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A209612
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Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations and reflections.
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6
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1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 8, 8, 3, 1, 1, 3, 12, 17, 12, 3, 1, 1, 4, 19, 41, 41, 19, 4, 1, 1, 4, 27, 78, 116, 78, 27, 4, 1, 1, 5, 38, 148, 298, 298, 148, 38, 5, 1, 1, 5, 50, 250, 680, 932, 680, 250, 50, 5, 1
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OFFSET
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1,8
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COMMENTS
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Like the Narayana triangle A001263 (and unlike A152176) this triangle is symmetric.
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins:
1;
1, 1;
1, 1, 1;
1, 2, 2, 1;
1, 2, 4, 2, 1;
1, 3, 8, 8, 3, 1;
1, 3, 12, 17, 12, 3, 1;
1, 4, 19, 41, 41, 19, 4, 1;
1, 4, 27, 78,116, 78, 27, 4, 1;
1, 5, 38,148,298,298,148, 38, 5, 1
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MATHEMATICA
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b[n_, k_] := Binomial[n - 1, n - k]*Binomial[n, n - k];
T[n_, k_] := (n*Binomial[Quotient[n - 1, 2], Quotient[k - 1, 2]]*Binomial[ Quotient[n, 2], Quotient[k, 2]] + DivisorSum[GCD[n, k], EulerPhi[#]* b[n/#, k/#]&] + DivisorSum[GCD[n, k - 1], EulerPhi[#]*b[n/#, (n + 1 - k)/#]&] - k*Binomial[n, k]^2/(n - k + 1))/(2*n);
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PROG
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(PARI)
b(n, k)=binomial(n-1, n-k)*binomial(n, n-k);
T(n, k)=(n*binomial((n-1)\2, (k-1)\2)*binomial(n\2, k\2) + sumdiv(gcd(n, k), d, eulerphi(d)*b(n/d, k/d)) + sumdiv(gcd(n, k-1), d, eulerphi(d)*b(n/d, (n+1-k)/d)) - k*binomial(n, k)^2/(n-k+1))/(2*n); \\ Andrew Howroyd, Nov 15 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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