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A259342
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Irregular triangle read by rows: T(n,k) = number of equivalence classes of binary sequences of length n containing exactly 2k ones.
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1
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1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 4, 1, 1, 4, 8, 4, 1, 1, 4, 10, 7, 1, 1, 5, 16, 16, 5, 1, 1, 5, 20, 26, 10, 1, 1, 6, 29, 50, 29, 6, 1, 1, 6, 35, 76, 57, 14, 1, 1, 7, 47, 126, 126, 47, 7, 1, 1, 7, 56, 185, 232, 111, 19, 1, 1, 8, 72, 280, 440, 280, 72, 8, 1
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OFFSET
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1,7
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LINKS
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FORMULA
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Theorem 1 of Hoskins-Street gives an explicit formula.
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EXAMPLE
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Triangle begins:
1;
1, 1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 3, 3, 1;
1, 3, 4, 1;
1, 4, 8, 4, 1;
1, 4, 10, 7, 1;
1, 5, 16, 16, 5, 1;
1, 5, 20, 26, 10, 1;
1, 6, 29, 50, 29, 6, 1;
1, 6, 35, 76, 57, 14, 1;
1, 7, 47, 126, 126, 47, 7, 1;
...
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MAPLE
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with(numtheory):
T:= (n, k)-> (add(`if`(irem(2*k*d, n)=0, phi(n/d)
*binomial(d, 2*k*d/n), 0), d=divisors(n))
+n*binomial(iquo(n, 2), k))/(2*n):
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MATHEMATICA
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T[n_, k_] := (DivisorSum[n, If[Mod[2k*#, n]==0, EulerPhi[n/#]*Binomial[#, 2k*#/n], 0]&] + n*Binomial[Quotient[n, 2], k])/(2n); Table[T[n, k], {n, 1, 20}, { k, 0, n/2}] // Flatten (* Jean-François Alcover, Feb 28 2017, after Alois P. Heinz *)
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CROSSREFS
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Row sums are (essentially) A000011.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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