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A059473
Triangle T(n, k) is coefficient of z^n*w^k in 1/(1 - 2*z - 2*w - 2*z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...
3
1, 2, 2, 4, 10, 4, 8, 32, 32, 8, 16, 88, 148, 88, 16, 32, 224, 536, 536, 224, 32, 64, 544, 1696, 2440, 1696, 544, 64, 128, 1280, 4928, 9344, 9344, 4928, 1280, 128, 256, 2944, 13504, 31936, 42256, 31936, 13504, 2944, 256, 512, 6656, 35456, 100736, 167072, 167072, 100736, 35456, 6656, 512
OFFSET
0,2
FORMULA
G.f.: 1/(1 - 2*z - 2*w - 2*z*w).
T(n, k) = Sum_{j=0..n} 2^(n + k - j)*binomial(n, j)*binomial(n + k - j, n). - G. C. Greubel, Oct 04 2017
T(n, k) = 2^n*binomial(n, k)*hypergeom([-k, k - n], [-n], -1/2). - Peter Luschny, Nov 26 2021
EXAMPLE
[0] 1;
[1] 2, 2;
[2] 4, 10, 4;
[3] 8, 32, 32, 8;
[4] 16, 88, 148, 88, 16;
[5] 32, 224, 536, 536, 224, 32;
[6] 64, 544, 1696, 2440, 1696, 544, 64;
...
MAPLE
read transforms; SERIES2(1/(1-2*z-2*w-2*z*w), x, y, 12): SERIES2TOLIST(%, x, y, 12);
# Alternative:
T := (n, k) -> 2^n*binomial(n, k)*hypergeom([-k, -n + k], [-n], -1/2):
for n from 0 to 10 do seq(simplify(T(n, k)), k = 0 .. n) end do; # Peter Luschny, Nov 26 2021
MATHEMATICA
T[n_, k_] := Sum[2^(n + k - j)*Binomial[n, j]*Binomial[n + k - j, n], {j, 0, n}]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 04 2017 *)
CROSSREFS
Column k = 0 gives A000079.
T(n, n) gives A098270.
Sequence in context: A220323 A295416 A202802 * A240629 A360313 A162508
KEYWORD
nonn,tabl,easy
AUTHOR
N. J. A. Sloane, Feb 03 2001
STATUS
approved