cp's OEIS Frontend

From Gary W. Adamson, Jul 29 2009: (Start)

Let (a,b) = roots to x^2 - 5*x + 1 = 0 = 4.79128... and 0.208712...

Then a(n) = (n + a) * (n + b). Example: a(5) = 51 = (5 + 4.79128...) * (5 + 0.208712...) (End)

For n > 0: a(n) = A176271(n+2,n). - Reinhard Zumkeller, Apr 13 2010

a(n-2) = n*(n+1) - 5, n >= 0, with a(-2) = -5 and a(-1) = -3, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 21 for b = 2*n + 1. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013

Numbers m > 0 such that 4m+21 is a square. - Bruce J. Nicholson, Jul 19 2017

Numbers represented as 151 in number base B. If 'digits' from B upwards are allowed then 151(2)=15, 151(3)=25, 151(4)=37, 151(5)=51 also. - Ron Knott, Nov 14 2017

If A and B are sequences satisfying the recurrence t(n) = 5*t(n-1) - t(n-2) with initial values A(0) = 1, A(1) = n+5 and B(0) = -1, B(1) = n, then a(n) = A(i)^2 - A(i-1)*A(i+1) = B(j)^2 - B(j-1)*B(j+1) for i, j > 0. - Klaus Purath, Oct 18 2020

The prime terms in this sequence are listed in A089376. The prime factors are given in A038893. With the exception of 3 and 7, each prime factor p divides exactly 2 out of any p consecutive terms. If a(i) and a(k) form such a pair that are divisible by p, then i + k == -5 (mod p). - Klaus Purath, Nov 24 2020