%I #18 Aug 25 2024 09:56:40
%S 1,4,9,49,2116,1104601,1220041748025,73506264463383837985201,
%T 152589000107917580345020742323132226398704361,
%U 1669769004292648133509475727812173973930459916514124965718261930975078855532062950740889
%N a(1) = 1, a(2) = 4; a(n) = smallest square of the form k*a(n-1) + a(n-2), k > 0.
%C The sequence is infinite. Proof: In a(n) = k*a(n-1)+a(n-2), one value of k is a(n-1)-2*a(n-2)^(1/2), which gives a(n) <= (a(n-1)-a(n-2)^(1/2))^2.
%C If k is allowed to be 0, the sequence would be 1, 4, 1, 4, ... - _Chai Wah Wu_, Mar 27 2020
%H Chai Wah Wu, <a href="/A086541/b086541.txt">Table of n, a(n) for n = 1..11</a>
%e a(3) = 4*2 +1 = 9, a(4) = 5*9 +4 = 49, a(5) = 43*49 + 9 = 2116= 46^2.
%o (PARI) A = vector(11); A[1] = 1; A[2] = 2; B = vector(11, i, A[i]^2); for (n = 3, 11, z = znstar(B[n - 1]); l = length(z[2]); c = vector(l, i, z[3][i]^(z[2][i]/2)); v = vector(2^l, i, A[n - 2]*prod(j = 1, l, c[j]^(i\2^(l - j)%2))); v = vecsort(lift(v)); print(B[n - 2],vector(2^l, i, v[i]^2%B[n - 1])); A[n] = if (v[1] == A[n - 2], v[2], v[1]); B[n] = A[n]^2; print(B[n])); \\ _David Wasserman_, Mar 21 2005
%Y Cf. A086542.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Aug 23 2003
%E More terms from _Rick L. Shepherd_, Aug 28 2003
%E More terms from _David Wasserman_, Mar 21 2005