cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A022168 Triangle of Gaussian binomial coefficients [ n,k ] for q = 4.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 21, 1, 1, 85, 357, 85, 1, 1, 341, 5797, 5797, 341, 1, 1, 1365, 93093, 376805, 93093, 1365, 1, 1, 5461, 1490853, 24208613, 24208613, 1490853, 5461, 1, 1, 21845, 23859109, 1550842085, 6221613541
Offset: 0

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Comments

The coefficients of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^n*A157784(n,k). - R. J. Mathar, Mar 12 2013

Examples

			Triangle begins:
  1;
  1,    1;
  1,    5,       1;
  1,   21,      21,        1;
  1,   85,     357,       85,        1;
  1,  341,    5797,     5797,      341,       1;
  1, 1365,   93093,   376805,    93093,    1365,    1;
  1, 5461, 1490853, 24208613, 24208613, 1490853, 5461, 1;
		

References

  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 698.

Crossrefs

Cf. A006118 (row sums), A002450 (k=1), A006105 (k=2), A006106 (k=3).

Programs

  • Maple
    A022168 := proc(n,m)
            A027637(n)/A027637(n-m)/A027637(m) ;
    end proc: # R. J. Mathar, Nov 14 2011
  • Mathematica
    gaussianBinom[n_, k_, q_] := Product[q^i - 1, {i, n}]/Product[q^j - 1, {j, n - k}]/Product[q^l - 1, {l, k}]; Column[Table[gaussianBinom[n, k, 4], {n, 0, 8}, {k, 0, n}], Center] (* Alonso del Arte, Nov 14 2011 *)
    Table[QBinomial[n,k,4], {n,0,10}, {k,0,n}]//Flatten (* or *) q:= 4; T[n_, 0]:= 1; T[n_,n_]:= 1; T[n_,k_]:= T[n,k] = If[k < 0 || n < k, 0, T[n-1, k -1] +q^k*T[n-1,k]]; Table[T[n,k], {n,0,10}, {k,0,n}] // Flatten  (* G. C. Greubel, May 27 2018 *)
  • PARI
    {q=4; T(n,k) = if(k==0,1, if (k==n, 1, if (k<0 || nG. C. Greubel, May 27 2018

Formula

T(n,k) = T(n-1,k-1) + q^k * T(n-1,k). - Peter A. Lawrence, Jul 13 2017
G.f. of column k: x^k * exp( Sum_{j>=1} f((k+1)*j)/f(j) * x^j/j ), where f(j) = 4^j - 1. - Seiichi Manyama, May 09 2025