A214398 -id:A214398 - cp's OEIS frontend

A215241 Unsigned matrix inverse of triangle A214398, as a triangle read by rows n >= 1.

1, 1, 1, 3, 4, 1, 18, 26, 9, 1, 172, 256, 99, 16, 1, 2313, 3489, 1416, 264, 25, 1, 40626, 61696, 25650, 5120, 575, 36, 1, 887326, 1352518, 569772, 117980, 14450, 1098, 49, 1, 23282964, 35566368, 15099042, 3193728, 410850, 34608, 1911, 64, 1, 715540140, 1094499820, 466865280, 100049120, 13259705, 1186857, 73696, 3104, 81, 1

Offset: 1

Paul D. Hanna, Aug 06 2012

			Triangle begins:
         1;
         1,        1;
         3,        4,        1;
        18,       26,        9,       1;
       172,      256,       99,      16,      1;
      2313,     3489,     1416,     264,     25,     1;
     40626,    61696,    25650,    5120,    575,    36,    1;
    887326,  1352518,   569772,  117980,  14450,  1098,   49,  1;
  23282964, 35566368, 15099042, 3193728, 410850, 34608, 1911, 64, 1;
  ...
The matrix inverse is a signed version of triangle A214398:
   1;
  -1,   1;
   1,  -4,     1;
  -1,  10,    -9,    1;
   1, -20,    45,  -16,     1;
  -1,  35,  -165,  136,   -25,   1;
   1, -56,   495, -816,   325, -36,   1;
  -1,  84, -1287, 3876, -2925, 666, -49, 1; ...
in which the g.f. of column k is 1/(1+x)^(k^2) for k >= 1.
ILLUSTRATE G.F. OF COLUMNS:
k=1: 1 = 1/(1+x) + 1*x/(1+x)^4 + 3*x^2/(1+x)^9 + 18*x^3/(1+x)^16 + 172*x^4/(1+x)^25 + 2313*x^5/(1+x)^36 + 40626*x^6/(1+x)^49 + ...
k=2: 1 = 1/(1+x)^4 + 4*x/(1+x)^9 + 26*x^2/(1+x)^16 + 256*x^3/(1+x)^25 + 3489*x^4/(1+x)^36 + 61696*x^5/(1+x)^49 + ...
k=3: 1 = 1/(1+x)^9 + 9*x/(1+x)^16 + 99*x^2/(1+x)^25 + 1416*x^3/(1+x)^36 + 25650*x^4/(1+x)^49 + ...
k=4: 1 = 1/(1+x)^16 + 16*x/(1+x)^25 + 264*x^2/(1+x)^36 + 5120*x^3/(1+x)^49 + ...

Paul D. Hanna, Table of n, a(n) for n = 1..1081

Cf. A177447 (column 1), A215242 (column 2), A215243 (column 3); A133316 (row sums).

Cf. A214398 (unsigned matrix inverse).

T[n_, k_] := Module[{M}, M = Table[Binomial[c^2 + r - c - 1, r - c], {r, 1, n}, {c, 1, n}]; (-1)^(n - k) Inverse[M][[n, k]]];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 05 2023, after PARI program *)

{T(n, k)=local(M=matrix(n,n,r,c,binomial(c^2+r-c-1, r-c)));(-1)^(n-k)*(M^-1)[n,k]}
for(n=1, 12, for(k=1, n, print1(T(n, k), ", ")); print(""))

G.f.: x*y/(1-x*y) = Sum_{n>=1} Sum_{k=1..n} T(n,k)*x^n*y^k/(1+x)^(n^2).
G.f. of column k: 1 = Sum_{n>=k} T(n,k)*x^(n-k)/(1+x)^(n^2).
Column 1 forms A177447.
Row sums form A133316.

A318795 Array read by antidiagonals: T(n,k) is the number of inequivalent nonnegative integer n X n matrices with sum of elements equal to k, under row and column permutations.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 5, 4, 1, 1, 11, 10, 4, 1, 1, 14, 24, 10, 4, 1, 1, 24, 51, 33, 10, 4, 1, 1, 30, 114, 78, 33, 10, 4, 1, 1, 45, 219, 224, 91, 33, 10, 4, 1, 1, 55, 424, 549, 277, 91, 33, 10, 4, 1, 1, 76, 768, 1403, 792, 298, 91, 33, 10, 4, 1, 1, 91, 1352, 3292, 2341, 881, 298, 91, 33, 10, 4, 1

Offset: 1

Andrew Howroyd, Sep 03 2018

			Array begins:
===========================================================
n\k| 1 2  3  4  5   6   7    8    9    10     11     12
---+-------------------------------------------------------
1  | 1 1  1  1  1   1   1    1    1     1      1      1 ...
2  | 1 4  5 11 14  24  30   45   55    76     91    119 ...
3  | 1 4 10 24 51 114 219  424  768  1352   2278   3759 ...
4  | 1 4 10 33 78 224 549 1403 3292  7677  16934  36581 ...
5  | 1 4 10 33 91 277 792 2341 6654 18802  51508 138147 ...
6  | 1 4 10 33 91 298 881 2825 8791 27947  87410 272991 ...
7  | 1 4 10 33 91 298 910 2974 9655 32287 108274 367489 ...
8  | 1 4 10 33 91 298 910 3017 9886 33767 116325 410298 ...
9  | 1 4 10 33 91 298 910 3017 9945 34124 118729 424498 ...
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Andrew Howroyd, Additional PARI Programs

Rows 2..7 are A053307, A052365, A052366, A052367, A052372, A052373.
Main diagonal is A007716.
Cf. A214398, A246106, A304942, A318805.

permcount[v_List] := Module[{m = 1, s = 0, k = 0, t}, For[i = 1, i <= Length[v], i++, t = v[[i]]; k = If[i > 1 && t == v[[i - 1]], k + 1, 1]; m *= t*k; s += t]; s!/m];
c[p_List, q_List, k_] := SeriesCoefficient[1/Product[(1 - x^LCM[p[[i]], q[[j]]])^GCD[p[[i]], q[[j]]], {j, 1, Length[q]}, {i, 1, Length[p]}], {x, 0, k}];
M[m_, n_, k_] := Module[{s=0}, Do[Do[s += permcount[p]*permcount[q]*c[p, q, k], {q, IntegerPartitions[n]}], {p, IntegerPartitions[m]}]; s/(m!*n!)];
Table[M[n-k+1, n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Sep 12 2018, after Andrew Howroyd *)

\\ see also link.
permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
K(q, t, k)={1/prod(j=1, #q, (1-y^lcm(t,q[j]) + O(y*y^k))^gcd(t, q[j]))}
M(m, n, k)={my(s=0); forpart(q=m, s+=permcount(q)*polcoef(polcoef(exp(sum(t=1, n, K(q, t, k)/t*x^t) + O(x*x^n)), n), k)); s/m!}
for(n=1, 10, for(k=1, 12, print1(M(n, n, k), ", ")); print); \\ updated Andrew Howroyd, Mar 29 2020

T(n,k) = T(k,k) for n > k.

A178325 G.f.: A(x) = Sum_{n>=0} x^n/(1-x)^(n^2).

Original entry on oeis.org

1, 1, 2, 6, 21, 83, 363, 1730, 8889, 48829, 284858, 1755325, 11374092, 77208275, 547261631, 4039201624, 30967330941, 246084049137, 2023030659970, 17175765057532, 150367445873108, 1355528352031358, 12566899017130088

Offset: 0

Paul D. Hanna, Dec 21 2010

nonn

Equals the row sums of triangle A214398.

a(n) is the number of weak compositions of n such that if the first part is equal to k then there are a total of k^2 + 1 parts. A weak composition is an ordered partition of the integer n into nonnegative parts. a(3) = 6 because we have: 1+2, 2+0+0+0+1, 2+0+0+1+0, 2+0+1+0+0, 2+1+0+0+0, 3+0+0+0+0+0+0+0+0+0. - Geoffrey Critzer, Oct 09 2013

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 83*x^5 + 363*x^6 +...
A(x) = 1 + (x-x^2)*((1-x)-x)/((1-x)^3-x) + (x-x^2)^2*((1-x)-x)*((1-x)^5-x)/(((1-x)^3-x)*((1-x)^7-x)) + (x-x^2)^3*((1-x)-x)*((1-x)^5-x)*((1-x)^9-x)/(((1-x)^3-x)*((1-x)^7-x)*((1-x)^11-x)) +...

Paul D. Hanna, Table of n, a(n) for n = 0..250

Cf. A214398, A230050, A227934, A227935.

nn=22;CoefficientList[Series[Sum[x^k/(1-x)^(k^2),{k,0,nn}],{x,0,nn}],x]  (* Geoffrey Critzer, Oct 09 2013 *)

{a(n)=sum(k=0,n,binomial((n-k)^2+k-1,k))}

{a(n)=polcoeff(sum(m=0,n,x^m/(1-x+x*O(x^n))^(m^2)),n)}

{a(n)=polcoeff(sum(m=0,n,(x-x^2)^m*prod(k=1,m,((1-x)^(4*k-3)-x)/((1-x)^(4*k-1)-x +x*O(x^n)))),n)}

a(n) = Sum_{k=0..n} C((n-k)^2 + k-1, k).
G.f.: A(x) = Sum_{n>=0} (x-x^2)^n*Product_{k=1..n} ((1-x)^(4*k-3) - x)/((1-x)^(4*k-1) - x) due to a q-series identity.
Let q = 1/(1-x), then g.f. A(x) equals the continued fraction:
. A(x) = 1/(1- q*x/(1- q*(q^2-1)*x/(1- q^5*x/(1- q^3*(q^4-1)*x/(1- q^9*x/(1- q^5*(q^6-1)*x/(1- q^13*x/(1- q^7*(q^8-1)*x/(1- ...)))))))))
due to an identity of a partial elliptic theta function.
log(a(n)) ~ n*(log(n) - 2) * (1 + log(4*n) - log((log(n) - 2)*log(n))) / log(n). - Vaclav Kotesovec, Jan 10 2023

A214400 a(n) = binomial(n^2 + 3*n, n).

Original entry on oeis.org

1, 4, 45, 816, 20475, 658008, 25827165, 1198774720, 64276915527, 3911395881900, 266401260897200, 20082459351180240, 1660305826125766950, 149389005978091284720, 14533945899753270066525, 1520398315196482557890304, 170190601112537814791748255

Offset: 0

Paul D. Hanna, Jul 15 2012

nonn

Equals the central terms of triangle A214398.

Robert Israel, Table of n, a(n) for n = 0..336

Cf. A054688, A178325, A214398.

seq(binomial(n^2+3*n,n),n=0..30); # Robert Israel, Mar 04 2022

PARI
```
a(n)=binomial(n^2+3*n, n)
```

a(n) = [x^n] 1/(1 - x)^((n+1)^2). - Ilya Gutkovskiy, Oct 04 2017

a(n) ~ n^(n-1/2)*exp(n+5/2)/sqrt(2*Pi). - Robert Israel, Mar 04 2022

A214403 Triangle, read by rows of terms T(n,k) for k=0..n^2, that starts with a '1' in row 0 with row n>0 consisting of 2*n-1 '1's followed by the partial sums of the prior row.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 3, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 8, 11, 15, 21, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 17, 22, 28, 36, 47, 62, 83, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 19, 24, 30, 37, 45, 55, 68, 85, 107, 135, 171, 218, 280, 363

Offset: 0

Paul D. Hanna, Jul 15 2012

Right border and row sums form A178325.

			Triangle begins:
  [1];
  [1, 1];
  [1,1,1, 1, 2];
  [1,1,1,1,1, 1,2,3, 4, 6];
  [1,1,1,1,1,1,1, 1,2,3,4,5, 6,8,11, 15, 21];
  [1,1,1,1,1,1,1,1,1, 1,2,3,4,5,6,7, 8,10,13,17,22, 28,36,47, 62, 83];
  ...
Row sums equal the row sums (A178325) of triangle A214398,
where A214398(n, k) = binomial(k^2+n-k-1, n-k):
  1;
  1, 1;
  1, 4, 1;
  1, 10, 9, 1;
  1, 20, 45, 16, 1;
  1, 35, 165, 136, 25, 1;
  1, 56, 495, 816, 325, 36, 1;
  1, 84, 1287, 3876, 2925, 666, 49, 1;
  ...

Paul D. Hanna, Rows n = 0..14, flattened.

Cf. A131338, A178325, A214398.

{T(n, k)=if(k>n^2||n<0||k<0, 0, if(n==0,1,if(k<=2*n-1, 1, sum(i=0, k-2*n+1, T(n-1, i)))))}
for(n=0,10,for(k=0,n^2,print1(T(n,k),", "));print(""))

cp's OEIS Frontend

A215241 Unsigned matrix inverse of triangle A214398, as a triangle read by rows n >= 1.

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A318795 Array read by antidiagonals: T(n,k) is the number of inequivalent nonnegative integer n X n matrices with sum of elements equal to k, under row and column permutations.

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A178325 G.f.: A(x) = Sum_{n>=0} x^n/(1-x)^(n^2).

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A214400 a(n) = binomial(n^2 + 3*n, n).

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A214403 Triangle, read by rows of terms T(n,k) for k=0..n^2, that starts with a '1' in row 0 with row n>0 consisting of 2*n-1 '1's followed by the partial sums of the prior row.

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PARI