OFFSET
1,8
COMMENTS
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
Andrew Howroyd, Table of n, a(n) for n = 1..1275
Tilman Piesk, Partition related number triangles (Wikiversity)
Tilman Piesk, Brute force MATLAB code used to calculate the rows (Pastebin)
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
KEYWORD
nonn,tabl
AUTHOR
Tilman Piesk, Mar 13 2012
STATUS
approved