reviewed
approved
reviewed
approved
proposed
reviewed
editing
proposed
Integers nj > 0 such that (2n2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.
A130284 = { P[k](m) ; k=1,2,3,..., m=2,3,4,... } where P[k] = (sqrt((X^2 Q[k]^2 - 1)/(X^2 - 1))-1)/2 and Q[0] = Q[ -1] = 1, Q[k+1] = (4X^2 -2)*Q[k] - Q[k-1]. Furthermore, (2P[k](m)+1)^2 (m^2 - 1)+1 = m^2 Q[k](m)^2, thus A130280(P[k](m)) <= m. So far, no case is known where we have strict inequality.
approved
editing
reviewed
approved
proposed
reviewed
editing
proposed
r[n_] := Reduce[m>1 && k>1 && (2n+1)^2*(m^2-1)+1 == k^2, {m, k}, Integers];
Reap[For[n=1, n <= 5000, n++, If[r[n] =!= False, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, May 12 2017 *)
approved
editing
_M. F. Hasler (Maximilian.Hasler(AT)gmail.com), _, May 24 2007, May 29 2007
a(n)=2*n^2+8*n+7 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 22 2009]
n=0, a(0)=7; n=1, a(1)=17; n=2, a(2)=31 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 22 2009]
nonn,new
nonn