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%I #16 Feb 14 2022 21:27:29
%S 1,1,2,1,3,3,1,4,7,4,1,5,11,15,5,1,6,15,26,31,6,1,7,19,37,57,63,7,1,8,
%T 23,48,83,120,127,8,1,9,27,59,109,177,247,255,9,1,10,31,70,135,234,
%U 367,502,511,10
%N Triangle generated from Eulerian numbers.
%C N-th diagonal starting from the right = binomial transform of [1, N, q, q, q, ...) where q = 2*N - 2. Given the infinite set of triangles "T" composed of partial column sums of the polygonal numbers, the N-th diagonal starting from the right = row sums of triangle "T": (T=3 = A104712; T=4 = A125165; T=5 = A125232; T=6 = A125233; T=7 = A125234, T=8 = A125235; and so on). For example, 3rd diagonal from the right = the offset Eulerian numbers, (1, 4, 11, 26, 57, 120, ...) = row sums of Triangle A104712 having partial column sums of the triangular numbers: 1; 3, 1; 6, 4, 1; 10, 10, 5, 1; 15, 20, 15, 6, 1; ... Row sums = A124671: (1, 3, 7, 16, 37, 85, 191, ...).
%H G. C. Greubel, <a href="/A126277/b126277.txt">Rows n=1..100 of triangle, flattened</a>
%F Given right border = (1,2,3,...), T(n,k) = A000295(k) + T(n-1,k); where A000295 = the Eulerian numbers starting (0, 1, 4, 11, 26, 57, ...).
%e First few rows of the triangle:
%e 1;
%e 1, 2;
%e 1, 3, 3;
%e 1, 4, 7, 4;
%e 1, 5, 11, 15, 5;
%e 1, 6, 15, 26, 31, 6;
%e 1, 7, 19, 37, 57, 63, 7;
%e 1, 8, 23, 48, 83, 120, 127, 8;
%e 1, 9, 27, 59, 109, 177, 247, 255, 9;
%e 1, 10, 31, 70, 135, 234, 367, 502, 511, 10;
%e ...
%e T(7,4) = 37 = A000295(4) + T(6,4) = 11 + 26.
%t T[n_,1]:=1; T[n_,n_]:=n; T[n_,k_]:= T[n-1,k] + 2^k - k - 1; Table[T[n,k], {n,1,15}, {k,1,n}]//Flatten (* _G. C. Greubel_, Oct 23 2018 *)
%o (PARI) {T(n,k) = if(k==1, 1, if(k==n, n, 2^k - k - 1 + T(n-1,k)))};
%o for(n=1,10, for(k=1,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Oct 23 2018
%Y Cf. A126268, A000295, A104712, A125165, A104712, A125232 - A125235.
%K nonn,tabl
%O 1,3
%A _Gary w. Adamson_, Dec 23 2006
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