A070780 -id:A070780 - cp's OEIS frontend

A228836 Triangle defined by T(n,k) = binomial(n^2, (n-k)*k), for n>=0, k=0..n, as read by rows.

1, 1, 1, 1, 4, 1, 1, 36, 36, 1, 1, 560, 1820, 560, 1, 1, 12650, 177100, 177100, 12650, 1, 1, 376992, 30260340, 94143280, 30260340, 376992, 1, 1, 13983816, 8217822536, 92263734836, 92263734836, 8217822536, 13983816, 1, 1, 621216192, 3284214703056, 159518999862720, 488526937079580, 159518999862720, 3284214703056, 621216192, 1

Offset: 0

Paul D. Hanna, Sep 05 2013

			The triangle of coefficients C(n^2, (n-k)*k), n>=k, k=0..n, begins:
  1;
  1, 1;
  1, 4, 1;
  1, 36, 36, 1;
  1, 560, 1820, 560, 1;
  1, 12650, 177100, 177100, 12650, 1;
  1, 376992, 30260340, 94143280, 30260340, 376992, 1;
  1, 13983816, 8217822536, 92263734836, 92263734836, 8217822536, 13983816, 1;
  ...

Paul D. Hanna, Rows 0..30 as a flattened table of n, a(n) for n = 0..495

Cf. A207136 (row sums), A228837 (antidiagonal sums), A070780 (column 1).

Cf. related triangles: A228900(exp), A209330, A226234, A228832.

T[n_,k_]:=Binomial[n^2, (n-k)*k]; Table[T[n,k],{n,0,8},{k,0,n}]//Flatten (* Stefano Spezia, Aug 02 2025 *)

{T(n,k)=binomial(n^2, (n-k)*k)}
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))

A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.

Original entry on oeis.org

1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, 130495143060, 8002598818976, 548083639109204, 41481408076217480, 3439242379130298624, 310088460754461374608, 30213233395705566500248, 3164041937175571224372118, 354458174755809132668847604, 42301583754989091491716706272, 5358139267082652649000120167456

Offset: 0

Paul D. Hanna, Jan 06 2018

nonn

			G.f.: A(x) = 1 + 2*x + 8*x^2 + 84*x^3 + 1522*x^4 + 38940*x^5 + 1278880*x^6 + 51136512*x^7 + 2407921070*x^8 + 130495143060*x^9 + 8002598818976*x^10 +...
such that
A(x) = 1 + 2*(x/A(x)) + 12*(x/A(x))^2 + 140*(x/A(x))^3 + 2530*(x/A(x))^4 + 62832*(x/A(x))^5 + 1997688*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)*(x/A(x))^n +...
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, ...];
  [1, 4, 20, 200, 3444, 85312, 2744928, 108267280, 5049708672, ...];
  [1, 6, 36, 356, 5862, 140508, 4424984, 172064160, 7946443350, ...];
  [1, 8, 56, 560, 8888, 206176, 6350112, 243284064, 11121338640, ...];
  [1, 10, 80, 820, 12650, 284252, 8556240, 322780800, 14599990830, ...];
  [1, 12, 108, 1144, 17292, 376992, 11084864, 411518448, 18410660208, ...];
  [1, 14, 140, 1540, 22974, 487004, 13983816, 510586400, 22584587382, ...];
  [1, 16, 176, 2016, 29872, 617280, 17308096, 621216192, 27156348512, ...];
  [1, 18, 216, 2580, 38178, 771228, 21120768, 744800256, 32164253550, ...]; ...
in which the main diagonal begins:
  [1, 4, 36, 560, 12650, 376992, 13983816, ..., binomial((n+1)^2,n), ...],
thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^((n+1)^2) for n>=0.

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A295765, A295763, A070780.

terms = 20; A[] = 1; Do[A[x] = Sum[Binomial[(n+1)^2, n]/(n+1)*x^n/A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)

{a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2,m) - V[m+1])/(m+1);); A[n+1]}
for(n=0,20,print1(a(n),", "))

G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n) for n>=0.

a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(3/2 - 2*exp(-2)) / sqrt(2*Pi) = 1.3639581816297942453935... - Vaclav Kotesovec, Aug 11 2021, updated Apr 20 2024

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A228836 Triangle defined by T(n,k) = binomial(n^2, (n-k)*k), for n>=0, k=0..n, as read by rows.

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