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A130284 Integers j > 0 such that (2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.

%I A130284 #13 Jun 13 2021 10:19:20
%S A130284 7,17,31,49,71,97,104,127,161,199,241,287,337,391,449,511,577,594,647,
%T A130284 721,799,881,967,1057,1151,1249,1351,1455,1457,1567,1681,1799,1921,
%U A130284 1952,2047,2177,2311,2449,2591,2737,2887,3041,3199,3361,3527,3697,3871,4049
%N A130284 Integers j > 0 such that (2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.
%C A130284 All terms > 4 in A130283 are odd squares, but not all odd squares are in that sequence: This sequence here gives the exceptions as (2a(n)+1)^2. The sequence consists mainly of the subsequences: (1) A056220(k) = 2k^2-1 with k>1: {7,17,31,49,...}, for which m=k gives (1+2*A056220(k))^2(k^2-1)+1 = k^2(4k^2-3)^2; (2) 2*A079414(k) = 2k^2(4k^2-3) with k>1: {104,594,1952,4850,...}, for which m=k gives (1+4*A079414(k))^2(k^2-1)+1 = k^2(16k^4-20k^2+5)^2. A third subsequence starts {1455,20195,...}; up to 20195, all terms are in one of these subsequences.
%F A130284 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.
%e A130284 Up to k=17, a(k)=P[1](k+1) with P[1] = 2x^2 - 1, A130280(a(k)) = k+1.
%e A130284 a(18) = P[2](2) < P[1](19) with P[2] = 2x^2*(4x^2 - 3), A130280(a(18)) = 2.
%e A130284 a(106) = P[1](100) < a(107) = P[3](3) < a(108) = P[4](2) < a(109) = P[1](101).
%t A130284 r[n_] := Reduce[m>1 && k>1 && (2n+1)^2*(m^2-1)+1 == k^2, {m, k}, Integers];
%t A130284 Reap[For[n=1, n <= 5000, n++, If[r[n] =!= False, Print[n]; Sow[n]]]][[2,1]] (* _Jean-François Alcover_, May 12 2017 *)
%o A130284 (PARI) A130284( LIM=9999, START=1 )={ local(N); for( n=START, LIM, N=(2*n+1)^2; for( m=2, sqrtint(n>>1+1), if(!issquare( N*(m^2-1)+1 ), next); print1(n", "); next(2))) }
%o A130284 (PARI) {Q(k,x=x)=if(m>0,(4*x^2-2)*Q(k-1,x)-Q(k-2,x),1)} {P(k,x=x)=if(type(x=(x^2*Q(k,x)^2-1)/(x^2-1))!="t_POL",sqrtint(x)\2,((-1)^k*Pol(sqrt(x))-1)/2)}
%Y A130284 Cf. A084702, A130280, A130284, A130288.
%Y A130284 Cf. A130280, A130283, A130281.
%K A130284 nonn
%O A130284 1,1
%A A130284 _M. F. Hasler_, May 24 2007, May 29 2007