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A209805
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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.
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5
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1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 10, 10, 3, 1, 1, 3, 15, 25, 15, 3, 1, 1, 4, 26, 64, 64, 26, 4, 1, 1, 4, 38, 132, 196, 132, 38, 4, 1, 1, 5, 56, 256, 536, 536, 256, 56, 5, 1, 1, 5, 75, 450, 1260, 1764, 1260, 450, 75, 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 A152175) this triangle is symmetric.
The diagonal entries are 1, 1, 4, 25, 196, 1764, ... which is probably sequence A001246 - the squares of the Catalan numbers.
The above conjecture about the diagonal entries T(2*n-1, n) is true since gcd(2*n-1, n) = gcd(2*n-1, n-1) = 1 and then T(2*n-1, n) simplifies to A001246(n-1) using the formula given below. - Andrew Howroyd, Nov 15 2017
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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, 10, 10, 3, 1;
1, 3, 15, 25, 15, 3, 1;
1, 4, 26, 64, 64, 26, 4, 1;
1, 4, 38, 132, 196, 132, 38, 4, 1;
1, 5, 56, 256, 536, 536, 256, 56, 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_] := (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))/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)=(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))/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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