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A258820
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Reversed rows of A178252 presented as diagonals of an irregular triangle.
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3
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1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 5, 2, 1, 1, 3, 10, 1, 1, 7, 5, 5, 1, 1, 4, 7, 5, 1, 1, 9, 28, 35, 3, 1, 1, 5, 12, 14, 7, 1, 1, 11, 15, 21, 14, 7, 1, 1, 6, 55, 30, 126, 28, 1, 1, 13, 22, 165, 42, 21, 4, 1
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OFFSET
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0,8
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COMMENTS
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The diagonals of T are the reversed rows of A178252. E.g., the fifth diagonal of T is (1,2,2,1,1) from the example below, which is the fifth reversed row of A178252.
Factoring out the greatest common divisor (gcd) of the coefficients of the sub-polynomials in the indeterminate q of the polynomials in s presented on p. 12 of the Alexeev et al. link and then evaluating the sub-polynomials at q=1 gives the first nine rows of T given in the example below. E.g., for k=6 (the seventh row), q*s^6 + (6*q + 9*q^2) s^4 + (15*q + 15*q^2) s^2 + 5 = q*s^6 + 3*(2*q + 3*q^2)*s^4 + 15*(q + q^2)*s^2 + 5 generates (1,2+3,1+1,1)=(1,5,2,1).
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 25 2015
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LINKS
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FORMULA
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T(n,k) = A178252(n-k,n-2k) = A055151(n,k) / A161642(n,k) = A007318(n,2k) * A000108(k) / A161642(n,k) = n! / [(n-2k)! k! (k+1)! A161642(n,k)] = A003989(n-k+1,k+1) * (n-k)! / [ (n-2k)! (k+1)! ], where A003989(j,k) = gcd(j,k).
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EXAMPLE
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The irregular triangle T(n,k) starts
n\k 0 1 2 3 4 5 ...
0: 1
1: 1
2: 1 1
3: 1 1
4: 1 3 1
5: 1 2 1
6: 1 5 2 1
7: 1 3 10 1
8: 1 7 5 5 1
9: 1 4 7 5 1
10: 1 9 28 35 3 1
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MATHEMATICA
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max = 15; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes] // Numerator; t[n_, k_] := inv[[n, k]]; t[n_, k_] /; k == n+1 = 1; Table[t[n-k+1, k], {n, 2, max+1}, {k, 2, Floor[n/2]+1}] // Flatten (* Jean-François Alcover, Jul 22 2015 *)
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CROSSREFS
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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