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%I #15 Nov 11 2019 21:47:11
%S 1,1,4,1,8,10,1,12,30,20,1,16,60,80,35,1,20,100,200,175,56,1,24,150,
%T 400,525,336,84,1,28,210,700,1225,1176,588,120,1,32,280,1120,2450,
%U 3136,2352,960,165,1,36,360,1680,4410,7056,7056,4320,1485,220,1,40,450,2400,7350
%N Triangle read by rows: T(n,k) = (k+1)*(k+2)*(k+3)*binomial(n,k)/6 (0 <= k <= n).
%C Sum of entries in row n = (2^n/48)*(n+4)*(n^2 + 11n + 12) = A049612(n+1).
%H Harvey P. Dale, <a href="/A124848/b124848.txt">Table of n, a(n) for n = 0..1000</a>
%e Triangle starts:
%e 1;
%e 1, 4;
%e 1, 8, 10;
%e 1, 12, 30, 20;
%e 1, 16, 60, 80, 35;
%e 1, 20, 100, 200, 175, 56;
%e 1, 24, 150, 400, 525, 336, 84;
%p T:=(n,k)->(k+1)*(k+2)*(k+3)*binomial(n,k)/6: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%t Flatten[Table[(k+1)(k+2)(k+3) Binomial[n,k]/6,{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, May 14 2012 *)
%Y Cf. A000292, A049612.
%K nonn,tabl
%O 0,3
%A _Gary W. Adamson_, Nov 10 2006
%E Edited by _N. J. A. Sloane_, Dec 02 2006
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