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A082907
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A modified Pascal's triangle, read by rows, and modified as follows: binomial(n,j) is replaced by gcd(2^n, binomial(n,j)), i.e., the largest power of 2 dividing binomial(n,j).
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10
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1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 2, 4, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 4, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 4, 8, 2, 8, 4, 8, 1, 1, 1, 4, 4, 2, 2, 4, 4, 1, 1, 1, 2, 1, 8, 2, 4, 2, 8, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 4, 2, 4, 1, 8, 4, 8, 1, 4, 2, 4, 1, 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1
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OFFSET
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0,5
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COMMENTS
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If N is a power of 2, then the first N rows are invariant under all 6 symmetries of an equilateral triangle. - Paul Boddington, Dec 17 2003
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LINKS
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Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
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FORMULA
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T(n, j) = c(n)/(c(j)*c(n-j)) where c(n)=A060818(n).
T(n, j) = (b(j)*b(n-j))/b(n) where b(n)=A001316(n) (Gould's sequence). (End)
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EXAMPLE
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Triangle read by rows:
1,
1,1,
1,2,1,
1,1,1,1,
1,4,2,4,1,
1,1,2,2,1,1,
1,2,1,4,1,2,1,
1,1,1,1,1,1,1,1,
1,8,4,8,2,8,4,8,1,
1,1,4,4,2,2,4,4,1,1,
...
For n = -1 + 2^k, such rows consist of all 1's since all binomial coefficients C(n,j) are odd.
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MATHEMATICA
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Flatten[Table[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}], {n, 0, 25}], 1]
f[n_] := Denominator[CatalanNumber[n - 1]/2^(n - 1)]; T[n_, k_] := f[n]/(f[k]*f[n - k]); Table[T[n, k], {n, 0, 7}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 24 2016 *)
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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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