A086541 a(1) = 1, a(2) = 4; a(n) = smallest square of the form k*a(n-1) + a(n-2), k > 0.

1, 4, 9, 49, 2116, 1104601, 1220041748025, 73506264463383837985201, 152589000107917580345020742323132226398704361, 1669769004292648133509475727812173973930459916514124965718261930975078855532062950740889

Offset

1,2

Comments

The sequence is infinite. Proof: In a(n) = k*a(n-1)+a(n-2), One ( and the largest) 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.

If k is allowed to be 0, the sequence would be 1, 4, 1, 4, ... - Chai Wah Wu, Mar 27 2020

Links

Chai Wah Wu, Table of n, a(n) for n = 1..11

Example

a(3) = 4*2 +1 = 9, a(4) = 5*9 +4 = 49, a(5) = 43*49 + 9 = 2116= 46^2.

Prog

(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

Crossrefs

Cf. A086542.

Sequence in context: A081069 A053967 A028945 * A053965 A058444 A053925

Adjacent sequences: A086538 A086539 A086540 * A086542 A086543 A086544

Keyword

nonn

Author

Amarnath Murthy, Aug 23 2003

Extensions

More terms from Rick L. Shepherd, Aug 28 2003

More terms from David Wasserman, Mar 21 2005

Status

approved