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(MAGMAMagma)
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From Philippe Deléham, Mar 15 2014: (Start)
As a number triangle, this is the Riordan array (1/(1-x), x*(1+4*x)/(1-x)). It has row sums A063727(n). - _Philippe Deléham_, Mar 15 2014
Sum_{k=0..n} T(n, k) = A063727(n). (End)
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 4. (End)
Sum_{k=0..n} T(n, k, 4) = A063727(n). (End)
Table[Hypergeometric2F1[-k, k-n, 1, 5], {n, 0, 1012}, {k, 0, n}]//Flatten (* Jean-François Alcover, May 24 2013 *)
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1, 1, 1, 1, 6, 1, 1, 11, 11, 1, 1, 16, 46, 16, 1, 1, 21, 106, 106, 21, 1, 1, 26, 191, 396, 191, 26, 1, 1, 31, 301, 1011, 1011, 301, 31, 1, 1, 36, 436, 2076, 3606, 2076, 436, 36, 1, 1, 41, 596, 3716, 9726, 9726, 3716, 596, 41, 1, 1, 46, 781, 6056, 21746, 33876, 21746, 6056, 781, 46, 1
Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) +4T 4*T(n-1, k-1) + T(n-1, k). Rows are the expansions of (1+4x)^k/(1-x)^(k+1).
Rows are the expansions of (1+4*x)^k/(1-x)^(k+1).
From G. C. Greubel, May 26 2021: (Start)
T(n, k, m) = Hypergeometric2F1([-k, k-n], [1], m+1), for m = 4.
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 4.
Sum_{k=0..n} T(n, k, 4) = A063727(n). (End)
Rows start
Square array begins as:
1 , 1 , 1 , 1 , 1 , ... A000012;
1 , 6 , 11 , 16 , 21 , ... A016861;
1 , 11 , 46 , 106 , 191 , ... A081587;
1 , 16 , 106 , 396 , 1011 , ... A081588;
1 , 21 , 191 , 1011 , 3606 , ...
1;
1, 1;
1, 6, 1;
1, 11, 11, 1;
1, 16, 46, 16, 1;
1, 21, 106, 106, 21, 1;
1, 26, 191, 396, 191, 26, 1;
1, 31, 301, 1011, 1011, 301, 31, 1;
1, 36, 436, 2076, 3606, 2076, 436, 36, 1;
1, 41, 596, 3716, 9726, 9726, 3716, 596, 41, 1;
1, 46, 781, 6056, 21746, 33876, 21746, 6056, 781, 46, 1; - Philippe Deléham, Mar 15 2014
Table[ Hypergeometric2F1[-k, k-n, 1, 5], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 24 2013 *)
(MAGMA)
A081579:= func< n, k, q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081579(n, k, 4): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021
(Sage) flatten([[hypergeometric([-k, k-n], [1], 5).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021
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