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.

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A306344 The q-analogs T(q; n,k) of the rascal-triangle, here q = 3.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 5, 1, 1, 14, 17, 14, 1, 1, 41, 53, 53, 41, 1, 1, 122, 161, 170, 161, 122, 1, 1, 365, 485, 521, 521, 485, 365, 1, 1, 1094, 1457, 1574, 1601, 1574, 1457, 1094, 1, 1, 3281, 4373, 4733, 4841, 4841, 4733, 4373, 3281, 1
Offset: 0

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Author

Werner Schulte, Feb 08 2019

Keywords

Comments

The formulas are given for the general case depending on some fixed integer q. The terms are valid for q = 3. For the special case q = 1 see A077028, for q = 2 see A306326. For q < 1 the terms might be negative.

Examples

			If q = 3 the triangle T(3; n,k) starts:
n\k:  0     1     2     3     4     5     6     7     8     9
=============================================================
  0:  1
  1:  1     1
  2:  1     2     1
  3:  1     5     5     1
  4:  1    14    17    14     1
  5:  1    41    53    53    41     1
  6:  1   122   161   170   161   122     1
  7:  1   365   485   521   521   485   365     1
  8:  1  1094  1457  1574  1601  1574  1457  1094     1
  9:  1  3281  4373  4733  4841  4841  4733  4373  3281     1
etc.
		

Crossrefs

Formula

T(q; n,k) = 1 + ((q^k-1)/(q-1))*((q^(n-k)-1)/(q-1)) for 0 <= k <= n.
T(q; n,k) = T(q; n,n-k) for 0 <= k <= n.
T(q; n,0) = T(q; n,n) = 1 for n >= 0.
T(q; n,1) = 1 + (q^(n-1)-1)/(q-1) for n > 0.
T(q; i,j) = 0 if i < j or j < 0.
The T(q; n,k) satisfy several recurrence equations:
(1) T(q; n,k) = q*T(q; n-1,k) + (q^k-1)/(q-1)-(q-1) for 0 <= k < n;
(2) T(q; n,k) = (T(q; n-1,k)*T(q; n-1,k-1) + q^(n-2))/T(q; n-2,k-1),
(3) T(q; n,k) = T(q; n,k-1) + T(q; n-1,k) + q^(n-k-1) - T(q; n-1,k-1),
(4) T(q; n,k) = T(q; n,k-1) + q*T(q; n-2,k-1) - q*T(q; n-2,k-2) for 0 < k < n;
(5) T(q; n,k) = T(q; n,k-2) + T(q; n-1,k) + (1+q)*q^(n-k-1) - T(q; n-1,k-2)
for 1 < k < n with initial values given above.
G.f. of column k >= 0: Sum_{n>=0} T(q; n+k,k)*t^n = (1+((q^k-1)/(q-1)-q)*t) / ((1-t)*(1-q*t)). Take account of lim_{q->1} (q^k-1)/(q-1) = k.
G.f.: Sum_{n>=0, k=0..n} T(q; n,k)*x^k*t^n = (1-q*t-q*x*t+(1+q^2)*x*t^2) / ((1-t)*(1-q*t)*(1-x*t)*(1-q*x*t)).
The row polynomials p(q; n,x) = Sum_{k=0..n} T(q; n,k)*x^k satisfy the recurrence equation p(q; n,x) = q*p(q; n-1,x) + x^n + Sum_{k=0..n-1} ((q^k-1)/(q-1)-(q-1))*x^k for n > 0 with initial value p(q; 0,x) = 1.
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