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%I #47 Jul 01 2018 13:56:10
%S 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,
%T 64,64,26,4,1,1,4,38,132,196,132,38,4,1,1,5,56,256,536,536,256,56,5,1,
%U 1,5,75,450,1260,1764,1260,450,75,5,1
%N Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations.
%C Like the Narayana triangle A001263 (and unlike A152175) this triangle is symmetric.
%C The diagonal entries are 1, 1, 4, 25, 196, 1764, ... which is probably sequence A001246 - the squares of the Catalan numbers.
%C 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
%H Andrew Howroyd, <a href="/A209805/b209805.txt">Table of n, a(n) for n = 1..1275</a>
%H Tilman Piesk, <a href="http://en.wikiversity.org/wiki/Partition_related_number_triangles#rot">Partition related number triangles</a> (Wikiversity)
%H Tilman Piesk, <a href="http://pastebin.com/kx1ZidZn">Brute force MATLAB code used to calculate the rows</a> (Pastebin)
%F 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
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 1, 2, 2, 1;
%e 1, 2, 4, 2, 1;
%e 1, 3, 10, 10, 3, 1;
%e 1, 3, 15, 25, 15, 3, 1;
%e 1, 4, 26, 64, 64, 26, 4, 1;
%e 1, 4, 38, 132, 196, 132, 38, 4, 1;
%e 1, 5, 56, 256, 536, 536, 256, 56, 5, 1;
%t b[n_, k_] := Binomial[n-1, n-k] Binomial[n, n-k];
%t 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;
%t Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jul 01 2018, after _Andrew Howroyd_ *)
%o (PARI)
%o b(n,k)=binomial(n-1,n-k)*binomial(n,n-k);
%o 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
%Y Cf. A054357 (row sums), A001246 (square Catalan numbers).
%Y Cf. A001263, A103371, A132812, A209612.
%K nonn,tabl
%O 1,8
%A _Tilman Piesk_, Mar 13 2012
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