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A062253
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2nd level triangle related to Eulerian numbers and binomial transforms (triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).
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5
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1, 3, 0, 7, 4, 0, 15, 30, 5, 0, 31, 146, 91, 6, 0, 63, 588, 868, 238, 7, 0, 127, 2136, 6126, 4096, 575, 8, 0, 255, 7290, 36375, 47400, 16929, 1326, 9, 0, 511, 23902, 193533, 434494, 306793, 64362, 2971, 10, 0, 1023, 76296, 956054, 3421902, 4169418
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OFFSET
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0,2
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COMMENTS
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Binomial transform of n^2*k^n is ((kn)^2 + kn)*(k + 1)^(n - 2); of n^3*k^n is ((kn)^3 + 3(kn)^2 + (1 - k)(kn))*(k + 1)^(n - 3); of n^4*k^n is ((kn)^4 + 6(kn)^3 + (7 - 4k)(kn)^2 + (1 - 4k + k^2)(kn))*(k + 1)^(n - 4); of n^5*k^n is ((kn)^5 + 10(kn)^4 + (25 - 10k)(kn)^3 + (15 - 30k + 5k^2)(kn)^2 + (1 - 11k + 11k^2 - k^3)(kn))*(k + 1)^(n - 5); of n^6*k^n is ((kn)^6 + 15(kn)^5 + (65 - 20k)(kn)^4 + (90 - 120k + 15k^2)(kn)^3 + (31 - 146k + 91k^2 - 6k^3)(kn)^2 + (1 - 26k + 66k^2 - 26k^3 + k^4)(kn))*(k + 1)^(n - 6). This sequence gives the (unsigned) polynomial coefficients of (kn)^2.
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LINKS
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FORMULA
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A(n, k)=(k+2)*A(n-1, k)+(n-k)*A(n-1, k-1)+E(n, k) where E(n, k)=(k+1)*E(n-1, k)+(n-k)*E(n-1, k-1) and E(0, 0)=1 is a triangle of Eulerian numbers, essentially A008292.
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EXAMPLE
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Rows start (1), (3,0), (7,4,0), (15,30,5,0) etc.
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CROSSREFS
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First column is A000225. Diagonals include A000007, A009056. Row sums are A000254. Taking all the levels together to create a pyramid, one face would be A010054 as a triangle with a parallel face which is Pascal's triangle (A007318) with two columns removed, another face would be a triangle of Stirling numbers of the second kind (A008277) and a third face would be A000007 as a triangle, with a triangle of Eulerian numbers (A008292), A062253, A062254 and A062255 as faces parallel to it. The row sums of this last group would provide a triangle of unsigned Stirling numbers of the first kind (A008275).
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KEYWORD
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AUTHOR
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STATUS
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approved
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