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A161642
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Triangle (by rows): T(n,k) = A007318(n,k) / A003989(n+1,k+1).
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2
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1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 2, 2, 2, 1, 1, 5, 10, 10, 5, 1, 1, 3, 15, 5, 15, 3, 1, 1, 7, 7, 35, 35, 7, 7, 1, 1, 4, 28, 28, 14, 28, 28, 4, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 5, 15, 30, 210, 42, 210, 30, 15, 5, 1
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OFFSET
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0,8
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COMMENTS
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Taking each row polynomial listed on p. 12 of the Alexeev et al. link and listing the GCD of each sub-polynomial in the indeterminate q gives the left half of this entry's symmetric/palindromic triangle. E.g., for k=6, 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 (1,3,15,5). See also A055151. - Tom Copeland, Jun 18 2015
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LINKS
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FORMULA
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EXAMPLE
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The triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 1 1
2: 1 1 1
3: 1 3 3 1
4: 1 2 2 2 1
5: 1 5 10 10 5 1
6: 1 3 15 5 15 3 1
7: 1 7 7 35 35 7 7 1
8: 1 4 28 28 14 28 28 4 1
9: 1 9 36 84 126 126 84 36 9 1
10: 1 5 15 30 210 42 210 30 15 5 1
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MATHEMATICA
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T[n_, k_] := Binomial[n, k]/GCD[n-k+1, k+1]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 06 2015, after R. J. Mathar *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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