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A264774
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Triangle T(n,k) = binomial(5*n - 4*k, 4*n - 3*k), 0 <= k <= n.
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3
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1, 5, 1, 45, 6, 1, 455, 55, 7, 1, 4845, 560, 66, 8, 1, 53130, 5985, 680, 78, 9, 1, 593775, 65780, 7315, 816, 91, 10, 1, 6724520, 736281, 80730, 8855, 969, 105, 11, 1, 76904685, 8347680, 906192, 98280, 10626, 1140, 120, 12, 1, 886163135, 95548245, 10295472, 1107568, 118755, 12650, 1330, 136, 13, 1
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OFFSET
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0,2
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COMMENTS
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Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + ... is the o.g.f. for A002294 and f(x) = g(x)/(5 - 4*g(x)) = 1 + 5*x + 45*x^2 + 455*x^3 + 4845*x^4 + ... is the o.g.f. for A001449.
More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 4 and b = 3. See A092392, A264772, A264773 and A113139 for further examples.
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LINKS
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FORMULA
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T(n,k) = binomial(5*n - 4*k, n - k).
O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(5*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(4*n + 1)*binomial(5*n,n)*x^n.
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EXAMPLE
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Triangle begins
n\k | 0 1 2 3 4 5 6 7
------+---------------------------------------------
0 | 1
1 | 5 1
2 | 45 6 1
3 | 455 55 7 1
4 | 4845 560 66 8 1
5 | 53130 5985 680 78 9 1
6 | 593775 65780 7315 816 91 10 1
7 | 6724520 736281 80730 8855 969 105 11 1
...
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MAPLE
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A264774:= proc(n, k) binomial(5*n - 4*k, 4*n - 3*k); end proc:
seq(seq(A264774(n, k), k = 0..n), n = 0..10);
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MATHEMATICA
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Table[Binomial[5 n - 4 k, 4 n - 3 k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2015 *)
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PROG
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(Magma) /* As triangle */ [[Binomial(5*n-4*k, 4*n-3*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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