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%I #22 Sep 08 2022 08:45:06
%S 1,1,1,1,2,2,1,3,5,6,1,4,9,16,19,1,5,14,31,54,63,1,6,20,52,111,188,
%T 219,1,7,27,80,197,405,676,787,1,8,35,116,320,752,1508,2492,2897,1,9,
%U 44,161,489,1276,2900,5712,9361,10869,1,10,54,216,714,2034,5095,11296,21933,35702,41414
%N Triangle read by rows giving numbers of paths in a lattice satisfying certain conditions.
%H G. C. Greubel, <a href="/A071944/b071944.txt">Rows n = 0..100 of triangle, flattened</a>
%H D. Merlini, D. G. Rogers, R. Sprugnoli and M. C. Verri, <a href="http://dx.doi.org/10.4153/CJM-1997-015-x">On some alternative characterizations of Riordan arrays</a>, Canad J. Math., 49 (1997), 301-320.
%F T(n, k) = ((n-k+1)/(n+1))*Sum_{i=0..k/3} binomial(n+1, i)*binomial(n+k -3*i, n), for k <= n.
%e Triangle begins with:
%e 1;
%e 1, 1;
%e 1, 2, 2;
%e 1, 3, 5, 6;
%e 1, 4, 9, 16, 19;
%e 1, 5, 14, 31, 54, 63;
%e 1, 6, 20, 52, 111, 188, 219;
%e 1, 7, 27, 80, 197, 405, 676, 787;
%e ...
%p a := proc(n,k) if k<=n then (n-k+1)*sum(binomial(n+1,i)*binomial(n+k-3*i,n),i=0..k/3)/(n+1) else 0 fi end;
%t Table[((n-k+1)/(n+1))*Sum[Binomial[n+1, j]*Binomial[n+k-3*j, n], {j, 0, k/3}], {n, 0, 10}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Mar 17 2019 *)
%o (PARI) {T(n,k) = ((n-k+1)/(n+1))*sum(j=0, floor(k/3), binomial(n+1, j)* binomial(n+k -3*j, n))}; \\ _G. C. Greubel_, Mar 17 2019
%o (Magma) [[((n-k+1)/(n+1))*(&+[Binomial(n+1, j)*Binomial(n+k -3*j, n): j in [0..Floor(k/3)]]): k in [0..n]]: n in [0..10]]; // _G. C. Greubel_, Mar 17 2019
%o (Sage) [[((n-k+1)/(n+1))*sum(binomial(n+1,j)*binomial(n+k-3*j,n) for j in (0..floor(k/3))) for k in (0..n)] for n in (0..10)] # _G. C. Greubel_, Mar 17 2019
%Y Diagonal entries form A071969.
%K nonn,easy,tabl
%O 0,5
%A _N. J. A. Sloane_, Jun 15 2002
%E More terms from _Emeric Deutsch_, Dec 19 2003
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