A083906 -id:A083906 - cp's OEIS frontend

A275216 Triangle read by rows, coefficients of the q-binomials (Gaussian binomial coefficients).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Peter Luschny, Jul 23 2016

			Row 8:
{1},
{1,1,1,1,1,1,1,1},
{1,1,2,2,3,3,4,3,3,2,2,1,1},
{1,1,2,3,4,5,6,6,6,6,5,4,3,2,1,1},
{1,1,2,3,5,5,7,7,8,7,7,5,5,3,2,1,1},
{1,1,2,3,4,5,6,6,6,6,5,4,3,2,1,1},
{1,1,2,2,3,3,4,3,3,2,2,1,1},
{1,1,1,1,1,1,1,1},
{1}.

Seiichi Manyama, Rows n = 0..21, flattened
NIST Digital Library of Mathematical Functions, Binomial Coefficients. (Release 1.0.11 of 2016-06-08)
Index entries related to Gaussian binomial coefficients.

See the link to the index.

Cf. A007318, A083906.

QPascalRow[n_] := Table[CoefficientList[QBinomial[n,k,q] // FunctionExpand, q], {k,0,n}]; Table[QPascalRow[n],{n,0,7}] // Flatten

A381899 Irregular triangular array read by rows. T(n,k) is the number of length n words x on {0,1} such that I(x) + W(x)*(n-W(x)) = k, where I(x) is the number of inversions in x and W(x) is the number of 1's in x, n >= 0, 0 <= k <= floor(n^2/2).

Original entry on oeis.org

1, 2, 2, 1, 1, 2, 0, 2, 2, 2, 2, 0, 0, 2, 3, 3, 4, 1, 1, 2, 0, 0, 0, 2, 2, 4, 4, 6, 4, 4, 2, 2, 2, 0, 0, 0, 0, 2, 2, 2, 4, 5, 7, 6, 9, 7, 7, 5, 4, 1, 1, 2, 0, 0, 0, 0, 0, 2, 2, 2, 2, 4, 4, 8, 6, 10, 12, 14, 12, 14, 10, 10, 6, 4, 2, 2

Offset: 0

Geoffrey Critzer, Mar 09 2025

Sum_{k>=0} T(n,k)*2^k = A132186(n).
Sum_{k>=0} T(n,k)*3^k = A053846(n).
Sum_{k>=0} T(n,k)*q^k = the number of idempotent n X n matrices over GF(q).
It appears that if n is even the n-th row converges to 2,0,0,...,21,13,9,5,4,1,1 which is A226622 reversed, and if n is odd the sequence is twice A226635.
From Alois P. Heinz, Mar 09 2025: (Start)
Sum_{k>=0} k * T(n,k) = 3*A001788(n-1) for n>=1.
Sum_{k>=0} (-1)^k * T(n,k) = A060546(n). (End)

			Triangle T(n,k) begins:
  1;
  2;
  2, 1, 1;
  2, 0, 2, 2, 2;
  2, 0, 0, 2, 3, 3, 4, 1, 1;
  2, 0, 0, 0, 2, 2, 4, 4, 6, 4, 4, 2, 2;
  ...
T(4,5) = 3 because we have: {0, 1, 0, 0}, {0, 1, 0, 1}, {1, 1, 0, 1}.

Alois P. Heinz, Rows n = 0..50, flattened

Row sums give A000079.

Cf. A001788, A060546, A226635, A226622, A132186, A053846, A083906.

b:= proc(i, j) option remember; expand(`if`(i+j=0, 1,
     `if`(i=0, 0, b(i-1, j))+`if`(j=0, 0, b(i, j-1)*z^i)))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(
         expand(add(b(n-j, j)*z^(j*(n-j)), j=0..n))):
seq(T(n), n=0..10);  # Alois P. Heinz, Mar 09 2025

nn = 7; B[n_] := FunctionExpand[QFactorial[n, q]]*q^Binomial[n, 2];e[z_] := Sum[z^n/B[n], {n, 0, nn}]; Map[CoefficientList[#, q] &, Table[B[n], {n, 0, nn}] CoefficientList[Series[e[z]^2, {z, 0, nn}],z]]

Sum_{n>=0} Sum_{k>=0} T(n,k)*q^k*x^n/(n_q!*q^binomial(n,2)) = e(x)^2 where e(x) = Sum_{n>=0} x^n/(n_q!*q^binomial(n,2)) where n_q! = Product{i=1..n} (q^n-1)/(q-1).

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A275216 Triangle read by rows, coefficients of the q-binomials (Gaussian binomial coefficients).

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