A048775 -id:A048775 - cp's OEIS frontend

A163765 Inverse binomial transform of A048775 (assuming offset zero in both sequences).

1, 6, 18, 48, 131, 363, 1017, 2873, 8169, 23349, 67024, 193086, 557949, 1616501, 4694034, 13657896, 39809649, 116218701, 339762942, 994553160, 2914608177, 8550424953, 25107964077, 73793368593, 217057617567, 638936722403, 1882096946232, 5547613247418

Offset: 1

Gary W. Adamson, Aug 03 2009

nonn

			a(4) = 48 = (-1, 3, -3, 1) dot (1, 7, 31, 121).

A048775

A007318^(-1) * A048775, where A048775 = (1, 7, 31, 121, 456, 1709,...).

A305161 Number A(n,k) of compositions of n into exactly n nonnegative parts <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 3, 1, 0, 1, 1, 3, 7, 1, 0, 1, 1, 3, 10, 19, 1, 0, 1, 1, 3, 10, 31, 51, 1, 0, 1, 1, 3, 10, 35, 101, 141, 1, 0, 1, 1, 3, 10, 35, 121, 336, 393, 1, 0, 1, 1, 3, 10, 35, 126, 426, 1128, 1107, 1, 0, 1, 1, 3, 10, 35, 126, 456, 1520, 3823, 3139, 1, 0

Offset: 0

Alois P. Heinz, Aug 17 2018

			A(3,1) = 1: 111.
A(3,2) = 7: 012, 021, 102, 111, 120, 201, 210.
A(3,3) = 10: 003, 012, 021, 030, 102, 111, 120, 201, 210, 300.
A(4,2) = 19: 0022, 0112, 0121, 0202, 0211, 0220, 1012, 1021, 1102, 1111, 1120, 1201, 1210, 2002, 2011, 2020, 2101, 2110, 2200.
A(4,3) = 31: 0013, 0022, 0031, 0103, 0112, 0121, 0130, 0202, 0211, 0220, 0301, 0310, 1003, 1012, 1021, 1030, 1102, 1111, 1120, 1201, 1210, 1300, 2002, 2011, 2020, 2101, 2110, 2200, 3001, 3010, 3100.
Square array A(n,k) begins:
  1, 1,    1,    1,    1,    1,    1,    1,    1, ...
  0, 1,    1,    1,    1,    1,    1,    1,    1, ...
  0, 1,    3,    3,    3,    3,    3,    3,    3, ...
  0, 1,    7,   10,   10,   10,   10,   10,   10, ...
  0, 1,   19,   31,   35,   35,   35,   35,   35, ...
  0, 1,   51,  101,  121,  126,  126,  126,  126, ...
  0, 1,  141,  336,  426,  456,  462,  462,  462, ...
  0, 1,  393, 1128, 1520, 1667, 1709, 1716, 1716, ...
  0, 1, 1107, 3823, 5475, 6147, 6371, 6427, 6435, ...

Alois P. Heinz, Antidiagonals n = 0..200

Columns k=0-10 give: A000007, A000012, A002426, A005725, A187925, A318113, A318114, A318115, A318116, A167403, A318117.
Rows n=0-1 give: A000012, A057427.
Main diagonal gives A088218 or A001700(n-1) for n>0.
A(n+1,n) gives A048775.
Cf. A180281.

A:= (n, k)-> coeff(series(((x^(k+1)-1)/(x-1))^n, x, n+1), x, n):
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
b:= proc(n, i, k) option remember; `if`(n=0, 1,
      `if`(i=0, 0, add(b(n-j, i-1, k), j=0..min(n, k))))
    end:
A:= (n, k)-> b(n$2, k):
seq(seq(A(n, d-n), n=0..d), d=0..12);

b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i == 0, 0, Sum[b[n - j, i - 1, k], {j, 0, Min[n, k]}]]];
A[n_, k_] := b[n, n, k];
Table[A[n, d - n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 05 2019, after Alois P. Heinz *)

A(n,k) = [x^n] ((x^(k+1)-1)/(x-1))^n.
A(n,k) - A(n,k-1) = A180281(n,k) for n,k > 0.
A(n,k) = A(n,n) for all k >= n.

A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)Sum_{j=0..n} binomial(n + nj + j, nj + j)x^j.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 31, 31, 1, 1, 121, 381, 121, 1, 1, 456, 3431, 3431, 456, 1, 1, 1709, 26769, 60691, 26769, 1709, 1, 1, 6427, 193705, 848443, 848443, 193705, 6427, 1, 1, 24301, 1343521, 10350421, 19610233, 10350421, 1343521, 24301, 1

Offset: 0

Paul D. Hanna, May 29 2005 and May 31 2005

G.f. of row n divided by (1-x)^(n+1) equals g.f. of row n of table A060543.
Matrix product of this triangle with Pascal's triangle (A007318) equals A108291.
Seeing each row as a polynomial, all roots seem to be negative reals. - F. Chapoton, Nov 01 2022
From Thomas Anton, Jan 05 2023: (Start)
Consider the set [m] := {1, 2, 3, ..., m} ordered cyclically, and then mapped into itself via f. Let us consider a in [m] as the (a-1)th m-th root of unity e^(2*Pi*i*(a-1)/m). Then f may be extended to a continuous map f':S^1 -> S^1 as follows:
For a immediately before b in the cyclic order, map the interval between a and b to S^1 so that a point in it moving clockwise at constant speed has a value moving clockwise at constant speed, and the map travels the shortest distance possible given this condition.
T(n, k) gives the number of f for m = n-1 such that f(1) = 1 and f' has degree k. This is trivially one n-th of the number of f with degree k when f(1) is arbitrary.
Equivalent to having degree k is that there are k values a immediately before b in the cyclic order such that f(a) > f(b) (in the standard order of N).
If we change things so that a immediately before b satisfies f(a) = f(b) corresponds to a full rotation (this is equivalent to using the condition f(a) >= f(b) in the last paragraph), then T(n, k) is the number of f with degree k+1.
T(n, k) is the (k+1)*(n-1)th (n-1)-nomial coefficient of power n - 1.
(End)

			Triangle begins:
  1;
  1,    1;
  1,    7,      1;
  1,   31,     31,      1;
  1,  121,    381,    121,      1;
  1,  456,   3431,   3431,    456,      1;
  1, 1709,  26769,  60691,  26769,   1709,    1;
  1, 6427, 193705, 848443, 848443, 193705, 6427, 1;
  ...
G.f. of row 3: (1 + 31*x + 31*x^2 + x^3) = (1-x)^4*(1 + 35*x + 165*x^2 + 455*x^3 + ... + C(4*j+3,4*j)*x^j + ...).

M. Bayer, B. Goeckner, S. J. Hong, T. McAllister, M. Olsen, C. Pinckney, J. Vega and M. Yip, Lattice polytopes from Schur and symmetric Grothendieck polynomials, Electronic Journal of Combinatorics, Volume 28, Issue 2 (2021). See Proposition 53 and Table 1.
Tanay Wakhare, Iterated Entropy Derivatives and Binary Entropy Inequalities, arXiv:2312.14743 [cs.IT], 2023.
Tanay Wakhare, Two Studies of Constraints in High Dimensions: Entropy Inequalities and the Randomized Symmetric Binary Perceptron, Master's Thesis, MIT (2024). See p. 22.
Raphael Yuster, Almost k-union closed set systems, arXiv:2302.12276 [math.CO], 2023, p. 8.

Cf. A108267, A000169, A048775, A060543, A108291, A108292, A056856.

p := n -> (1-x)^(n+1)*add(binomial(n + n*j + j, n*j + j)*x^j, j = 0..n):
seq(print(seq(coeff(p(n), x, k), k = 0..n)), n = 0..8); # Peter Luschny, Nov 02 2022

T[n_, k_] := Coefficient[(1 - x)^(n + 1)*
     Sum[Binomial[n + n*j + j, n*j + j]*x^j, {j, 0, n}], x, k];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 23 2021 *)

T(n,k)=polcoeff((1-x)^(n+1)*sum(j=0,n,binomial(n+n*j+j,n*j+j)*x^j),k)

T(n, 1) = A048775(n) = binomial(2*n + 1, n + 1) - (n + 1).
Sum_{k=0..n} T(n, k) = A000169(n) = (n + 1)^n.
Sum_{k=0..n} T(n, k)*2^k = A108292(n).
From Thomas Anton, Jan 05 2023: (Start)
T(n, k) = Sum_{i=0..k} (-1)^i*binomial(n + 1, i)*binomial(n+(n+1)*(k-i), n).
T(n, k) = T(n, n-k).
(End)

A222555 T(n,k)=Number of length (n+k)X1 arrays of occupancy after each element moves up to +-k places but not 0.

Original entry on oeis.org

1, 7, 2, 31, 20, 4, 121, 102, 64, 8, 456, 427, 359, 208, 16, 1709, 1668, 1562, 1279, 651, 32, 6427, 6372, 6220, 5761, 4537, 2056, 64, 24301, 24230, 24024, 23341, 21239, 15929, 6496, 128, 92368, 92279, 92011, 91052, 87752, 78245, 56041, 20483, 256, 352705

Offset: 1

R. H. Hardin Feb 24 2013

Table starts
...1.....7.....31....121....456...1709...6427..24301..92368.352705
...2....20....102....427...1668...6372..24230..92279.352596
...4....64....359...1562...6220..24024..92011.352258
...8...208...1279...5761..23341..91052.350967
..16...651...4537..21239..87752.346091
..32..2056..15929..78245.330274
..64..6496..56041.287858
.128.20483.197313
.256.64627
.512

			Some solutions for n=3 k=4
..0....1....3....1....0....1....0....2....0....2....0....0....2....0....1....0
..2....0....0....0....1....2....0....2....3....2....1....0....0....1....0....1
..1....1....0....0....0....1....3....2....0....3....0....2....1....2....3....0
..1....0....1....4....1....1....1....1....1....0....0....2....0....0....3....0
..0....2....1....0....1....2....0....0....0....0....0....0....3....4....0....1
..3....2....1....1....0....0....1....0....1....0....4....2....0....0....0....4
..0....1....1....1....4....0....2....0....2....0....2....1....1....0....0....1

R. H. Hardin, Table of n, a(n) for n = 1..55

Column 1 is A000079(n-1)

Row 1 is A048775

A144510 Array T(n,k) (n >= 1, k >= 0) read by downwards antidiagonals: T(n,k) = total number of partitions of [1, 2, ..., i] into exactly k nonempty blocks, each of size at most n, for any i in the range n <= i <= k*n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 7, 3, 1, 1, 37, 31, 4, 1, 1, 266, 842, 121, 5, 1, 1, 2431, 45296, 18252, 456, 6, 1, 1, 27007, 4061871, 7958726, 405408, 1709, 7, 1, 1, 353522, 546809243, 7528988476, 1495388159, 9268549, 6427, 8, 1

Offset: 1

David Applegate and N. J. A. Sloane, Dec 15 2008, Jan 30 2009

			Array begins:
1, 1,    1,       1,            1,                 1,                       1, ...
1, 2,    7,      37,          266,              2431,                   27007, ...
1, 3,   31,     842,        45296,           4061871,               546809243, ...
1, 4,  121,   18252,      7958726,        7528988476,          13130817809439, ...
1, 5,  456,  405408,   1495388159,    15467641899285,      361207016885536095, ...
1, 6, 1709, 9268549, 295887993624, 34155922905682979, 10893033763705794846727, ...
...

Seiichi Manyama, Antidiagonals n = 1..50, flattened

For the transposed array see A144512.
Rows include A001515, A144416, A144508, A144509.
Columns include A048775, A144511.
A(n+1,n) gives A281901.
A(n,n) gives A308296.
Cf. A308292.

b := proc(n, i, k) local r;
option remember;
if n = i then 1;
elif i < n then 0;
elif n < 1 then 0;
else add( binomial(i-1,r)*b(n-1,i-1-r,k), r=0..k);
end if;
end proc;
T:=proc(n,k); add(b(n,i,k),i=0..(k+1)*n); end proc;
# Peter Luschny, Apr 26 2011
A144510 := proc(n, k) local m;
add(m!*coeff(expand((exp(x)*GAMMA(n+1,x)/GAMMA(n+1)-1)^k),x,m),m=k..k*n)/k! end: for row from 1 to 6 do seq(A144510(row, col), col = 0..5) od;

multinomial[n_, k_List] := n!/Times @@ (k!); t[n_, k_] := Module[{i, ik}, ik = Array[i, k]; 1/k!* Sum[multinomial[Total[ik], ik], Evaluate[Sequence @@ Thread[{ik, 1, n}]]]]; Table[t[n-k, k], {n, 1, 10}, {k, n-1, 0, -1}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)

T(n,k) = (1/k!)*Sum_{i_1=1..n} Sum_{i_2=1..n} ... Sum_{i_k=1..n} multinomial(i_1+i_2+...+i_k; i_1, i_2, ..., i_k).

T(n,k) = (1/k!)*Sum_{m=k..k*n} m! [x^m](e^x Gamma(n+1,x)/Gamma(n+1)-1)^k. Here [x^m]f(x) is the coefficient of x^m in the series expansion of f(x). - Peter Luschny, Apr 26 2011

A144512 Array read by upwards antidiagonals: T(n,k) = total number of partitions of [1, 2, ..., k] into exactly n blocks, each of size 1, 2, ..., k+1, for 0 <= k <= (k+1)*n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 7, 1, 1, 4, 31, 37, 1, 1, 5, 121, 842, 266, 1, 1, 6, 456, 18252, 45296, 2431, 1, 1, 7, 1709, 405408, 7958726, 4061871, 27007, 1, 1, 8, 6427, 9268549, 1495388159, 7528988476, 546809243, 353522, 1, 1, 9, 24301, 216864652, 295887993624, 15467641899285

Offset: 0

David Applegate and N. J. A. Sloane, Dec 15 2008, Dec 21 2008

			Array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 2, 7, 37, 266, 2431, 27007, 353522, 5329837, ...
1, 3, 31, 842, 45296, 4061871, 546809243, 103123135501, ...
1, 4, 121, 18252, 7958726, 7528988476, 13130817809439, ...
1, 5, 456, 405408, 1495388159, 15467641899285, 361207016885536095, ...
1, 6, 1709, 9268549, 295887993624, 34155922905682979, 10893033763705794846727, ...
...

Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394, 2017.
David Applegate and N. J. A. Sloane, The Gift Exchange Problem (arXiv:0907.0513, 2009)

See A144510 for Maple code.
Rows include A001515, A144416, A144508, A144509, A149187.
Columns include A048775, A144511, A144662, A147984.
Transpose of array in A144510.
Main diagonal gives A281901.

b := proc(n, i, k) local r;
option remember;
if n = i then 1;
elif i < n then 0;
elif n < 1 then 0;
else add( binomial(i-1,r)*b(n-1,i-1-r,k), r=0..k);
end if;
end proc;
T:=proc(n,k); add(b(n,i,k),i=0..(k+1)*n); end proc;

multinomial[n_, k_List] := n!/Times @@ (k!); t[n_, k_] := Module[{i, ik}, ik = Array[i, k]; 1/k!* Sum[multinomial[Total[ik], ik], Evaluate[Sequence @@ Thread[{ik, 1, n}]]]]; Table[t[n-k, k], {n, 1, 10}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)

A307349 a(n) = Sum_{i=1..n} Sum_{j=1..n} (-1)^(i+j) * (i+j)!/(2!i!j!).

Original entry on oeis.org

0, 1, 1, 5, 15, 56, 203, 757, 2839, 10736, 40821, 155948, 598065, 2301118, 8878591, 34340085, 133100055, 516851528, 2010358061, 7831136920, 30546063745, 119291436738, 466379022561, 1825168170620, 7149316835465, 28027993191706, 109965636641173

Offset: 0

Seiichi Manyama, Apr 03 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1665

Cf. A048775, A120305, A307350, A307351.

Table[Sum[Sum[(-1)^(i + j)*(i + j)!/(2*i!*j!), {i, 1, n}], {j, 1, n}], {n, 0, 30}] (* Vaclav Kotesovec, Apr 03 2019 *)

{a(n) = sum(i=1, n, sum(j=1, n, (-1)^(i+j)*(i+j)!/(2*i!*j!)))}

{a(n) = sum(i=2, 2*n, (-1)^i*i!*polcoef(sum(j=1, n, x^j/j!)^2, i))/2} \\ Seiichi Manyama, May 20 2019

a(n) = (A120305(n) - (-1)^n)/2. - Vaclav Kotesovec, Apr 03 2019
a(n) ~ 2^(2*n+1) / (9*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 03 2019
G.f.: (1/sqrt(1-4*z)-1+2*z/(1-z^2))/(2*(2+z)). - Sergey Perepechko, Jul 11 2019

A331672 Sum of all base-n numbers with digit sum n and length at most n.

Original entry on oeis.org

3, 91, 2635, 94501, 4254936, 234572213, 15403880115, 1176838159861, 102631111100848, 10063085278250005, 1095923297151849530, 131253123286275198027, 17145216226230367266330, 2425892898650501790637545, 369599184391990522425455939, 60326656013944234430010524773

Offset: 2

Alois P. Heinz, Feb 22 2020

The cardinality of these numbers is given by A048775(n-1).

			a(2) = 3 = 11_2.
a(3) = 91 = 5 + 7 + 11 + 13 + 15 + 19 + 21 = 12_3 + 21_3 + 102_3 + 111_3 + 120_3 + 201_3 + 210_3.
a(10) = A130835(10) = 102631111100848.

Alois P. Heinz, Table of n, a(n) for n = 2..322

Cf. A007953, A023037, A048775, A130835, A331673.

b:= proc(n, i, k) option remember; `if`(n=0, [1, 0],
      `if`(i=0, 0, add((p->[p[1], p[2]*k+p[1]*d])(
         b(n-d, i-1, k)), d=0..min(n, k-1))))
    end:
a:= n-> b(n$3)[2]:
seq(a(n), n=2..17);
# second Maple program:
a:= n-> (binomial(2*n-1, n)-n)*(n^n-1)/(n-1):
seq(a(n), n=2..17);

a(n) = A048775(n-1)*A023037(n) = (binomial(2*n-1,n)-n)*(n^n-1)/(n-1).

A120279 a(n) = Sum[Sum[(i+j)!/i!/j!,{i,1,j}],{j,1,n}].

Original entry on oeis.org

2, 11, 45, 170, 631, 2346, 8780, 33089, 125466, 478181, 1830258, 7030557, 27088856, 104647615, 405187809, 1571990918, 6109558567, 23782190466, 92705454875, 361834392094, 1413883873953, 5530599237752, 21654401079301, 84859704298176

Offset: 1

Alexander Adamchuk, Jul 05 2006

p divides a(p-1) and a(p-2) for prime p=5,11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1.
p divides a([(2p-1)/2]) for prime p=5,11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1.
p divides a((p-5)/2) for prime p=17,29,41,53,89,101.. =A040115[n] Primes of form 12n+5. Primes congruent to 5 (mod 12) excluding 5.
p divides a((p-5)/3) for prime p=11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1 excluding 5.
p divides a([(p-3)/3]) for prime p=11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1 excluding 5.

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A007528, A007528, A040115, A048775, A079309, A079309, A066796.

Table[Sum[Sum[(i+j)!/i!/j!,{i,1,j}],{j,1,n}],{n,1,50}]

a(n) = Sum[Sum[(i+j)!/i!/j!,{i,1,j}],{j,1,n}]. a(n) = A079309(n+1) - (n+1). a(n) = A066796(n+1)/2 - (n+1).
Recurrence: (n+1)*(3*n-2)*a(n) = 6*(3*n^2-1)*a(n-1) - 3*(9*n^2-n-2)*a(n-2) + 2*(2*n-1)*(3*n+1)*a(n-3). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(2*n+3)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012
a(n) = Sum_{k=1..n} Sum_{i=1..k} C(k+i,i). - Wesley Ivan Hurt, Sep 19 2017

A144657 a(n) = Sum_{j = 1..n} Sum_{i = 1..n} (i + j)! / (i! * j!).

Original entry on oeis.org

0, 2, 14, 62, 242, 912, 3418, 12854, 48602, 184736, 705410, 2704132, 10400574, 40116572, 155117490, 601080358, 2333606186, 9075135264, 35345263762, 137846528780, 538257874398, 2104098963676, 8233430727554, 32247603683052, 126410606437702, 495918532948052, 1946939425648058

Offset: 0

N. J. A. Sloane, Jan 30 2009

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Suggested by a formula in A048775.

Table[Sum[Sum[(i+j)!/i!/j!,{i,1,n}],{j,1,n}],{n,0,20}] (* corrected by Vaclav Kotesovec, Oct 20 2012 *)

Recurrence: (n+1)*(12*n-19)*a(n) = 2*(30*n^2 - 24*n - 19)*a(n-1) - (48*n^2 - 9*n - 7)*a(n-2) - 2*(2*n-3)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 4^(n+1)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 20 2012
a(n) = 2*A048775(n) for n>0. - Hugo Pfoertner, Mar 13 2024

cp's OEIS Frontend

A163765 Inverse binomial transform of A048775 (assuming offset zero in both sequences).

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A305161 Number A(n,k) of compositions of n into exactly n nonnegative parts <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)*Sum_{j=0..n} binomial(n + n*j + j, n*j + j)*x^j.

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A222555 T(n,k)=Number of length (n+k)X1 arrays of occupancy after each element moves up to +-k places but not 0.

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A144510 Array T(n,k) (n >= 1, k >= 0) read by downwards antidiagonals: T(n,k) = total number of partitions of [1, 2, ..., i] into exactly k nonempty blocks, each of size at most n, for any i in the range n <= i <= k*n.

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Maple

Mathematica

Formula

A144512 Array read by upwards antidiagonals: T(n,k) = total number of partitions of [1, 2, ..., k] into exactly n blocks, each of size 1, 2, ..., k+1, for 0 <= k <= (k+1)*n.

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A307349 a(n) = Sum_{i=1..n} Sum_{j=1..n} (-1)^(i+j) * (i+j)!/(2!*i!*j!).

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Mathematica

PARI

PARI

Formula

A331672 Sum of all base-n numbers with digit sum n and length at most n.

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A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)Sum_{j=0..n} binomial(n + nj + j, nj + j)x^j.

A307349 a(n) = Sum_{i=1..n} Sum_{j=1..n} (-1)^(i+j) * (i+j)!/(2!i!j!).