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 1-3 of 3 results.

A193685 5-Stirling numbers of the second kind.

Original entry on oeis.org

1, 5, 1, 25, 11, 1, 125, 91, 18, 1, 625, 671, 217, 26, 1, 3125, 4651, 2190, 425, 35, 1, 15625, 31031, 19981, 5590, 740, 45, 1, 78125, 201811, 170898, 64701, 12250, 1190, 56, 1, 390625, 1288991, 1398097, 688506, 174951, 24150, 1806, 68, 1, 1953125, 8124571, 11075670, 6906145, 2263065, 416451, 44016, 2622, 81, 1, 9765625, 50700551, 85654261, 66324830, 27273730, 6427575, 900627, 75480, 3675, 95, 1
Offset: 0

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Author

Wolfdieter Lang, Oct 06 2011

Keywords

Comments

This is the lower triangular Sheffer matrix (exp(5*x),exp(x)-1). For Sheffer matrices see the W. Lang link under A006232 with references, and the rules for the conversion to the umbral notation of S. Roman's book.
The general case is Sheffer (exp(r*x),exp(x)-1), r=0,1,..., corresponding to r-Stirling2 numbers with row and column offsets 0. See the Broder link for r-Stirling2 numbers with offset [r,r].
a(n,m), n >= m >= 0, gives the number of partitions of the set {1.2....,n+5} into m+5 nonempty distinct subsets such that 1,2,3,4 and 5 belong to distinct subsets.
a(n,m) appears in the following normal ordering of Bose operators a and a* satisfying the Lie algebra [a,a*]=1: (a*a)^n (a*)^5 = Sum_{m=0..n} a(n,m)*(a*)^(5+m)*a^m, n >= 0. See the Mikhailov papers, where a(n,m) = S(n+5,m+5,5).
With a->D=d/dx and a*->x we also have
(xD)^n x^5 = Sum_{m=0..n} a(n,m)*x^(5+m)*D^m, n >= 0.

Examples

			n\m  0    1    2   3  4  5 ...
0    1
1    5    1
2   25   11    1
3  125   91   18   1
4  625  671  217  26  1
5 3125 4651 2190 425 35  1
...
5-restricted S2: a(1,0)=5 from 1,6|2|3|4|5, 2,6|1|3|4|5,
3,6|1|2|4|5, 4,6|1|2|3|5 and 5,6|1|2|3|4.
Recurrence: a(4,2) = (5+2)*a(3,2)+ a(3,1) = 7*18 + 91 = 217.
Normal ordering (n=1): (xD)^1 x^5 = Sum_{m=0..1} a(1,m)*x^(5+m)*D^m = 5*x^5 + 1*x^6*D.
a(2,1) = Sum_{j=0..1} S1(5,5-j)*S2(7-j,6) = 1*21 - 10*1 = 11.
		

Crossrefs

Cf. A196834 (row sums), A196835 (alternating row sums).
Columns: A000351 (m=0), A005062 (m=1), A019757 (m=2), A028165 (m=3), ...

Programs

  • Mathematica
    a[n_, m_] := Sum[ StirlingS1[5, 5-j]*StirlingS2[n+5-j, m+5], {j, 0, Min[5, n-m]}]; Flatten[ Table[ a[n, m], {n, 0, 10}, {m, 0, n}] ] (* Jean-François Alcover, Dec 02 2011, after Wolfdieter Lang *)

Formula

E.g.f. of row polynomials s(n,x):=Sum_{m=0..n} a(n,m)*x^m: exp(5*z + x(exp(z)-1)).
E.g.f. of column no. m (with leading zeros):
exp(5*x)*((exp(x)-1)^m)/m!, m >= 0 (Sheffer).
O.g.f. of column no. m (without leading zeros):
1/Product_{j=0..m} (1-(5+j)*x), m >= 0. (Compute the first derivative of the column e.g.f. and compare its Laplace transform with the partial fraction decomposition of the o.g.f. x^(m-1)/Product_{j=0..m} (1-(5+j)*x). This works for every r-restricted Stirling2 triangle.)
Recurrence: a(n,m) = (5+m)*a(n-1,m) + a(n-1,m-1), a(0,0)=1, a(n,m)=0 if n < m, a(n,-1)=0.
a(n,m) = Sum_{j=0..min(5,n-m)} S1(5,5-j)*S2(n+5-j,m+5), n >= m >= 0, with S1 and S2 the Stirling1 and Stirling2 numbers A008275 and A048993, respectively (see the Mikailov papers).
Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k>=0} k*(4+k)^(n-1)*x^(k-1)/k!. - Peter Bala, Jun 23 2014

A028165 Expansion of 1/((1-5x)*(1-6x)*(1-7x)*(1-8x)).

Original entry on oeis.org

1, 26, 425, 5590, 64701, 688506, 6906145, 66324830, 616252901, 5580303586, 49508360265, 432061044870, 3720287489101, 31681154472266, 267320885100785, 2238337148081710, 18621251375573301, 154069635600426546
Offset: 0

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Author

Keywords

Comments

This is the column m=2 sequence (without leading zeros) of the Sheffer triangle (exp(5*x), exp(x)-1) of the 5-restricted Stirling2 numbers A193685. For a proof see the column o.g.f. formula there. - Wolfdieter Lang, Oct 07 2011

Crossrefs

Programs

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,5), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = 26*a(n-1) - 251*a(n-2) + 1066*a(n-3) - 1680*a(n-4), n >= 4. - Vincenzo Librandi, Mar 19 2011
a(n) = 15*a(n-1) - 56*a(n-2) + 6^(n+1) - 5^(n+1), a(0)=1, a(1)=26. - Vincenzo Librandi, Mar 19 2011
E.g.f.: (d^3/dx^3)(exp(5*x)*((exp(x)-1)^3)/3!). See the Sheffer triangle comment above. - Wolfdieter Lang, Oct 07 2011
a(n) = -125*5^n/6 + 108*6^n - 343*7^n/2 + 256*8^n/3. - R. J. Mathar, Jun 23 2013

A372118 Square array A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2 for k, n >= 0 read by ascending antidiagonals.

Original entry on oeis.org

1, 3, 1, 7, 6, 1, 15, 25, 9, 1, 31, 90, 55, 12, 1, 63, 301, 285, 97, 15, 1, 127, 966, 1351, 660, 151, 18, 1, 255, 3025, 6069, 4081, 1275, 217, 21, 1, 511, 9330, 26335, 23772, 9751, 2190, 295, 24, 1, 1023, 28501, 111645, 133057, 70035, 19981, 3465, 385, 27, 1
Offset: 0

Views

Author

Werner Schulte, Apr 19 2024

Keywords

Comments

Depending on some fixed integer m >= 0 we define a family of square arrays A(m; n, k) = (Sum_{i=0..m} (-1)^i * binomial(m, i) * (k + m - i)^(n+m)) / m! for k, n >= 0. Special cases are: A004248 (m=0), A343237 (m=1) and this array (m=2). The A(m; n, k) satisfy: A(m; n, k) = (k+m) * A(m; n-1, k) + A(m-1; n, k) with initial values A(0; n, k) = k^n and A(m; 0, k) = 1.
Further properties are conjectures:
(1) O.g.f. of column k is Prod_{i=k..k+m} 1 / (1 - i * t);
(2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+m, m) * A048993(n+m, k+m) * x^k);
(3) The LU decompositions of these arrays are given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(m; n, k) = A048993(n+m, k+m) * (k+m)! / m!, i.e., A(m; n, k) = Sum_{i=0..k} L(m; n, i) * binomial(k, i).
The three conjectures are true, see links. - Sela Fried, Jul 07 2024

Examples

			Square array A(n, k) starts:
n\k :    0     1       2       3        4         5         6         7
=======================================================================
  0 :    1     1       1       1        1         1         1         1
  1 :    3     6       9      12       15        18        21        24
  2 :    7    25      55      97      151       217       295       385
  3 :   15    90     285     660     1275      2190      3465      5160
  4 :   31   301    1351    4081     9751     19981     36751     62401
  5 :   63   966    6069   23772    70035    170898    365001    706104
  6 :  127  3025   26335  133057   481951   1398097   3463615   7628545
  7 :  255  9330  111645  724260  3216795  11075670  31794105  79669320
  etc.
		

Crossrefs

Rows: A000012 (n=0), A008585 (n=1), A227776 (n=2).
Columns: A000225 (k=0), A000392 (k=1), A016269 (k=2), A016753 (k=3), A016103 (k=4), A019757 (k=5), A020570 (k=6), A020782 (k=7).
Main diagonal: A281596(n+2).

Programs

  • Mathematica
    A372118[n_, k_] := ((k+2)^(n+2) - 2*(k+1)^(n+2) + k^(n+2))/2;
    Table[A372118[n-k, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jul 10 2024 *)
  • PARI
    A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2

Formula

A(n, k) = (k+2) * A(n-1, k) + (k+1)^(n+1) - k^(n+1) for n > 0.
Conjectures:
(1) O.g.f. of column k is Prod_{i=k..k+2} 1 / (1 - i * t);
(2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+2, 2) * A048993(n+2, k+2) * x^k);
(3) The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(n, k) = A048993(n+2, k+2) * (k+2)! / 2!, i.e., A(n, k) = Sum_{i=0..k} L(n, i) * binomial(k, i).
The three conjectures are true. See comments. - Sela Fried, Jul 09 2024
Showing 1-3 of 3 results.