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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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%I #8 Apr 16 2021 02:35:44

%S 1,1,1,1,6,1,1,24,24,1,1,70,230,70,1,1,90,881,881,90,1,1,231,2790,

%T 7060,2790,231,1,1,295,8383,28270,28270,8383,295,1,1,684,21441,181680,

%U 242172,181680,21441,684,1,1,750,58320,378009,882549,882549,378009,58320,750,1

%N 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.

%C Row sums are: {1, 2, 8, 50, 372, 1944, 13104, 73898, 649784, 2639258, ...}.

%H G. C. Greubel, <a href="/A174045/b174045.txt">Rows n = 1..50 of the triangle, flattened</a>

%F 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.

%e The triangle begins as:

%e 1;

%e 1, 1;

%e 1, 6, 1;

%e 1, 24, 24, 1;

%e 1, 70, 230, 70, 1;

%e 1, 90, 881, 881, 90, 1;

%e 1, 231, 2790, 7060, 2790, 231, 1;

%e 1, 295, 8383, 28270, 28270, 8383, 295, 1;

%e 1, 684, 21441, 181680, 242172, 181680, 21441, 684, 1;

%e 1, 750, 58320, 378009, 882549, 882549, 378009, 58320, 750, 1;

%t T[n_, k_, q_]:= Sum[q^j*Floor[Binomial[n-1, k-1]*Binomial[n+1,k]/(2^j*(n+1))], {j, 0, 10}];

%t Table[T[n,k,3], {n,12}, {k,n}]//Flatten (* modified by _G. C. Greubel_, Apr 16 2021 *)

%o (Magma)

%o 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]]) >;

%o [T(n,k,3): k in [1..n], n in [1..12]]; // _G. C. Greubel_, Apr 16 2021

%o (Sage)

%o 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))

%o flatten([[T(n,k,3) for k in (1..n)] for n in (1..12)]) # _G. C. Greubel_, Apr 16 2021

%Y Cf. A174043 (q=1), A174044 (q=2), this sequence (q=3).

%K nonn,tabl,easy,less

%O 1,5

%A _Roger L. Bagula_, Mar 06 2010

%E Edited by _G. C. Greubel_, Apr 16 2021