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A126277
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Triangle generated from Eulerian numbers.
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5
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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, 23, 48, 83, 120, 127, 8, 1, 9, 27, 59, 109, 177, 247, 255, 9, 1, 10, 31, 70, 135, 234, 367, 502, 511, 10
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OFFSET
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1,3
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COMMENTS
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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, ...).
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LINKS
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FORMULA
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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, ...).
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EXAMPLE
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First few rows of the triangle:
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, 23, 48, 83, 120, 127, 8;
1, 9, 27, 59, 109, 177, 247, 255, 9;
1, 10, 31, 70, 135, 234, 367, 502, 511, 10;
...
T(7,4) = 37 = A000295(4) + T(6,4) = 11 + 26.
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MATHEMATICA
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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 *)
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PROG
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(PARI) {T(n, k) = if(k==1, 1, if(k==n, n, 2^k - k - 1 + T(n-1, k)))};
for(n=1, 10, for(k=1, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Oct 23 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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_Gary w. Adamson_, Dec 23 2006
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STATUS
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approved
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