A297924 -id:A297924 - cp's OEIS frontend

A121207 Triangle read by rows. The definition is by diagonals. The r-th diagonal from the right, for r >= 0, is given by b(0) = b(1) = 1; b(n+1) = Sum_{k=0..n} binomial(n+2,k+r)*a(k).

1, 1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 9, 15, 1, 1, 5, 14, 31, 52, 1, 1, 6, 20, 54, 121, 203, 1, 1, 7, 27, 85, 233, 523, 877, 1, 1, 8, 35, 125, 400, 1101, 2469, 4140, 1, 1, 9, 44, 175, 635, 2046, 5625, 12611, 21147, 1, 1, 10, 54, 236, 952, 3488, 11226, 30846, 69161, 115975

Offset: 0

N. J. A. Sloane, based on email from R. J. Mathar, Dec 11 2006

From Paul D. Hanna, Dec 12 2006: (Start)
Consider the row reversal, which is A124496 with an additional left column (A000110 = Bell numbers). The matrix inverse of this triangle is very simple:
   1;
  -1,   1;
  -1,  -1,   1;
  -1,  -2,  -1,   1;
  -1,  -3,  -3,  -1,   1;
  -1,  -4,  -6,  -4,  -1,   1;
  -1,  -5, -10, -10,  -5,  -1,   1;
  -1,  -6, -15, -20, -15,  -6,  -1,   1;
  -1,  -7, -21, -35, -35, -21,  -7,  -1,   1;
  -1,  -8, -28, -56, -70, -56, -28,  -8,  -1,   1; ...
This gives the recurrence and explains why the Bell numbers appear. (End)
Triangle A160185 = reversal then deletes right border of 1's. - Gary W. Adamson, May 03 2009

			Triangle begins (compare also table 9.2 in the Gould-Quaintance reference):
  1;
  1, 1;
  1, 1,  2;
  1, 1,  3,  5;
  1, 1,  4,  9,  15;
  1, 1,  5, 14,  31, 52;
  1, 1,  6, 20,  54, 121, 203;
  1, 1,  7, 27,  85, 233, 523,  877;
  1, 1,  8, 35, 125, 400,1101, 2469,  4140;
  1, 1,  9, 44, 175, 635,2046, 5625, 12611, 21147;
  1, 1, 10, 54, 236, 952,3488,11226, 30846, 69161, 115975;
  1, 1, 11, 65, 309,1366,5579,20425, 65676,180474, 404663, 678570;
  1, 1, 12, 77, 395,1893,8494,34685,126817,407787,1120666,2512769,4213597;

Alois P. Heinz, Rows n = 0..140, flattened
Robert Dougherty-Bliss, Gosper's algorithm and Bell numbers, arXiv:2210.13520 [cs.SC], 2022.
Robert Dougherty-Bliss, Experimental Methods in Number Theory and Combinatorics, Ph. D. Dissertation, Rutgers Univ. (2024). See pp. 69-70.
H. W. Gould and Jocelyn Quaintance, A linear binomial recurrence and the Bell numbers and polynomials, Applicable Analysis and Discrete Mathematics, 1 (2007), 371-385.

Diagonals, reading from the right, are A000110, A040027, A045501, A045499, A045500.
A124496 is a very similar triangle, obtained by reversing the rows and appending a rightmost diagonal which is A000110, the Bell numbers. See also A046936, A298804, A186020, A160185.
T(2n,n) gives A297924.

function Gould_diag(diag, size)
    size < 1 && return []
    size == 1 && return [1]
    L = [1, 1]
    accu = ones(BigInt, diag)
    for _ in 1:size-2
        accu = cumsum(vcat(accu[end], accu))
        L = vcat(L, accu[end])
    end
L end # Peter Luschny, Mar 30 2022

# This is the Jovovic formula with general index 'd'
# where A040027, A045499, etc. use one explicit integer
# Index n+1 is shifted to n from the original formula.
Gould := proc(n, d) local k;
    if n <= 1 then return 1 else
    return add(binomial(n-1+d, k+d)*Gould(k, d), k=0..n-1);
    fi
end:
# row and col refer to the extrapolated super-table:
# working up to row, not row-1, shows also the Bell numbers
# at the end of each row.
for row from 0 to 13 do
    for col from 0 to row do
       # 'diag' is constant for one of A040027, A045499 etc.
       diag := row - col;
       printf("%4d, ", Gould(col, diag));
    od;
    print();
od; # R. J. Mathar
# second Maple program:
T:= proc(n, k) option remember; `if`(k=0, 1,
      add(T(n-j, k-j)*binomial(n-1, j-1), j=1..k))
    end:
seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Jan 08 2018

g[n_ /; n <= 1, ] := 1; g[n, d_] := g[n, d] = Sum[ Binomial[n-1+d, k+d]*g[k, d], {k, 0, n-1}]; Flatten[ Table[ diag = row-col; g[col, diag], {row, 0, 13}, {col, 0, row}]] (* Jean-François Alcover, Nov 25 2011, after R. J. Mathar *)
T[n_, k_] := T[n, k] = If[k == 0, 1, Sum[T[n-j, k-j] Binomial[n-1, j-1], {j, 1, k}]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2018, after Alois P. Heinz *)

# Computes the n-th diagonal of the triangle reading from the right.
from itertools import accumulate
def Gould_diag(diag, size):
    if size < 1: return []
    if size == 1: return [1]
    L, accu = [1,1], [1]*diag
    for _ in range(size-2):
        accu = list(accumulate([accu[-1]] + accu))
        L.append(accu[-1])
    return L # Peter Luschny, Apr 24 2016

A124496 Triangle read by rows: T(n,k) is the number of set partitions of {1,2,...,n} in which the size of the last block is k, 1<=k<=n; the blocks are ordered with increasing least elements.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 9, 4, 1, 1, 31, 14, 5, 1, 1, 121, 54, 20, 6, 1, 1, 523, 233, 85, 27, 7, 1, 1, 2469, 1101, 400, 125, 35, 8, 1, 1, 12611, 5625, 2046, 635, 175, 44, 9, 1, 1, 69161, 30846, 11226, 3488, 952, 236, 54, 10, 1, 1, 404663, 180474, 65676, 20425, 5579, 1366, 309, 65, 11, 1, 1

Offset: 1

Emeric Deutsch, Nov 14 2006

Number of restricted growth functions of length n with a multiplicity k of the maximum value. RGF's are here defined as f(1)=1, f(i) <= 1+max_{1<=jR. J. Mathar, Mar 18 2016

This is table 9.2 in the Gould-Quaintance reference. - Peter Luschny, Apr 25 2016

			T(4,2) = 4 because we have 13|24, 14|23, 12|34 and 1|2|34.
Triangle starts:
  1;
  1,1;
  3,1,1;
  9,4,1,1;
  31,14,5,1,1;
  121,54,20,6,1,1;
  523,233,85,27,7,1,1;
  2469,1101,400,125,35,8,1,1;
  12611,5625,2046,635,175,44,9,1,1;
  69161,30846,11226,3488,952,236,54,10,1,1;
  404663,180474,65676,20425,5579,1366,309,65,11,1,1;
  2512769,1120666,407787,126817,34685,8494,1893,395,77,12,1,1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
H. W. Gould and Jocelyn Quaintance, A linear binomial recurrence and the Bell numbers and polynomials. Applicable Analysis and Discrete Mathematics, 1 (2007), 371-385.

Row sums are the Bell numbers (A000110). It seems that T(n, 1), T(n, 2), T(n, 3) and T(n, 4) are given by A040027, A045501, A045499 and A045500, respectively. A121207 gives a very similar triangle.
T(2n,n) gives A297924.
Cf. A000110, A040027, A045501, A045499, A045500, A186020, A160185.

Q[1]:=t*s: for n from 2 to 12 do Q[n]:=expand(t*s*subs(t=1,Q[n-1])+s*diff(Q[n-1],s)+t*Q[n-1]-Q[n-1]) od:for n from 1 to 12 do P[n]:=sort(subs(s=1,Q[n])) od: for n from 1 to 12 do seq(coeff(P[n],t,j),j=1..n) od;
# second Maple program:
T:= proc(n, k) option remember; `if`(n=k, 1,
      add(T(n-j, k)*binomial(n-1, j-1), j=1..n-k))
    end:
seq(seq(T(n, k), k=1..n), n=1..12);  # Alois P. Heinz, Jul 05 2016

T[n_, k_] := T[n, k] = If[n == k, 1, Sum[T[n-j, k]*Binomial[n-1, j-1], {j, 1, n-k}]];
Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten; (* Jean-François Alcover, Jul 21 2016, after Alois P. Heinz *)

The row enumerating polynomial P[n](t)=Q[n](t,1), where Q[1](t,s)=ts and Q[n](t,s)=s*dQ[n-1](t,s)/ds +(t-1)Q[n-1](t,s)+tsQ[n-1](1,s) for n>=2.
A008275^-1*ONES*A008275 or A008277*ONES*A008277^-1 where ONES is a triangle with all entries = 1. [From Gerald McGarvey, Aug 20 2009]
Conjectures: T(n,n-3) = A000096(n). T(n,n-4)= A055831(n+1). - R. J. Mathar, Mar 13 2016

A276961 Number of set partitions of [2n] with largest set of size n.

Original entry on oeis.org

1, 1, 9, 90, 1015, 12978, 187110, 3008148, 53275365, 1028142830, 21426984722, 478684639524, 11394222257054, 287518726261900, 7658231720886900, 214521099685649640, 6299407928673657135, 193373975592937777770, 6189939300880260745050, 206159811915115686404700

Offset: 0

Alois P. Heinz, Sep 22 2016

nonn

The blocks are ordered with increasing least elements.

a(0) = 1 by convention.

			a(1) = 1: 1|2.
a(2) = 9: 12|34, 12|3|4, 13|24, 13|2|4, 14|23, 1|23|4, 14|2|3, 1|24|3, 1|2|34.

Alois P. Heinz, Table of n, a(n) for n = 0..445
Wikipedia, Partition of a set

Cf. A000110, A080510, A276923, A297924, A297926, A327884, A328156.

b:= proc(n, k) option remember; `if`(n=0, 1, add(
      b(n-i, k)*binomial(n-1, i-1), i=1..min(n, k)))
    end:
a:= n-> `if`(n=0, 1, b(2*n, n)-b(2*n, n-1)):
seq(a(n), n=0..20);

b[n_, k_] := b[n, k] = If[n == 0, 1, Sum[b[n - i, k]*Binomial[n - 1, i - 1], {i, 1, Min[n, k]}]];
a[n_] := If[n == 0, 1, b[2*n, n] - b[2*n, n - 1]];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 20 2018, translated from Maple *)

a(n) = A080510(2n,n).
a(n) = A327884(2n,n).
a(n) = ceiling(C(2n,n)*(A000110(n)-1/2)). - Ludovic Schwob, Jan 15 2022

A297926 Number of set partitions of [2n] in which the size of the first block is n.

Original entry on oeis.org

1, 1, 6, 50, 525, 6552, 93786, 1504932, 26640900, 514083570, 10713538550, 239342496120, 5697111804566, 143759365731100, 3829115870472600, 107260549881604200, 3149703964487098665, 96686987797052290440, 3094969650442399156350, 103079905957566679518300

Offset: 0

Alois P. Heinz, Jan 08 2018

nonn

The blocks are ordered with increasing least elements.

a(0) = 1 by convention.

			a(1) = 1: 1|2.
a(2) = 6: 12|34, 12|3|4, 13|24, 13|2|4, 14|23, 14|2|3.

Alois P. Heinz, Table of n, a(n) for n = 0..445
Wikipedia, Partition of a set

Cf. A000110, A056857, A056860, A276961, A297924.

b:= proc(n) option remember; `if`(n=0, 1,
      add(b(n-j)*binomial(n-1, j-1), j=1..n))
    end:
a:= n-> binomial(2*n-1, n-1)*b(n):
seq(a(n), n=0..25);

b[n_] := b[n] = If[n == 0, 1, Sum[b[n-j]*Binomial[n-1, j-1], {j, 1, n}]];
a[n_] := Binomial[2*n-1, n-1] * b[n];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, May 20 2018, translated from Maple *)

a(n) = binomial(2*n-1,n-1) * Bell(n).

a(n) = A056857(2n,n) = A056860(2n,n).

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A121207 Triangle read by rows. The definition is by diagonals. The r-th diagonal from the right, for r >= 0, is given by b(0) = b(1) = 1; b(n+1) = Sum_{k=0..n} binomial(n+2,k+r)*a(k).

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A124496 Triangle read by rows: T(n,k) is the number of set partitions of {1,2,...,n} in which the size of the last block is k, 1<=k<=n; the blocks are ordered with increasing least elements.

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A276961 Number of set partitions of [2n] with largest set of size n.

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Maple

Mathematica

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A297926 Number of set partitions of [2n] in which the size of the first block is n.

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Programs

Maple

Mathematica

Formula