1,1

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.

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.

a(n)=2*n^2+8*n+7 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 22 2009]

Up to k=17, a(k)=P[1](k+1) with P[1]=2x^2 -1, A130280(a(k))=k+1.

a(18) = P[2](2) < P[1](19) with P[2]=2x^2 (4x^2 -3), A130280(a(18))=2.

a(106) = P[1](100) < a(107) = P[3](3) < a(108) = P[4](2) < a(109) = P[1](101).

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]

(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))) }

(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)}

Cf. A084702, A130280, A130284, A130288.

Cf. A130280, A130283, A130281.

Sequence in context: A094080 A046118 A120092 this_sequence A056220 A024840 A024835

Adjacent sequences: A130281 A130282 A130283 this_sequence A130285 A130286 A130287

nonn

M. F. Hasler (Maximilian.Hasler(AT)gmail.com), May 24 2007, May 29 2007