cp's OEIS Frontend

Partial sums of A077422. - R. J. Mathar, Nov 27 2011

Indices of centered pentagonal numbers (A005891) which are also centered hexagonal numbers (A003215). - Colin Barker, Feb 07 2015

The problem is to find p and r such that 6*(2*p-1)^2 = 5*(2*r+1)^2 + 1 equivalent to 3*p^2 - 3*p + 1 = (5*r^2 + 5*r + 2)/2. The Diophantine equation (6*X)^2 = 30*Y^2 + 6 is such that

X is given by 1, 21, 461, 10121, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2 - 20);

Y is given by 1, 23, 805, 11087, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2+24);

r is given by 0, 11, 252, 5543, 121704, ... with a(n+2) = 22*a(n+1) - a(n) + 10 and also a(n+1) = 11*a(n) + 5 + sqrt(120*a(n)^2 + 120*a(n) + 36);

p is given by 1, 11, 231, 5061, ... with a(n+2) = 22*a(n+1) - a(n) - 10 and also a(n+1) = 11*a(n) - 5 + sqrt(120*a(n)^2 - 120*a(n) + 25).