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A174045
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Triangle T(n, k, q) = Sum_{j=0..10} q^j * floor( binomial(n+1,k)*binomial(n-1,k-1)/(2^j*(n+1)) ) for q = 3, read by rows.
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3
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1, 1, 1, 1, 6, 1, 1, 24, 24, 1, 1, 70, 230, 70, 1, 1, 90, 881, 881, 90, 1, 1, 231, 2790, 7060, 2790, 231, 1, 1, 295, 8383, 28270, 28270, 8383, 295, 1, 1, 684, 21441, 181680, 242172, 181680, 21441, 684, 1, 1, 750, 58320, 378009, 882549, 882549, 378009, 58320, 750, 1
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OFFSET
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1,5
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COMMENTS
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Row sums are: {1, 2, 8, 50, 372, 1944, 13104, 73898, 649784, 2639258, ...}.
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LINKS
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FORMULA
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T(n, k, q) = Sum_{j=0..10} q^j * floor( binomial(n+1,k)*binomial(n-1,k-1)/(2^j*(n+1)) ) for q = 3.
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EXAMPLE
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The triangle begins as:
1;
1, 1;
1, 6, 1;
1, 24, 24, 1;
1, 70, 230, 70, 1;
1, 90, 881, 881, 90, 1;
1, 231, 2790, 7060, 2790, 231, 1;
1, 295, 8383, 28270, 28270, 8383, 295, 1;
1, 684, 21441, 181680, 242172, 181680, 21441, 684, 1;
1, 750, 58320, 378009, 882549, 882549, 378009, 58320, 750, 1;
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MATHEMATICA
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T[n_, k_, q_]:= Sum[q^j*Floor[Binomial[n-1, k-1]*Binomial[n+1, k]/(2^j*(n+1))], {j, 0, 10}];
Table[T[n, k, 3], {n, 12}, {k, n}]//Flatten (* modified by G. C. Greubel, Apr 16 2021 *)
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PROG
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(Magma)
T:= func< n, k, q | (&+[ q^j*Floor( Binomial(n+1, k)*Binomial(n-1, k-1)/(2^j*(n+1)) ): j in [0..10]]) >;
[T(n, k, 3): k in [1..n], n in [1..12]]; // G. C. Greubel, Apr 16 2021
(Sage)
def T(n, k, q): return sum( q^j*( (binomial(n+1, k)*binomial(n-1, k-1)//(2^j*(n+1))) ) for j in (0..10))
flatten([[T(n, k, 3) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Apr 16 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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