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Triangle read by rows: T(n, m) = floor(binomial(n+1, m)* binomial(n+2, m)/(2*m+2)), 1 <= m <= n.
2

%I #25 Sep 08 2022 08:45:51

%S 1,3,3,5,10,5,7,25,25,7,10,52,87,52,10,14,98,245,245,98,14,18,168,588,

%T 882,588,168,18,22,270,1260,2646,2646,1260,270,22,27,412,2475,6930,

%U 9702,6930,2475,412,27,33,605,4537,16335,30492,30492,16335,4537,605,33

%N Triangle read by rows: T(n, m) = floor(binomial(n+1, m)* binomial(n+2, m)/(2*m+2)), 1 <= m <= n.

%C Row sums are {1, 6, 20, 64, 211, 714, 2430, 8396, 29390, 104004, 371448, 1337216, ...}.

%H G. C. Greubel, <a href="/A174102/b174102.txt">Rows n = 1..100 of triangle, flattened</a>

%F T(n, m) = floor(binomial(n+1, m-1)*binomial(n+2, m-1)/(2*m)).

%e Triangle begins as:

%e 1;

%e 3, 3;

%e 5, 10, 5;

%e 7, 25, 25, 7;

%e 10, 52, 87, 52, 10;

%e 14, 98, 245, 245, 98, 14;

%e 18, 168, 588, 882, 588, 168, 18;

%e 22, 270, 1260, 2646, 2646, 1260, 270, 22;

%e 27, 412, 2475, 6930, 9702, 6930, 2475, 412, 27;

%t T[n_, k_] = Floor[Binomial[n+1, k]*Binomial[n+2, k]/(2*(k+1))];

%t Table[T[n, k], {n,1,12}, {k,1,n}]//Flatten (* modified by _G. C. Greubel_, Apr 13 2019 *)

%o (PARI) {T(n,k) = (binomial(n+1,k)*binomial(n+2,k)/(2*k+2))\1};

%o for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Apr 13 2019

%o (Magma) [[Floor(Binomial(n+1, k)*Binomial(n+2, k)/(2*k+2)): k in [1..n]]: n in [1..12]]; // _G. C. Greubel_, Apr 13 2019

%o (Sage) [[floor(binomial(n+1,k)*binomial(n+2,k)/(2*k+2)) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, Apr 13 2019

%Y Cf. A166454.

%Y Cf. A011848 (right diagonal).

%K nonn,tabl

%O 1,2

%A _Roger L. Bagula_, Mar 07 2010

%E Partially edited by _Jon E. Schoenfield_, Dec 02 2013

%E Edited by _G. C. Greubel_, Apr 13 2019