, 1, 3, 4, 1, 15, 23, 9, 1, 105, 176, 86, 16, 1, 945, 1689, 950, 230, 25, 1, 10395, 19524, 12139, 3480, 505, 36, 1, 135135, 264207, 177331, 57379, 10045, 973, 49, 1, 2027025, 4098240, 2924172, 1038016, 208054, 24640, 1708, 64, 1, 34459425, 71697105, 53809164, 20570444, 4574934, 626934, 53676, 2796, 81, 1
Offset: 0

Views

Author

Keywords

Comments

Exponential Riordan array (1/sqrt(1-2*x), log(1/sqrt(1-2*x))). - Paul Barry, May 09 2011
The o.g.f.s D(d, x) of the column sequences, for d, d >= 0,(d=0 for the main diagonal) are P(d, x)/(1 - x)^(2*d+1), with the row polynomial P(d, x) = Sum_{m=0..d} A288875(d, m)*x^m. See A288875 for details. - Wolfdieter Lang, Jul 21 2017

Examples

			G.f. for n = 4: (x + 1)*(x + 3)*(x + 5)*(x + 7) = 105 + 176*x + 86*x^2 + 16*x^3 + x^4.
The triangle T(n, k) begins:
n\k       0        1        2        3       4      5     6    7  8  9
0:        1
1:        1        1
2:        3        4        1
3:       15       23        9        1
4:      105      176       86       16       1
5:      945     1689      950      230      25      1
6:    10395    19524    12139     3480     505     36     1
7:   135135   264207   177331    57379   10045    973    49    1
8:  2027025  4098240  2924172  1038016  208054  24640  1708   64  1
9: 34459425 71697105 53809164 20570444 4574934 626934 53676 2796 81  1
...
row n = 10: 654729075 1396704420 1094071221 444647600 107494190 16486680 1646778 106800 4335 100 1.
...  reformatted and extended. - _Wolfdieter Lang_, May 09 2017
O.g.f.s of diagonals d >= 0: D(2, x) = (3 + 8*x + x^2)/(1 - x)^5 generating [3, 23, 86, ...] = A024196(n+1), from the row d=2 entries of A288875 [3, 8, 1]. - _Wolfdieter Lang_, Jul 21 2017
Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(2*(1+4*(5/12))*T(2,2)/2! + 1*(1 + 4*(1/2))*T(3,2)/3!) = (4!/2)*(8/3*1 + 3*9/3!) = 86. - _Wolfdieter Lang_, Aug 11 2017
		

Crossrefs

A039757 is signed version.
Row sums: A000165.
Diagonals: A000012, A000290(n+1), A024196(n+1), A024197(n+1), A024198(n+1).
A161198 is a scaled triangle version and A109692 is a transposed triangle version.
Central terms: A293318.
Cf. A286718, A002208(n+1)/A002209(n+1).

Programs

  • Maple
    nmax:=8; for n from 0 to nmax do a(n, 0) := doublefactorial(2*n-1) od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := (2*n-1)*a(n-1, m) + a(n-1, m-1) od; od: seq(seq(a(n, m), m=0..n), n=0..nmax); # Johannes W. Meijer, Jun 08 2009, revised Nov 25 2012
  • Mathematica
    T[n_, k_] := Sum[(-2)^(n-i) Binomial[i, k] StirlingS1[n, i], {i, k, n}] (* Woodhouse *)
    Join[{1},Flatten[Table[CoefficientList[Expand[Times@@Table[x+i,{i,1,2n+1,2}]],x],{n,0,10}]]] (* Harvey P. Dale, Jan 29 2013 *)

Formula

Triangle T(n, k), read by rows, given by [1, 2, 3, 4, 5, 6, 7, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 20 2005
T(n, k) = Sum_{i=k..n} (-2)^(n-i) * binomial(i, k) * s(n, i) where s(n, k) are signed Stirling numbers of the first kind. - Francis Woodhouse (fwoodhouse(AT)gmail.com), Nov 18 2005
G.f. of row polynomials in y: 1/(1-(x+x*y)/(1-2*x/(1-(3*x+x*y)/(1-4*x/(1-(5*x+x*y)/(1-6*x*y/(1-... (continued fraction). - Paul Barry, Feb 07 2009
T(n, m) = (2*n-1)*T(n-1,m) + T(n-1,m-1) with T(n, 0) = (2*n-1)!! and T(n, n) = 1. - Johannes W. Meijer, Jun 08 2009
From Wolfdieter Lang, May 09 2017: (Start)
E.g.f. of row polynomials in y: (1/sqrt(1-2*x))*exp(-y*log(sqrt(1-2*x))) = exp(-(1+y)*log(sqrt(1-2*x))) = 1/sqrt(1-2*x)^(1+y).
E.g.f. of column m sequence: (1/sqrt(1-2*x))* (-log(sqrt(1-2*x)))^m/m!. For the special Sheffer, also known as exponential Riordan array, see a comment above. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 2^(n-1-p)*(1 + 2*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). See a comment and references in A286718. - Wolfdieter Lang, Aug 09 2017

A282629 Sheffer triangle (exp(x), exp(3*x) - 1). Named S2[3,1].

Original entry on oeis.org

1, 1, 3, 1, 15, 9, 1, 63, 108, 27, 1, 255, 945, 594, 81, 1, 1023, 7380, 8775, 2835, 243, 1, 4095, 54729, 109890, 63180, 12393, 729, 1, 16383, 395388, 1263087, 1151010, 387828, 51030, 2187, 1, 65535, 2816865, 13817034, 18752391, 9658278, 2133054, 201204, 6561, 1, 262143, 19914660, 146620935, 285232185, 210789621, 69502860, 10825650, 767637, 19683
Offset: 0

Views

Author

Wolfdieter Lang, Apr 03 2017

Keywords

Comments

For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references).
The e.g.f. for the sequence of column m is (Sheffer property) exp(x)*(exp(3*x) - 1)^m/m!.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0].
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).
The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(1/3)) and is A284857(n) / A284858(n).
The main diagonal gives A000244.
The row sums give A284859. The alternating row sums give A284860.
The triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 3*m)^n}{m>=0}, as G(n, x) = Sum{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 3*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225117 with row reversion.
The first column k sequences divided by 3^k are A000012, A002450 (with a leading 0), A016223, A021874. For the e.g.f.s and o.g.f.s see below. - Wolfdieter Lang, Apr 09 2017
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,m)*x^m of the Sheffer triangle S2[d,a] satisfy, as special polynomials of the Boas-Buck class, the 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*a*R(d,a;n-1,x) - Sum_{k=0..n-1} binomial(n, k+1)*(-d)^(k+1)*Bernoulli(k+1)*E_x*R(d,a;n-1-k,x), with E_x = x*d/dx (Euler operator).
This entails a recurrence for the sequence of column m, for n > m:
T(d,a;n,m) = (1/(n - m))*[(n/2)*(2*a + d*m)*T(d,a;n-1,m) + m*Sum_{p=m..n-2} binomial(n,p)(-d)^(n-p)*Bernoulli(n-p)*T(d,a;p,m)], with input T(d,a;n,n) = d^n. For the present [d,a] = [3,1] case see the formula and example sections below. - Wolfdieter Lang, Aug 09 2017 (End)
The inverse of this triangular Sheffer matrix S2[3,1] is S1[3,1] with rational elements S1[3,1](n, k) = (-1)^(n-k)*A286718(n, k)/3^k. - Wolfdieter Lang, Nov 15 2018
Named after the American mathematician Isador Mitchell Sheffer (1901-1992). - Amiram Eldar, Jun 19 2021

Examples

			The triangle T(n, m) begins:
  n\m 0      1        2         3         4         5        6        7      8     9
  0:  1
  1:  1      3
  2:  1     15        9
  3:  1     63      108        27
  4:  1    255      945       594        81
  5:  1   1023     7380      8775      2835       243
  6:  1   4095    54729    109890     63180     12393      729
  7:  1  16383   395388   1263087   1151010    387828    51030     2187
  8:  1  65535  2816865  13817034  18752391   9658278  2133054   201204   6561
  9:  1 262143 19914660 146620935 285232185 210789621 69502860 10825650 767637 19683
  ...
------------------------------------------------------------------------------------
Nontrivial recurrence for m=0 column from z-sequence: T(4,0) = 4*(1*1 + 63*(-1/6) + 108*(11/54) + 27*(-49/108)) = 1.
Recurrence for m=2 column from a-sequence: T(4, 2) = (4/2)*(1*63*3 + 2*108*(3/2) + 3*27*(-3/6)) = 945.
Recurrence for row polynomial R(3, x) (Meixner type): ((3*x + 1) + 3*x*d_x)*(1 + 15*x + 9*x^2) = 1 + 63*x + 108*x^2 + 27*x^3.
E.g.f. and o.g.f. of n = 1 powers {(1 + 3*m)^1}_{m>=0} A016777: E(1, x) = exp(x) * (T(1, 0) + T(1, 1)*x) = exp(x)*(1+3*x). O.g.f.: G(1, x) = T(1, 0)*0!/(1-x) + T(1, 1)*1!*x/(1-x)^2 = (1+2*x)/(1-x)^2.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (1/2)*(2*(2 + 3*2)*T(3, 2) + 2*6*(-3)^2*bernoulli(2)*T(2, 2)) = (1/2)*(16*108 + 12*9*(1/6)*9) = 945. - _Wolfdieter Lang_, Aug 09 2017
		

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

Programs

  • Mathematica
    Table[Sum[Binomial[m, k] (-1)^(k - m) (1 + 3 k)^n/m!, {k, 0, m}], {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 08 2017 *)
  • PARI
    T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
    for(n=0, 9, for(m=0, n, print1(T(n, m),", ");); print();) \\ Indranil Ghosh, Apr 08 2017

Formula

A nontrivial recurrence for the column m=0 entries T(n, 0) = 1 from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+1) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 3*k)^n/m!, 0 <= m <= n.
E.g.f. of triangle: exp(z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column m is exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
From Wolfdieter Lang, Apr 09 2017: (Start)
Standard three-term recurrence: T(n, m) = 0 if n < m, T(n,-1) = 0, T(0, 0) = 1, T(n, m) = 3*T(n-1, m-1) + (1+3*m)*T(n-1, m) for n >= 1. From the T(n, m) formula. Compare with the recurrence of S2[3,2] given in A225466.
The o.g.f. for sequence of column m is 3^m*x^m/Product_{j=0..m} (1 - (1+3*j)*x). (End)
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^k*Stirling2(k, m), 0 <= m <= n. - Wolfdieter Lang, Apr 13 2017
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 3*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n,p)*(-3)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m, m) = 3^m. See a comment above. - 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

A024212 2nd elementary symmetric function of first n+1 positive integers congruent to 1 mod 3.

Original entry on oeis.org

4, 39, 159, 445, 1005, 1974, 3514, 5814, 9090, 13585, 19569, 27339, 37219, 49560, 64740, 83164, 105264, 131499, 162355, 198345, 240009, 287914, 342654, 404850, 475150, 554229, 642789, 741559, 851295, 972780, 1106824, 1254264, 1415964, 1592815, 1785735
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Magma
    [n*(n+1)*(9*n^2+9*n-2)/8: n in [1..40]]; // Vincenzo Librandi, Oct 10 2011
  • Mathematica
    Table[n(n+1)(9n^2+9n-2)/8,{n,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{4,39,159,445,1005},40] (* Harvey P. Dale, Oct 16 2023 *)

Formula

a(n) = n*(n+1)*(9*n^2+9*n-2)/8.
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Clark Kimberling, Aug 18 2012
G.f.: (4 + 19*x + 4*x^2)/(1 - x)^5. - Clark Kimberling, Aug 18 2012
From Wolfdieter Lang, Jul 30 2017: (Start)
E.g.f.: exp(x)*x*(32+124*x+72*x^2+9*x^3)/8 = exp(x)*x*(2 + x)*(16 + 54*x + 9*x^2)/8.
a(n) = A286718(n+1, n-1), n >= 1. (End)

A225471 Triangle read by rows, 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, 1, 21, 10, 1, 231, 131, 21, 1, 3465, 2196, 446, 36, 1, 65835, 45189, 10670, 1130, 55, 1, 1514205, 1105182, 290599, 36660, 2395, 78, 1, 40883535, 31354119, 8951355, 1280419, 101325, 4501, 105, 1, 1267389585, 1012861224, 308846124, 48644344, 4421494, 240856, 7756, 136, 1
Offset: 0

Views

Author

Peter Luschny, May 17 2013

Keywords

Comments

The Stirling-Frobenius cycle numbers are defined in A225470.
Triangle T(n,k), read by rows, given by (3, 4, 7, 8, 11, 12, 15, 16, 19, 20, ... (A014601)) 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

			[n\k][    0,       1,      2,     3,    4,  5,  6 ]
[0]       1,
[1]       3,       1,
[2]      21,      10,      1,
[3]     231,     131,     21,     1,
[4]    3465,    2196,    446,    36,    1,
[5]   65835,   45189,  10670,  1130,   55,  1,
[6] 1514205, 1105182, 290599, 36660, 2395, 78,  1.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (4*4 - 1)*T(3, 2) = 131 +15*21 = 446.
Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(4*(3+8*(5/12)) *T(2, 2)/2! + 1*(3 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(4*(19/3)/2  + 7*21/3!) =  446.
(End)
		

Crossrefs

Columns k=0..3 give A008545, A286723(n-1), A383702, A383703.
Cf. A132393 (m=1), A028338 (m=2), A225470 (m=3).

Programs

  • Mathematica
    T[0, 0] = 1; T[n_, k_] := Sum[Binomial[n - j, k]*Abs[StirlingS1[n, n - j]]* 3^(n - k - j)*4^j, {j, 0, n - k}];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Wolfdieter Lang *)
  • Sage
    @CachedFunction
    def SF_C(n, k, m):
        if k > n or k < 0 : return 0
        if n == 0 and k == 0: return 1
        return SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m)
    for n in (0..8): [SF_C(n, k, 4) for k in (0..n)]

Formula

For a recurrence see the Sage program.
T(n, 0) ~ A008545; T(n, n) ~ A000012; T(n, n-1) = A014105.
Row sums ~ A047053; alternating row sums ~ A001813.
From Wolfdieter Lang, May 29 2017: (Start)
This is the Sheffer triangle (1/(1 - 4*x)^{-3/4}, -(1/4)*log(1-4*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(4,0,3)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 4*z)^{-(3+x)/4}.
E.g.f. of column k: (1-4*x)^(-3/4)*(-(1/4)*log(1-4*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+3)*R(n-1,x+4), with R(0, x) = 1.
R(n, x) = risefac(4,3;x,n) := Product_{j=0..(n-1)} (x + (3 + 4*j)). (See the P. Bala link, eq. (16) for the signed s_{4,0,3} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*3^(n-k-j)*4^j, with S1p(n, m) = A132393(n, m).
T(n, k) = sigma[4,3]^{(n)}_{n-k}, with the elementary symmetric functions sigma[4,3]^{(n)}_m of degree m in the n numbers 3, 7, 11, ..., 3+4*(n-1), with sigma[4,3]^{(n)}_0 := 1. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(3 + 8*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

A024216 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 1 mod 3.

Original entry on oeis.org

1, 5, 39, 418, 5714, 95064, 1864456, 42124592, 1077459120, 30777463360, 971142388160, 33547112941440, 1259204418129280, 51032742579123200, 2220990565060377600, 103308619261574809600, 5114702794181847910400
Offset: 0

Views

Author

Keywords

Comments

Comment by R. J. Mathar, Oct 01 2016: (Start)
The k-th elementary symmetric functions of the integers 1+j*3, j=0..n-1, form a triangle T(n,k), 0 <= k <= n, n >= 0:
1
1 1
1 5 4
1 12 39 28
1 22 159 418 280
1 35 445 2485 5714 3640
1 51 1005 9605 45474 95064 58240
1 70 1974 28700 227969 959070 1864456 1106560
1 92 3514 72128 859369 5974388 22963996 42124592 24344320
This here is the first subdiagonal. The diagonal seems to be A007559. The first columns are A000012, A000326, A024212, A024213, A024214. (End)

Examples

			From _Gheorghe Coserea_, Dec 24 2015: (Start)
For n = 1 we have a(1) = 1*4*(1/1 + 1/4) = 5.
For n = 2 we have a(2) = 1*4*7*(1/1 + 1/4 + 1/7) = 39.
For n = 3 we have a(3) = 1*4*7*10*(1/1 + 1/4 + 1/7 + 1/10) = 418.
(End)
		

Crossrefs

Cf. A024395, A024382, A286718 (first column).

Programs

  • Magma
    I:=[5,39]; [1] cat [n le 2 select I[n] else (6*n-1) * Self(n-1) - (3*n-2)^2 * Self(n-2) : n in [1..30]]; // Vincenzo Librandi, Aug 30 2015
  • Maple
    f:= gfun:-rectoproc({-(3*n+1)^2*a(n-1)+(6*n+5)*a(n)-a(n+1), a(0) = 1, a(1) = 5, a(2) = 39}, a(n), remember):
    map(f, [$0..30]); # Robert Israel, Aug 30 2015
  • Mathematica
    Rest[CoefficientList[Series[-(1/3)*Log[1-3*x]/(1-3*x)^(1/3), {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Oct 07 2013 *)
  • PARI
    n = 33; a = vector(n); a[1] = 5; a[2] = 39;
    for (k = 2, n-1, a[k+1] = (6*k+5) * a[k] - (3*k+1)^2 * a[k-1]);
    print(concat(1,a));  \\ Gheorghe Coserea, Aug 29 2015
    

Formula

E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(1/3). - Vladeta Jovovic, Sep 26 2003
For n >= 1, a(n-1) = 3^(n-1)*n!*Sum_{k=0..n-1} binomial(k-2/3, k)/(n-k). - Milan Janjic, Dec 14 2008, corrected by Peter Bala, Oct 08 2013
a(n) ~ (n+1)! * GAMMA(2/3) * 3^(n+3/2) * (log(n) + gamma + Pi*sqrt(3)/6 + 3*log(3)/2) / (6*Pi*n^(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 07 2013
a(n+1) = (6*n+5) * a(n) - (3*n+1)^2 * a(n-1). - Gheorghe Coserea, Aug 29 2015
E.g.f.: (3 - log(1-3*x))/(3*(1-3*x)^(4/3)). - Robert Israel, Aug 30 2015
a(n) = A286718(n+1, 1), n >= 0.
Boas-Buck type recurrence: a(0) = 1 and for n >= 1: a(n) = ((n+1)!/n) * Sum_{p=1..n} 3^(n-p)*(1 + 3*beta(n-p))*a(p-1)/p!, with beta(k) = A002208(k+1) / A002209(k+1). Proof from a(n) = A286718(n+1, 1). - Wolfdieter Lang, Aug 09 2017

Extensions

More terms from Vladeta Jovovic, Sep 26 2003
Showing 1-10 of 15 results. Next