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 21 results. Next

A284859 Row sums of the Sheffer triangle (exp(x), exp(3*x)-1) given in A282629.

Original entry on oeis.org

1, 4, 25, 199, 1876, 20257, 245017, 3266914, 47450923, 743935375, 12497579698, 223619318215, 4240423494685, 84855613320004, 1785410320771933, 39373503608087299, 907548770965519660, 21810536356271794549, 545305573054110017125, 14155835044848094831018
Offset: 0

Views

Author

Wolfdieter Lang, Apr 05 2017

Keywords

Comments

See A282629 for details. These are generalized Bell numbers (A000110) because A282629 is a generalized Stirling2 triangle.

Crossrefs

Programs

  • Mathematica
    T[n_, m_]:= Sum[Binomial[m, k] (-1)^(k - m) (1 + 3k)^n/m!, {k, 0, m}]; Table[Sum[T[n, m], {m, 0, n}], {n, 0, 20}] (* Indranil Ghosh, Apr 10 2017 *)
    Table[Sum[3^k*Binomial[n,k]*BellB[k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 22 2022 *)
  • PARI
    T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
    a(n) = sum(m=0, n, T(n, m)); \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import binomial, factorial
    def T(n, m): return sum([binomial(m, k) * (-1)**(k - m) * (1 + 3*k)**n for k in range(m + 1)])//factorial(m)
    def a(n): return sum([T(n, k) for k in range(n + 1)])
    print([a(n) for n in range(20)]) # Indranil Ghosh, Apr 10 2017

Formula

a(n) = Sum_{m=0..n} A282629(n, m).
E.g.f.: exp(x)*exp(exp(3*x) -1).
a(n) = (1/e)*Sum_{m>=0} (1/m!)*(1+3*m)^n, n >= 0. (Dobiński type formula from the A282629(n,m) sum formula, interchanging summations).
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
a(n) ~ Bell(n) * (3 + LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 3^n * n^(n + 1/3) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/3)). - Vaclav Kotesovec, Jun 27 2022

A284861 Triangle read by rows: T(n, k) = S2[3,1](n, k)*k! with the Sheffer triangle S2[3,1] = (exp(x), exp(3*x) -1) given in A282629.

Original entry on oeis.org

1, 1, 3, 1, 15, 18, 1, 63, 216, 162, 1, 255, 1890, 3564, 1944, 1, 1023, 14760, 52650, 68040, 29160, 1, 4095, 109458, 659340, 1516320, 1487160, 524880, 1, 16383, 790776, 7578522, 27624240, 46539360, 36741600, 11022480, 1, 65535, 5633730, 82902204, 450057384, 1158993360, 1535798880, 1014068160, 264539520
Offset: 0

Views

Author

Wolfdieter Lang, Apr 09 2017

Keywords

Comments

This is a generalization of triangle A131689(n, k) = Stirling2(n, k)*k!, because S2[3,1] is a generalization of the Stirling2 triangle written as S2[1,0].
This triangle appears in the o.g.f. G(3,1;n,x) of the powers {(1+3*m)^n}{m>=0} as G(3,1;n,x) = Sum{k>=0..n} T(n, k)*x^k / (1-x)^k.
This triangle is also related to the generalized row reversed Euler triangle rEu[3,1] with row polynomial rEu(3,1;n,x) = Sum_{m=0..n} rEu(3,1;n,m)*x^m with rEu(3,1;n,m) = Sum_{j=0..m} (-1)^(m-j)*binomial(n-j, m-j)*T(n, m). This follows from the above given o.g.f. of powers G(3,1;n,x) = rEu(3,1;n,x)/(1-x)^(n+1). The Euler triangle E[3,1] (row reversed rEu[3,1] is given in A225117. See a formula below.
The e.g.f. of the row polynomials R(3,1;n,x) = Sum_{m=0..n} T(n, m)*x^m follows from the e.g.f. of the row polynomials of the Sheffer triangle A282629. See the formula section.
The diagonal sequence is A032031(k) = k!*3^k.
The row sums give unsigned A151919, and the alternating row sums give A122803.
The first column k sequences divided by A032031(k) are A000012, A002450 (with a leading 0), A016223, A021874. For the e.g.f.s and o.g.f.s see below.

Examples

			The triangle T(n, k) begins
n\k 0     1      2       3        4        5        6        7 ...
0:  1
1:  1     3
2:  1    15     18
3:  1    63    216     162
4:  1   255   1890    3564     1944
5:  1  1023  14760   52650    68040    29160
6:  1  4095 109458  659340  1516320  1487160   524880
7:  1 16383 790776 7578522 27624240 46539360 36741600 11022480
...
row n=8: 1 65535 5633730 82902204 450057384 1158993360 1535798880 1014068160 264539520,
row n=9: 1 262143 39829320 879725610 6845572440 25294754520 50042059200 54561276000 30951123840 7142567040,
row n=10: 1 1048575 280378098 9155719980 99549149040 507399658920 1406104706160 2251231315200 2083248720000 1035672220800 214277011200.
------------------------------------------------------------------
T(2, 1) =  -1 + 4^2 = 15 = 2*A225117(2,2) + 1*A225117(2,1) = 2*1 + 1*13.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[k, m] (-1)^(k - m) (1 + 3m)^n, {m, 0, k}], {n, 0, 10}, {k, 0, n}]// Flatten (* Indranil Ghosh, Apr 09 2017 *)
  • PARI
    for(n=0, 10, for(k=0, n, print1(sum(m=0, k, binomial(k, m) * (-1)^(k - m)*(1 + 3*m)^n),", "); ); print();) \\ Indranil Ghosh, Apr 09 2017
    
  • Python
    # Indranil Ghosh, Apr 09 2017
    from sympy import binomial
    for n in range(11):
        print([sum([binomial(k, m)*(-1)**(k - m)*(1 + 3*m)**n for m in range(k + 1)]) for k in range(n + 1)])

Formula

E.g.f. of the row polynomials R(n, x) (see a comment above) is exp(z)/(1 - x*(exp(3*z) - 1)). This is the e.g.f. for the triangle.
T(n, k) = Sum_{m=0..k} binomial(k, m)*(-1)^(k-m)*(1+ 3*m)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..k} binomial(n-m, k-m)*A225117(n,n-m), 0 <= k <= n.
Three term recurrence: T(n, k) = 0 if n < k, T(n,-1) = 0, T(0, 0) = 1, T(n, k) = 3*k*T(n-1, k-1) + (1+3*k)*T(n-1, k) for n >= 1. See A282629.
The column k sequence has e.g.f. exp(x)*(exp(3*x) - 1)^k (from the Sheffer property of A282629).
The o.g.f. is A032031(k)*x^k/Product_{j=0..k} (1 - (1+3*j)*x).
From Peter Bala, Jan 12 2018: (Start)
n-th row polynomial R(n,x) = (1 + 3*x) o (1 + 3*x) o ... o (1 + 3*x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E14 in the Bala link. Cf. A145901.
R(n,x) = Sum_{k = 0..n} binomial(n,k)*3^k*F(k,x) where F(k,x) is the Fubini polynomial of order k, the k-th row polynomial of A019538. (End)

A284860 Alternating row sums of the Sheffer triangle (exp(x), exp(3*x) - 1) given in A282629.

Original entry on oeis.org

1, -2, -5, 19, 178, 175, -7739, -72056, -33179, 6899311, 87861076, 215532301, -11151014291, -222077806202, -1563185592617, 22953386817343, 878911293113026, 12330887396253691, 1416506544326449, -4284948239134152536
Offset: 0

Views

Author

Wolfdieter Lang, Apr 05 2017

Keywords

Comments

See A282629 for details. This is a generalization of A000587.

Crossrefs

Programs

  • Mathematica
    Fold[#2 - #1 &, Reverse@ #] & /@ Table[Sum[Binomial[m, k] (-1)^(k - m) (1 + 3 k)^n/m!, {k, 0, m}], {n, 0, 19}, {m, 0, n}] (* 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!);
    a(n) = sum(m=0, n, (-1)^m*T(n, m)); \\ Indranil Ghosh, Apr 10 2017

Formula

a(n) = Sum_{m=0..n} (-1)^m*A282629(n, m), n >= 0.
E.g.f.: exp(x)*exp(1 - exp(3*x)).
a(n) = (1/e)*Sum_{m>=0} ((-1)^m / m!)*(1 + 3*m)^n, n >= 0, (Dobiński type formula).- Wolfdieter Lang, Apr 10 2017
a(0) = 1; a(n) = a(n-1) - Sum_{k=1..n} binomial(n-1,k-1) * 3^k * a(n-k). - Ilya Gutkovskiy, Nov 29 2023

A321329 One third of the numerators of a Boas-Buck sequence for the triangular Sheffer matrix S2[3,1] = A282629.

Original entry on oeis.org

1, 1, 0, -3, 0, 9, 0, -81, 0, 81, 0, -167913, 0, 2187, 0, -23731137, 0, 287811387, 0, -10310604939, 0, 13761310401, 0, -125613568885131, 0, 3146863577139, 0, -5409187422305481, 0, 8241860346410471769
Offset: 0

Views

Author

Wolfdieter Lang, Nov 15 2018

Keywords

Comments

The denominators are given in A321330.
The general Boas-Buck type recurrence for lower triangular Sheffer matrices S(n, m) is: S(n, m) = (n!/(n-m))*Sum_{k=m..n-1} (1/k!)*(alpha(n-1-k) + m*beta(n-1-k))*S(k, m), for n >= m + 1 >= 1, and inputs S(n, n).
See the Boas-Buck type recurrence for the columns of S2[3,1] = A282629.
For S2[3,1] the Boas-Buck sequence alpha is {1, repeat(0)}.

Examples

			The rationals beta begin: {3/2, 3/4, 0, -9/80, 0, 27/1120, 0, -243/44800, 0, 243/197120, 0, -503739/1793792000, 0, 6561/102502400, 0, -71193411/4879114240000, 0, 863434161/259568877568000, 0, -30931814817/40789395046400000, 0, ...}.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Numerator[(-3)^(n+1)*BernoulliB[n+1]/(n+1)!/3]; Array[a, 30, 0] (* Amiram Eldar, Nov 15 2018 *)

Formula

a(n) = (1/3)*numerator(beta(n)), with beta(n) = (-3)^{n+1}* B(n+1)/(n+1)!, where B(n) = A027641(n)/A027642(n) (Bernoulli).
G.f. for rationals {beta(n)}_{n >= 0} is d/dx(log((exp(3*x) - 1)/x)) = (3*x*e^(3*x) - e^(3*x) + 1)/(x*(e^(3*x)-1)).

A321330 Denominators of a Boas-Buck sequence for the triangular Sheffer matrix S2[3,1] = A282629.

Original entry on oeis.org

2, 4, 1, 80, 1, 1120, 1, 44800, 1, 197120, 1, 1793792000, 1, 102502400, 1, 4879114240000, 1, 259568877568000, 1, 40789395046400000, 1, 238803367362560000, 1, 9561686829196902400000, 1, 1050734816395264000000, 1, 7922540515620290560000000, 1, 52951091790199773986816000000
Offset: 0

Views

Author

Wolfdieter Lang, Nov 15 2018

Keywords

Comments

The numerators are given A321329.
For the rationals beta see the example in A321329. The sequence alpha = {1, repeat(0)}.
For the Boas-Buck recurrence see A282629.

Crossrefs

Programs

  • Mathematica
    a[n_] := Denominator[(-3)^(n+1)*BernoulliB[n+1]/(n+1)!]; Array[a, 30, 0] (* Amiram Eldar, Nov 15 2018 *)

Formula

a(n) = denominator(beta(n)), with beta(n) = (-3)^{n+1}* B(n+1)/(n+1)!, where B(n) = A027641(n)/A027642(n) (Bernoulli).
The g.f. for {beta(n)}_{n>=0} is given in A321329.

A290316 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 A282629 (S2[3,1] generalized Stirling2), for n >= 0.

Original entry on oeis.org

1, 1, 6, 1, 48, 90, 1, 234, 2214, 2160, 1, 996, 27432, 114588, 71280, 1, 4062, 260748, 2791800, 6770628, 2993760, 1, 16344, 2178630, 48256344, 280652364, 454137840, 152681760, 1, 65490, 16966530, 691711920, 7846782660, 29157089832, 34236464400, 9160905600, 1, 262092, 126820980, 8851303620, 174637926180, 1219804572672, 3187159638984, 2871984146400, 632102486400, 1, 1048518, 924701832, 105253405560, 3359003385600, 39425596747272, 188635513271256, 369150976563264, 265665182896800, 49303993939200
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 A282629 = (e^x, e^(3*x) - 1), called S2[3,1], is GS2(3,1;n,x) = P(n, x)/(1 - 3*x)^(2*n+1), with the row polynomials P(n, x) = Sum_{k=0..n} T(n, k)*x^k, n >= 0.
For the general case Sheffer S2[d,a] = (e^(a*x), e^(d*x) - 1) (with gcd(d,a) = 1, d >=0, a >= 0, and for d = 1 one takes a = 0) see a comment in A290315.
For the computation of the exponential generating function (e.g.f.) of the o.g.f.s of the diagonal sequences of a Sheffer triangle (lower triangular matrix) via Lagrange's theorem see a comment and link in A290311.

Examples

			The triangle T(n, k) begins:
n\k 0     1        2         3          4           5           6          7 ...
0:  1
1:  1     6
2:  1    48       90
3:  1   234     2214      2160
4:  1   996    27432    114588      71280
5:  1  4062   260748   2791800    6770628     2993760
6:  1 16344  2178630  48256344  280652364   454137840   152681760
7:  1 65490 16966530 691711920 7846782660 29157089832 34236464400 9160905600
...
n = 8: 1 262092 126820980 8851303620 174637926180 1219804572672 3187159638984 2871984146400 632102486400,
n = 9: 1 1048518 924701832 105253405560 3359003385600 39425596747272 188635513271256 369150976563264 265665182896800 49303993939200.
...
n = 3: The o.g.f. of the 4th diagonal sequence of A282629, [1, 255, 7380, ...], is P(3, x) = (1 + 234*x + 2214*x^2 + 2160*x^3)/(1 - 3*x)^7.
		

Crossrefs

Formula

T(n, k) = [x^k] P(n, x) with the numerator polynomials of the o.g.f. of the (n+1)-th diagonal sequence of the triangle A282629. See a comment above.

A048993 Triangle of Stirling numbers of 2nd kind, S(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 7, 6, 1, 0, 1, 15, 25, 10, 1, 0, 1, 31, 90, 65, 15, 1, 0, 1, 63, 301, 350, 140, 21, 1, 0, 1, 127, 966, 1701, 1050, 266, 28, 1, 0, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 0, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Also known as Stirling set numbers.
S(n,k) enumerates partitions of an n-set into k nonempty subsets.
The o.g.f. for the sequence of diagonal k (k=0 for the main diagonal) is G(k,x) = ((x^k)/(1-x)^(2*k+1))*Sum_{m=0..k-1} A008517(k,m+1)*x^m. A008517 is the second-order Eulerian triangle. - Wolfdieter Lang, Oct 14 2005
From Philippe Deléham, Nov 14 2007: (Start)
Sum_{k=0..n} S(n,k)*x^k = B_n(x), where B_n(x) = Bell polynomials.
The first few Bell polynomials are:
B_0(x) = 1;
B_1(x) = 0 + x;
B_2(x) = 0 + x + x^2;
B_3(x) = 0 + x + 3x^2 + x^3;
B_4(x) = 0 + x + 7x^2 + 6x^3 + x^4;
B_5(x) = 0 + x + 15x^2 + 25x^3 + 10x^4 + x^5;
B_6(x) = 0 + x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6;
(End)
This is the Sheffer triangle (1, exp(x) - 1), an exponential (binomial) convolution triangle. The a-sequence is given by A006232/A006233 (Cauchy sequence). The z-sequence is the zero sequence. See the link under A006232 for the definition and use of these sequences. The row sums give A000110 (Bell), and the alternating row sums give A000587 (see the Philippe Deléham formulas and crossreferences below). - Wolfdieter Lang, Oct 16 2014
Also the inverse Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015
From Wolfdieter Lang, Feb 21 2017: (Start)
The transposed (trans) of this lower triagonal Sheffer matrix of the associated type S = (1, exp(x) - 1) (taken as N X N matrix for arbitrarily large N) provides the transition matrix from the basis {x^n/n!}, n >= 0, to the basis {y^n/n!}, n >= 0, with y^n/n! = Sum_{m>=n} S^{trans}(n, m) x^m/m! = Sum_{m>=0} x^m/m!*S(m, n).
The Sheffer transform with S = (g, f) of a sequence {a_n} to {b_n} for n >= 0, in matrix notation vec(b) = S vec(a), satisfies, with e.g.f.s A and B, B(x) = g(x)*A(f(x)) and B(x) = A(y(x)) identically, with vec(xhat) = S^{trans,-1} vec(yhat) in symbolic notation with vec(xhat)_n = x^n/n! (similarly for vec(yhat)).
(End)
Number of partitions of {1, 2, ..., n+1} into k+1 nonempty subsets such that no subset contains two adjacent numbers. - Thomas Anton, Sep 26 2022

Examples

			The triangle S(n,k) begins:
  n\k 0 1    2     3      4       5       6      7      8     9   10 11 12
  0:  1
  1:  0 1
  2:  0 1    1
  3:  0 1    3     1
  4:  0 1    7     6      1
  5:  0 1   15    25     10       1
  6:  0 1   31    90     65      15       1
  7:  0 1   63   301    350     140      21      1
  8:  0 1  127   966   1701    1050     266     28      1
  9:  0 1  255  3025   7770    6951    2646    462     36     1
 10:  0 1  511  9330  34105   42525   22827   5880    750    45    1
 11:  0 1 1023 28501 145750  246730  179487  63987  11880  1155   55  1
 12:  0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66  1
 ... reformatted and extended - _Wolfdieter Lang_, Oct 16 2014
Completely symmetric function S(4, 2) = h^{(2)}_2 = 1^2 + 2^2 + 1^1*2^1 = 7; S(5, 2) = h^{(2)}_3 = 1^3 + 2^3 + 1^2*2^1 + 1^1*2^2 = 15. - _Wolfdieter Lang_, May 26 2017
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: S(5, 3) = S(4, 2) + 2*S(4, 3) = 7 + 3*6 = 25.
Boas-Buck recurrence for column m = 3, and n = 5: S(5, 3) = (3/2)*((5/2)*S(4, 3) + 10*Bernoulli(2)*S(3, 3)) = (3/2)*(15 + 10*(1/6)*1) = 25. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

Crossrefs

See especially A008277 which is the main entry for this triangle.
A000110(n) = sum(S(n, k)) k=0..n, n >= 0. Cf. A085693.
Cf. A084938, A106800 (mirror image), A138378, A213061 (mod 2).

Programs

  • Haskell
    a048993 n k = a048993_tabl !! n !! k
    a048993_row n = a048993_tabl !! n
    a048993_tabl = iterate (\row ->
       [0] ++ (zipWith (+) row $ zipWith (*) [1..] $ tail row) ++ [1]) [1]
    -- Reinhard Zumkeller, Mar 26 2012
  • Maple
    for n from 0 to 10 do seq(Stirling2(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, Nov 01 2006
  • Mathematica
    t[n_, k_] := StirlingS2[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v *)
  • Maxima
    create_list(stirling2(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    for(n=0, 22, for(k=0, n, print1(stirling(n, k, 2), ", ")); print()); \\ Joerg Arndt, Apr 21 2013
    

Formula

S(n, k) = k*S(n-1, k) + S(n-1, k-1), n > 0; S(0, k) = 0, k > 0; S(0, 0) = 1.
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Sum_{k = 0..n} x^k*S(n, k) = A213170(n), A000587(n), A000007(n), A000110(n), A001861(n), A027710(n), A078944(n), A144180(n), A144223(n), A144263(n) respectively for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. - Philippe Deléham, May 09 2004, Feb 16 2013
S(n, k) = Sum_{i=0..k} (-1)^(k+i)binomial(k, i)i^n/k!. - Paul Barry, Aug 05 2004
Sum_{k=0..n} k*S(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Equals the inverse binomial transform of A008277. - Gary W. Adamson, Jan 29 2008
G.f.: 1/(1-xy/(1-x/(1-xy/(1-2x/(1-xy/1-3x/(1-xy/(1-4x/(1-xy/(1-5x/(1-... (continued fraction equivalent to Deléham DELTA construction). - Paul Barry, Dec 06 2009
G.f.: 1/Q(0), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
Inverse of padded A008275 (padded just as A048993 = padded A008277). - Tom Copeland, Apr 25 2014
E.g.f. for the row polynomials s(n,x) = Sum_{k=0..n} S(n,k)*x^k is exp(x*(exp(z)-1)) (Sheffer property). E.g.f. for the k-th column sequence with k leading zeros is ((exp(x)-1)^k)/k! (Sheffer property). - Wolfdieter Lang, Oct 16 2014
G.f. for column k: x^k/Product_{j=1..k} (1-j*x), k >= 0 (with the empty product for k = 0 put to 1). See Abramowitz-Stegun, p. 824, 24.1.4 B. - Wolfdieter Lang, May 26 2017
Boas-Buck recurrence for column sequence m: S(n, k) = (k/(n - k))*(n*S(n-1, k)/2 + Sum_{p=k..n-2} (-1)^(n-p)*binomial(n,p)*Bernoulli(n-p)*S(p, k)), for n > k >= 0, with input T(k,k) = 1. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the white diamond multiplication operator defined in Bala - see Example E4. - Peter Bala, Jan 07 2018
Sum_{k=1..n} k*S(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
S(n,k) = Sum_{j=k..n} (-1)^(j-k)*A059297(n,j)*A354794(j,k). - Mélika Tebni, Jan 27 2023

A111577 Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 12, 1, 1, 85, 105, 22, 1, 1, 341, 820, 325, 35, 1, 1, 1365, 6081, 4070, 780, 51, 1, 1, 5461, 43932, 46781, 14210, 1596, 70, 1, 1, 21845, 312985, 511742, 231511, 39746, 2926, 92, 1, 1, 87381, 2212740, 5430405, 3521385, 867447, 95340, 4950, 117, 1
Offset: 1

Views

Author

Gary W. Adamson, Aug 07 2005

Keywords

Comments

In triangles of analogs to Stirling numbers of the second kind, the multipliers of T(n-1,k) in the recurrence are terms in arithmetic sequences: in Pascal's triangle A007318, the multiplier = 1. In triangle A008277, the Stirling numbers of the second kind, the multipliers are in the set (1,2,3...). For this sequence here, the multipliers are from A016777.
Riordan array [exp(x), (exp(3x)-1)/3]. - Paul Barry, Nov 26 2008
From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the triangle of connection constants between the polynomial basis sequences {x^n}, n>=0 and {n!*3^n*binomial((x - 1)/3,n)}, n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318, then P * M = A225468, P^2 * M = A075498. Also P^(-1) * M is a shifted version of A075498.
This triangle is the particular case a = 3, 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)
Named after the English scientist Francis Galton (1822-1911). - Amiram Eldar, Jun 13 2021
This is the array of (r, β)-Stirling numbers for r = 1 and β = 3. See Corcino. - Peter Bala, Feb 26 2025

Examples

			T(5,3) = T(4,2) + 7*T(4,3) = 21 + 7*12 = 105.
The triangle starts in row n = 1 as:
  1;
  1,  1;
  1,  5,   1;
  1, 21,  12,  1;
  1, 85, 105, 22, 1;
Connection constants: Row 4: [1, 21, 12, 1] so
x^3 = 1 + 21*(x - 1) + 12*(x - 1)*(x - 4) + (x - 1)*(x - 4)*(x - 7). - _Peter Bala_, Jan 27 2015
From _Peter Bala_, Feb 26 2025: (Start)
The array factorizes as
/1                \     /1               \/1              \/1             \
|1   1            |     |1   1           ||0  1           ||0  1          |
|1   5    1       |  =  |1   4   1       ||0  1   1       ||0  0  1       | ...
|1  21   12   1   |     |1  13   7   1   ||0  1   4  1    ||0  0  1  1    |
|1  85  105  22  1|     |1  44  34  10  1||0  1  13  7  1 ||0  0  1  4  1 |
|...              |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - x), x/(1 - 3*x)). Cf. A193843. (End)
		

Crossrefs

Programs

  • Maple
    A111577 := proc(n,k) option remember; if k = 1 or k = n then 1; else procname(n-1,k-1)+(3*k-2)*procname(n-1,k) ; fi; end:
    seq( seq(A111577(n,k),k=1..n), n=1..10) ; # R. J. Mathar, Aug 22 2009
  • Mathematica
    T[, 1] = 1; T[n, n_] = 1;
    T[n_, k_] := T[n, k] = T[n-1, k-1] + (3k-2) T[n-1, k];
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* Jean-François Alcover, Jun 13 2019 *)

Formula

T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k).
E.g.f.: exp(x)*exp((y/3)*(exp(3x)-1)). - Paul Barry, Nov 26 2008
Let f(x) = exp(1/3*exp(3*x) + x). Then, with an offset of 0, the row polynomials R(n,x) are given by R(n,exp(3*x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A105794, A143494 and A154537. - Peter Bala, Mar 01 2012
T(n, k) = 1/(3^k*k!)*Sum_{j=0..k}((-1)^(k-j)*binomial(k,j)*(3*j+1)^n). - Peter Luschny, May 20 2013
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n-1} 3^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
O.g.f. for n-th diagonal: exp(-x/3)*Sum_{k >= 0} (3*k + 1)^(k+n-1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k (with offset 0): 1/( (1 - x)*(1 - 4*x)*...*(1 - (3*k + 1)*x) ). (End)

Extensions

Edited and extended by R. J. Mathar, Aug 22 2009

A154537 Triangle T(n,m) read by rows: let p(n,x) = exp(-x) * Sum_{m >= 0} (2*m + 1)^n * x^m/m!; then T(n,m) = [x^m] p(n,x).

Original entry on oeis.org

1, 1, 2, 1, 8, 4, 1, 26, 36, 8, 1, 80, 232, 128, 16, 1, 242, 1320, 1360, 400, 32, 1, 728, 7084, 12160, 6320, 1152, 64, 1, 2186, 36876, 99288, 81200, 25312, 3136, 128, 1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256, 1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512
Offset: 0

Views

Author

Roger L. Bagula, Jan 11 2009

Keywords

Comments

Row sums are A126390.
These numbers are related to Stirling numbers of the second kind as MacMahon numbers A060187 are related to Eulerian numbers.
Let p and q denote operators acting on a function f(x) by pf(x) = x*f(x) and qf(x) = d/dx(f(x)). Let A be the anticommutator operator qp + pq. Then A^n = Sum_{k = 0..n} T(n,k) p^k q^k. For example, A^3(f) = f + 26*x*df/dx + 36*x^2*d^2(f)/dx^2 + 8*x^3*d^3(f)/dx^3. - Peter Bala, Jul 24 2014
From Peter Bala, May 21 2023: (Start)
Compare the definition of the polynomial p(n,x) with Dobiński's formula for the Bell polynomials (row polynomials of A008277 for n >= 1): Bell(n,x) = exp(-x) * Sum_{m >= 0} m^n * x^m/m!.
Boyadzhiev has shown that Bell(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 2^n + ... + (m-1)^n) * x^m/m! ). The corresponding result for this table is that the n-th row polynomial p(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 3^n + ... + (2*m-1)^n) * x^m/m! ). (End)

Examples

			Triangle begins:
  {1},
  {1, 2},
  {1, 8, 4},
  {1, 26, 36, 8},
  {1, 80, 232, 128, 16},
  {1, 242, 1320, 1360, 400, 32},
  {1, 728, 7084, 12160, 6320, 1152, 64},
  {1, 2186, 36876, 99288, 81200, 25312, 3136, 128},
  {1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256},
  {1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512},
  ...
Boas-Buck recurrence for column m = 2, and n = 4: T(4,2) = (1/2)*[4*3*T(3, 2) + 2*6*(-2)^2*Bernoulli(2)*T(2,2)] = (1/2)*(12*36 + 12*4*(1/6)*4) = 232. - _Wolfdieter Lang_, Aug 11 2017
		

Crossrefs

Programs

  • Mathematica
    p[x_, n_] = Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);
    Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]
    Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
    Flatten[%]

Formula

From Peter Bala, Oct 28 2011: (Start)
T(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
Recurrence relation: T(n,k) = 2*T(n-1,k-1) + (2*k+1)*T(n-1,k).
T(n,k) = (2^k)*A039755(n,k).
E.g.f.: exp(x + y*(exp(2*x) - 1)) = 1 + (1 + 2*y)*x + (1 + 8*y + 4*y^2)*x^2/2! + .... (End)
T(n, k) = Sum_{m=0..n} binomial(n, m)*2^m*Stirling2(m, k), 0 <= k <= n, where Stirling2 is A048993. - Wolfdieter Lang, Apr 13 2017
Boas-Buck recurrence for column sequence m: T(n,k) = (1/(n - k))*[n*(1 + m)*T(n-1,k) + k*Sum_{p=m..n-2} binomial(n,p)*(-2)^(n-p)*Bernoulli(n-p)*T(p,k)], for n > m >= 0, with input T(m,m) = 2^m. See a comment in A282629, also for references, and an example below. - Wolfdieter Lang, Aug 11 2017

Extensions

Edited by N. J. A. Sloane, Jan 12 2009

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
Showing 1-10 of 21 results. Next