A126347 -id:A126347 - cp's OEIS frontend

A126348 Limit of reversed rows of triangle A126347, in which row sums equal Bell numbers (A000110).

1, 1, 2, 4, 7, 12, 20, 33, 53, 84, 131, 202, 308, 465, 695, 1030, 1514, 2209, 3201, 4609, 6596, 9386, 13284, 18705, 26211, 36561, 50776, 70226, 96742, 132765, 181540, 247369, 335940, 454756, 613689, 825698, 1107755, 1482038, 1977465, 2631664

Offset: 0

Paul D. Hanna, Dec 31 2006

nonn

In triangle A126347, row n lists coefficients of q in B(n,q) that satisfies: B(n,q) = Sum_{k=0..n-1} C(n-1,k)*B(k,q)*q^k for n>0, with B(0,q) = 1; row sums equal the Bell numbers: B(n,1) = A000110(n).

Row sums of A253830. a(n) equals the number of colored compositions of n, as defined in A253830, whose associated color partition has distinct parts. An example is given below. - Peter Bala, Jan 20 2015

			a(5) = 12: The colored compositions (defined in A253830) of 5 whose color partitions have distinct parts are
5(c1), 5(c2), 5(c3), 5(c4), 5(c5),
1(c1) + 4(c2), 1(c1) + 4(c3), 1(c1) + 4(c4),
3(c1) + 2(c2),
2(c1) + 3(c2), 2(c1) + 3(c3), 2(c2) + 3(c3). - _Peter Bala_, Jan 20 2015

Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..500 from Seiichi Manyama)

Cf. A126347, A126349; factorial variant: A126471. A253830, A307599, A307601, A307602.

nmax = 50; CoefficientList[Series[Product[(1 - x + x^k)/(1 - x), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 16 2019 *)

{B(n,q)=if(n==0,1,sum(k=0,n-1,binomial(n-1,k)*B(k,q)*q^k))}
{a(n)=Vec(B(n+1,'q)+O('q^(n*(n-1)/2+1)))[n*(n-1)/2+1]}

{a(n) = local(t); if( n<0, 0, t = 1; polcoeff( sum(k=1, (sqrtint(8*n + 1) - 1)\2, t = t * x^k / (1 - x) / (1 - x^k) + x * O(x^n), 1), n))} /* Michael Somos, Aug 17 2008 */

1 + Sum_{k>0} x^(k * (k + 1) / 2) / ((1 - x)^k * (1 - x) * (1 - x^2) ... (1 - x^k)). - Michael Somos, Aug 17 2008
G.f.: Product_{k>0} (1+x^k/(1-x)). - Vladeta Jovovic, Oct 05 2008
G.f.: exp(Sum_{k>=1} x^k * Sum_{d|k} (-1)^(d+1)/(d*(1 - x)^d)). - Ilya Gutkovskiy, Apr 19 2019

A126349 Largest term in rows of triangle A126347, in which row sums equal Bell numbers (A000110).

Original entry on oeis.org

1, 1, 1, 2, 4, 10, 29, 101, 392, 1721, 8180, 42234, 232893, 1376660, 8608564, 56880860, 395780193, 2887365769, 22022011448, 175148070545, 1451626482840, 12489021006783, 111407399288300, 1028527130941484, 9818171726308337

Offset: 0

Paul D. Hanna, Dec 31 2006

nonn

In triangle A126347, row n lists coefficients of q in B(n,q) that satisfies: B(n,q) = Sum_{k=0..n-1} C(n-1,k)*B(k,q)*q^k for n>0, with B(0,q) = 1.

Cf. A126347, A126348.

PARI

A126443 a(n) = Sum_{k=0..n-1} C(n-1,k)a(k)2^k for n>0, with a(0)=1.

Original entry on oeis.org

1, 1, 3, 17, 179, 3489, 127459, 8873137, 1195313043, 315321098561, 164239990789571, 169810102632595281, 349630019758589841523, 1436268949679165936016097, 11784559509424676876673518499, 193243076262167105764611875139569

Offset: 0

Paul D. Hanna, Jan 01 2007

nonn

Generated by a generalization of a recurrence for the Bell numbers (A000110).

Starting with offset 1 = eigensequence of triangle A013609. - Gary W. Adamson, Sep 04 2009

Seiichi Manyama, Table of n, a(n) for n = 0..81

Cf. A126347, A000110.
Cf. A013609. - Gary W. Adamson, Sep 04 2009
Column k=2 of A306245.

a(n)=if(n==0,1,sum(k=0,n-1,binomial(n-1,k)*a(k)*2^k))

a(n) = Sum_{k=0..n*(n-1)/2} A126347(n,k)*2^k.
G.f. A(x) satisfies: A(x) = 1 + x*A(2*x/(1 - x))/(1 - x). - Ilya Gutkovskiy, Sep 02 2019
a(n) ~ c * 2^(n*(n-1)/2), where c = A081845 = 4.7684620580627434482997985... - Vaclav Kotesovec, Sep 16 2019

A126470 Triangle, read by rows, where row n lists coefficients of q in F(n,q) that satisfies: F(n,q) = Sum_{k=0..n-1} C(n-1,k)F(k,q)F(n-k-1,q)*q^k for n>0, with F(0,q) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 6, 7, 5, 3, 1, 1, 1, 10, 25, 25, 26, 11, 12, 5, 3, 1, 1, 1, 15, 65, 110, 136, 117, 92, 70, 43, 32, 17, 12, 5, 3, 1, 1, 1, 21, 140, 385, 616, 784, 694, 687, 478, 411, 255, 222, 127, 91, 50, 39, 17, 12, 5, 3, 1, 1, 1, 28, 266, 1106, 2471, 4032, 4887, 5189

Offset: 0

Paul D. Hanna, Dec 31 2006

Row sums equal the factorials: F(n,1) = n!.

Limit of reversed rows equals A126471. Largest term in rows equal A126472.

			Number of terms in row n is: n*(n-1)/2 + 1.
Row functions B(n,q) begin:
F(0,q) = F(1,q) = 1;
F(2,q) = 1 + q;
F(3,q) = 1 + 3*q + q^2 + q^3;
F(4,q) = 1 + 6*q + 7*q^2 + 5*q^3 + 3*q^4 + q^5 + q^6.
Triangle begins:
1;
1;
1, 1;
1, 3, 1, 1;
1, 6, 7, 5, 3, 1, 1;
1, 10, 25, 25, 26, 11, 12, 5, 3, 1, 1;
1, 15, 65, 110, 136, 117, 92, 70, 43, 32, 17, 12, 5, 3, 1, 1;
1, 21, 140, 385, 616, 784, 694, 687, 478, 411, 255, 222, 127, 91, 50, 39, 17, 12, 5, 3, 1, 1;
1, 28, 266, 1106, 2471, 4032, 4887, 5189, 4832, 4240, 3426, 2658, 2143, 1534, 1143, 790, 575, 351, 262, 151, 99, 58, 39, 17, 12, 5, 3, 1, 1;
1, 36, 462, 2730, 8589, 17892, 28519, 35613, 40639, 39200, 37934, 31508, 28076, 21570, 18288, 13451, 11009, 7747, 6120, 4089, 3106, 2056, 1530, 943, 683, 396, 289, 160, 108, 58, 39, 17, 12, 5, 3, 1, 1;
1, 45, 750, 6000, 25977, 70497, 141499, 220500, 291877, 336945, 357638, 347396, 323795, 288162, 247473, 207630, 170336, 139565, 109967, 87581, 66534, 51411, 37845, 28948, 20626, 15284, 10727, 7810, 5169, 3731, 2446, 1700, 1063, 733, 426, 299, 170, 108, 58, 39, 17, 12, 5, 3, 1, 1;
...
E.g.f.: A(x,q) = 1 + x + x^2*(1+q)/2! + x^3*(1+3*q+q^2+q^3)/3! +...
where A(x,q) = exp( Integral A(q*x,q) dx ),
A(q*x,q) = exp( q * Integral A(q^2*x,q) dx ),
A(q^2*x,q) = exp( q^2 * Integral A(q^3*x,q) dx ), ...
A(q^n*x,q) = exp( q^n * Integral A(q^(n+1)*x,q) dx ) for n>=0.
Here the Integral is always in the limits 0..x.

Paul D. Hanna, Rows n = 0..40, flattened.
Miguel A. Mendez, Combinatorial differential operators in: Faà di Bruno formula, enumeration of ballot paths, enriched rooted trees and increasing rooted trees, arXiv:1610.03602 [math.CO], 2016.

Cf. A126471, A126472; Bell number variant: A126347.

F[0, ] = 1; F[n, q_] := F[n, q] = Sum[Binomial[n-1, k] F[k, q] F[n-k-1, q] q^k, {k, 0, n-1}];
row[n_] := CoefficientList[F[n, q], q];
Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jul 26 2018 *)

F(n,q)=if(n==0,1,sum(k=0,n-1,binomial(n-1,k)*F(k,q)*F(n-k-1,q)*q^k))
{T(n,k)=Vec(F(n,q)+O(q^(n*(n-1)/2+1)))[k+1]}
for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print(""))

T(n,k)=local(A=vector(n+2, j, 1+j*x)); for(i=0, n+1, for(j=0, n, m=n+1-j; A[m]=exp(q^(m-1)*intformal(A[m+1]+x*O(x^n))))); polcoeff(n!*polcoeff(A[1], n, x),k,q) \\ From Paul D. Hanna, Oct 04 2008
for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print(""))

/* Faster to use: A(x,q) = 1 + Integral A(x,q)*A(qx,q) dx */
{T(n,k)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+intformal(A*subst(A,x,q*x))); polcoeff(n!*polcoeff(A,n,x),k,q)}
for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Oct 04 2008

From Paul D. Hanna, Oct 04 2008: (Start)
E.g.f. satisfies: A(x,q) = exp( Integral A(q*x,q) dx ); further,
A(q^n*x,q) = exp( q^n * Integral A(q^(n+1)*x,q) dx ) for n>=0, where A(x,q) = Sum_{n>=0} x^n*[Sum_{k=0..n(n-1)/2} T(n,k)*q^k]/n!. (End)
E.g.f. satisfies: d/dx A(x,q) = A(x,q) * A(q*x,q) with A(0,q)=1; i.e., the logarithmic derivative of A(x,q) with respect to x equals A(q*x,q). - Paul D. Hanna, Oct 04 2008

A346772 Total sum of block indices of the elements over all partitions of [n].

Original entry on oeis.org

0, 1, 5, 22, 100, 482, 2475, 13527, 78476, 481687, 3117962, 21218851, 151387882, 1129430737, 8790433999, 71222812912, 599577147056, 5235054113412, 47331036294905, 442462325254995, 4270909302907430, 42514043248222709, 435920900603529954, 4599155199953703373

Offset: 0

Alois P. Heinz, Aug 02 2021

nonn

			a(3) = 22 = 3 + 4 + 4 + 5 + 6, summing block indices 111, 112, 121, 122, 123 of the 5 partitions of [3]: 123, 12|3, 13|2, 1|23, 1|2|3.

Alois P. Heinz, Table of n, a(n) for n = 0..574
Wikipedia, Partition of a set

Row sums of A120057.

Cf. A000110, A070071, A105488, A126347, A270236.

b:= proc(n, m) option remember; `if`(n=0, [1, 0], add(
     (p-> p+[0, p[1]*j])(b(n-1, max(m, j))), j=1..m+1))
    end:
a:= n-> b(n, 0)[2]:
seq(a(n), n=0..25);

b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[
     Function[p, p+{0, p[[1]]*j}][b[n-1, Max[m, j]]], {j, 1, m+1}]];
a[n_] := b[n, 0][[2]];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 27 2022, after Alois P. Heinz *)

a(n) = Sum_{k=1..n} A120057(n,k).
a(n) = Sum_{k=0..n*(n-1)/2} (n+k) * A126347(n,k).
a(n) = Sum_{k=1..n} k * A270236(n,k).

A383253 Number of compositions of n with parts in standard order.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 9, 16, 29, 53, 98, 182, 340, 638, 1202, 2273, 4312, 8204, 15650, 29925, 57344, 110101, 211771, 407987, 787174, 1520851, 2942030, 5697842, 11046881, 21438881, 41645541, 80967881, 157547508, 306791828, 597847686, 1165828440, 2274890125

Offset: 0

John Tyler Rascoe, May 06 2025

A composition with parts in standard order satisfies the condition that for any part p > 1, the part p - 1 has already appeared. All compositions of this kind have first part 1.

			a(6) = 9 counts: (1,1,1,1,1,1), (1,1,1,1,2), (1,1,1,2,1), (1,1,2,1,1), (1,2,1,1,1), (1,1,2,2), (1,2,1,2), (1,2,2,1), (1,2,3).

Alois P. Heinz, Table of n, a(n) for n = 0..3333

Cf. A000110, A011782, A047998, A107429, A126347, A278984, (row sums of A383713).

b:= proc(n, i) option remember; `if`(n=0, 1, add(
      b(n-j, max(i, j)), j=1..min(n, i+1)))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..36);  # Alois P. Heinz, May 08 2025

A_x(N) = {my(x='x+O('x^(N+1))); Vec(1 + sum(i=1,(N/2)+1, x^(i*(i+1)/2)/prod(j=1,i, 1 - (x-x^(j+1))/(1-x))))}
A_x(40)

G.f.: 1 + Sum_{i>0} x^(i*(i+1)/2) / Product_{j=1..i} 1 - (x - x^(j+1))/(1 - x).

A380822 Triangle read by rows: T(n,k) is the number of compositions of n with k pairs of equal adjacent parts and all parts in standard order.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 3, 1, 0, 1, 2, 1, 4, 1, 0, 1, 3, 3, 3, 5, 1, 0, 1, 2, 10, 5, 4, 6, 1, 0, 1, 5, 9, 17, 8, 5, 7, 1, 0, 1, 8, 16, 22, 26, 10, 6, 8, 1, 0, 1, 10, 35, 33, 37, 37, 12, 7, 9, 1, 0, 1, 19, 44, 80, 59, 56, 48, 14, 8, 10, 1, 0, 1

Offset: 1

John Tyler Rascoe, May 08 2025

A composition with parts in standard order satisfies the condition that for any part p > 1, the part p - 1 has already appeared. All compositions of this kind have first part 1.

			Triangle begins:
     k=0   1   2   3   4  5  6  7  8  9
 n=1  [1],
 n=2  [0,  1],
 n=3  [1,  0,  1],
 n=4  [1,  1,  0,  1],
 n=5  [0,  3,  1,  0,  1],
 n=6  [2,  1,  4,  1,  0, 1],
 n=7  [3,  3,  3,  5,  1, 0, 1],
 n=8  [2, 10,  5,  4,  6, 1, 0, 1],
 n=9  [5,  9, 17,  8,  5, 7, 1, 0, 1],
 n=10 [8, 16, 22, 26, 10, 6, 8, 1, 0, 1],
...
Row n = 6 counts:
 T(6,0) = 2: (1,2,1,2), (1,2,3).
 T(6,1) = 1: (1,2,2,1).
 T(6,2) = 4: (1,1,1,2,1), (1,1,2,1,1), (1,1,2,2), (1,2,1,1,1).
 T(6,3) = 1: (1,1,1,1,2).
 T(6,4) = 0: .
 T(6,5) = 1: (1,1,1,1,1,1).

Alois P. Heinz, Rows n = 1..200, flattened

Cf. A000110, A003242, A047998, A106356, A107429, A126347, A278984, A383253 (row sums).

b:= proc(n, l, i) option remember; expand(`if`(n=0, 1, add(
      `if`(j=l, x, 1)*b(n-j, j, max(i, j)), j=1..min(n, i+1))))
    end:
T:= (n, k)-> coeff(b(n, 0$2), x, k):
seq(seq(T(n, k), k=0..n-1), n=1..12);  # Alois P. Heinz, May 08 2025

G(k,N) = {my(x='x+O('x^N)); if(k<1,1, G(k-1,N) * x^k * (1 + (1/(1 - sum(j=1,k, x^j/(1-(z-1)*x^j)))) * sum(j=1,k, z^(if(j==k,1,0)) * x^j/(1-(z-1)*x^j))))}
T_xz(max_row) = {my(N = max_row+1, x='x+O('x^N), h = sum(i=1,N/2+1, G(i,N))); vector(N-1, n, Vecrev(polcoeff(h, n)))}
T_xz(10)

G.f.: Sum_{i>0} G(i) where G(k) = G(k-1) * x*k * (1 + 1/(1 - Sum_{j=1..k} ( x^j/(1 - (z-1)*x^j) )) * Sum_{j=1..k} ( z^[j=k] * x^j/(1 - (z-1)*x^j) )) and G(0) = 1.

A383713 Triangle read by rows: T(n,k) is the number of compositions of n with k parts all in standard order.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 1, 0, 0, 0, 1, 3, 1, 0, 0, 0, 1, 3, 4, 1, 0, 0, 0, 0, 4, 6, 5, 1, 0, 0, 0, 0, 2, 10, 10, 6, 1, 0, 0, 0, 0, 1, 9, 20, 15, 7, 1, 0, 0, 0, 0, 1, 7, 25, 35, 21, 8, 1, 0, 0, 0, 0, 0, 7, 26, 55, 56, 28, 9, 1, 0, 0, 0, 0, 0, 4, 29, 71, 105, 84, 36, 10, 1

Offset: 0

John Tyler Rascoe, May 06 2025

A composition with parts in standard order satisfies the condition that for any part p > 1, the part p - 1 has already appeared. All compositions of this kind have first part 1.

			Triangle begins:
     k=0  1  2  3  4   5   6   7   8  9 10
 n=0  [1],
 n=1  [0, 1],
 n=2  [0, 0, 1],
 n=3  [0, 0, 1, 1],
 n=4  [0, 0, 0, 2, 1],
 n=5  [0, 0, 0, 1, 3,  1],
 n=6  [0, 0, 0, 1, 3,  4,  1],
 n=7  [0, 0, 0, 0, 4,  6,  5,  1],
 n=8  [0, 0, 0, 0, 2, 10, 10,  6,  1],
 n=9  [0, 0, 0, 0, 1,  9, 20, 15,  7, 1],
 n=10 [0, 0, 0, 0, 1,  7, 25, 35, 21, 8, 1],
 ...
Row n = 6 counts:
 T(6,3) = 1: (1,2,3).
 T(6,4) = 3: (1,1,2,2), (1,2,1,2), (1,2,2,1).
 T(6,5) = 4: (1,1,1,1,2), (1,1,1,2,1), (1,1,2,1,1), (1,2,1,1,1).
 T(6,6) = 1: (1,1,1,1,1,1).

Cf. A000110 (column sums), A047998, A107429, A126347 (triangle transposed with no zeros), A278984, A383253 (row sums).

T_xy(max_row) = {my(N = max_row+1, x='x+O('x^N), h = 1 + sum(i=1,1+(N/2), y^i * x^(i*(i+1)/2)/prod(j=1,i, 1 - y*(x-x^(j+1))/(1-x)))); vector(N, n, Vecrev(polcoeff(h, n-1)))}
T_xy(10)

G.f.: 1 + Sum_{i>0} y^i * x^(i*(i+1)/2) / Product_{j=1..i} 1 - y*(x - x^(j+1))/(1 - x).

A383751 Number of Carlitz compositions of n with parts in standard order.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 2, 3, 2, 5, 8, 10, 19, 31, 44, 73, 123, 193, 315, 524, 847, 1392, 2317, 3810, 6303, 10506, 17451, 29066, 48603, 81223, 135965, 228153, 383014, 643756, 1083693, 1825640, 3078574, 5197246, 8780823, 14847669, 25128385, 42558687, 72131730, 122343844

Offset: 0

John Tyler Rascoe, May 08 2025

A composition with parts in standard order satisfies the condition that for any part p > 1, the part p - 1 has already appeared. All compositions of this kind have first part 1.

			a(9) = 5 counts: (1,2,1,2,1,2), (1,2,1,2,3), (1,2,1,3,2), (1,2,3,1,2), (1,2,3,2,1).

Alois P. Heinz, Table of n, a(n) for n = 0..4135

Cf. A000110, A003242, A011782, A047998, A107429, A126347, A278984, (column k=0 of A380822), A383253, A383713.

b:= proc(n, l, i) option remember; `if`(n=0, 1, add(
      `if`(j=l, 0, b(n-j, j, max(i, j))), j=1..min(n, i+1)))
    end:
a:= n-> b(n, 0$2):
seq(a(n), n=0..43);  # Alois P. Heinz, May 09 2025

G(k,N) = {my(x='x+O('x^N)); if(k<1,1, G(k-1,N) * x^k * (1 + (1/(1 - sum(j=1,k, x^j/(1+x^j)))) * sum(j=1,k-1, x^j/(1+x^j))))}
A_x(N) = {my(x='x+O('x^N)); Vec(sum(i=0,N/2+1, G(i,N+1)))}
A_x(50)

G.f.: Sum_{i>=0} G(i) where G(k) = G(k-1) * x*k * (1 + 1/(1 - Sum_{j=1..k} ( x^j/(1+x^j) )) * Sum_{j=1..k-1} ( x^j/(1+x^j) )) and G(0) = 1.

cp's OEIS Frontend

A126348 Limit of reversed rows of triangle A126347, in which row sums equal Bell numbers (A000110).

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A126349 Largest term in rows of triangle A126347, in which row sums equal Bell numbers (A000110).

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A126443 a(n) = Sum_{k=0..n-1} C(n-1,k)*a(k)*2^k for n>0, with a(0)=1.

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A126470 Triangle, read by rows, where row n lists coefficients of q in F(n,q) that satisfies: F(n,q) = Sum_{k=0..n-1} C(n-1,k)*F(k,q)*F(n-k-1,q)*q^k for n>0, with F(0,q) = 1.

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A346772 Total sum of block indices of the elements over all partitions of [n].

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Maple

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A383253 Number of compositions of n with parts in standard order.

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A380822 Triangle read by rows: T(n,k) is the number of compositions of n with k pairs of equal adjacent parts and all parts in standard order.

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PARI

A126443 a(n) = Sum_{k=0..n-1} C(n-1,k)a(k)2^k for n>0, with a(0)=1.

A126470 Triangle, read by rows, where row n lists coefficients of q in F(n,q) that satisfies: F(n,q) = Sum_{k=0..n-1} C(n-1,k)F(k,q)F(n-k-1,q)*q^k for n>0, with F(0,q) = 1.