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A209805
Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations.
5
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
OFFSET
1,8
COMMENTS
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
LINKS
FORMULA
T(n,k) = (1/n)*((Sum_{d|gcd(n,k)} phi(d)*A103371(n/d-1,k/d-1)) + (Sum_{d|gcd(n,k-1)} phi(d)*A103371(n/d-1,(n+1-k)/d-1)) - A132812(n,k)). - Andrew Howroyd, Nov 15 2017
EXAMPLE
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;
MATHEMATICA
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;
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 01 2018, after Andrew Howroyd *)
PROG
(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
CROSSREFS
Cf. A054357 (row sums), A001246 (square Catalan numbers).
Sequence in context: A339788 A122085 A209612 * A238453 A376991 A066287
KEYWORD
nonn,tabl
AUTHOR
Tilman Piesk, Mar 13 2012
STATUS
approved