A022174 Triangle of Gaussian binomial coefficients [ n,k ] for q = 10.
1, 1, 1, 1, 11, 1, 1, 111, 111, 1, 1, 1111, 11211, 1111, 1, 1, 11111, 1122211, 1122211, 11111, 1, 1, 111111, 112232211, 1123333211, 112232211, 111111, 1, 1, 1111111, 11223332211, 1123445443211, 1123445443211, 11223332211, 1111111, 1
Offset: 0
Examples
Triangle begins: 1; 1, 1; 1, 11, 1; 1, 111, 111, 1; 1, 1111, 11211, 1111, 1; 1, 11111, 1122211, 1122211, 11111, 1; 1, 111111, 112232211, 1123333211, 112232211, 111111, 1; 1, 1111111, 11223332211, 1123445443211, 1123445443211, 11223332211, 1111111, 1;
References
- F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 698.
Links
- G. C. Greubel, Rows n=0..50 of triangle, flattened
- Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.
- Index entries for sequences related to Gaussian binomial coefficients
Crossrefs
Row sums give A015196.
Programs
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Maple
A027878 := proc(n) mul(10^i-1,i=1..n) ; end proc: A022174 := proc(n,m) A027878(n)/A027878(m)/A027878(n-m) ; end proc:# R. J. Mathar, Jul 19 2017
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Mathematica
a027878[n_]:=Times@@ Table[10^i - 1, {i, n}]; T[n_, m_]:=a027878[n]/( a027878[m] a027878[n - m]); Table[T[n, m], {n, 0, 10}, {m, 0, n}]//Flatten (* Indranil Ghosh, Jul 20 2017, after Maple code *) Table[QBinomial[n,k,10], {n,0,10}, {k,0,n}]//Flatten (* or *) q:= 10; 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 28 2018 *) -
PARI
{q=10; T(n,k) = if(k==0,1, if (k==n, 1, if (k<0 || n -
Python
from operator import mul def a027878(n): return 1 if n==0 else reduce(mul, [10**i - 1 for i in range(1, n + 1)]) def T(n, m): return a027878(n)/(a027878(m)*a027878(n - m)) for n in range(11): print([T(n, m) for m in range(n + 1)]) # Indranil Ghosh, Jul 20 2017, after Maple code
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) = 10^j - 1. - Seiichi Manyama, May 09 2025