OFFSET
1,3
COMMENTS
Numbers k such that there exists m such that m^2 <= k^2*phi < m^2+1. For k > 0, m/k is a lower convergent to sqrt(phi) since m/k < sqrt(phi) < sqrt(m^2+1)/k, so |sqrt(phi) - m/k| < (sqrt(m^2+1)-m)/k < 1/(2*k^2) (see the Mathematics Stack Exchange link). As a result, this is a subsequence of {A225205(2*n): n>=0}. The terms > 0 are b(0), b(2), b(4), b(10), b(12), b(14), ... for b = A225205. Note that sqrt(phi) - A225204(2*r)/A225205(2*r) < 1/(A225205(2*r)*A225205(2*r+1)) (by Theorem 5 of the Wikipedia link), so A225205(2*r) is a term if sqrt(A225204(2*r)^2+1) + A225204(2*r) < A225205(2*r+1).
Consider the numbers s such that A035513(s,0) = A000201(s) = floor(s*phi) and A035513(s,1) = A003622(s) = floor(floor(s*phi)*phi) are both squares. If s has this property, then clearly floor(s*phi) is the square of a term. However, for k > 0 being a term, k^2 is not always in A000201. This happens if and only if {k^2*phi} < phi^(-2), where {} denotes the fractional part (see A000201); for example k = 2799936925, for which floor(k^2*phi) = 3561574786^2. In this case, floor(k^2*phi) is not in A003622 since n -> floor(n*phi) is an injection. Suppose that A035513(p,q) = floor(k^2*phi) for p >= 1, q >= 2, then A035513(p,q-1) = floor(k^2*phi) since floor((k^2+1)*phi) = floor(k^2*phi)+1 in this case.
LINKS
Jianing Song, Table of n, a(n) for n = 1..140
Mathematics Stack Exchange, If |x - p/q| < 1/(2*q^2) then p/q is necessarily one of the convergents
Wikipedia, Continued fraction
EXAMPLE
4 is a term since floor(4^2*phi) = 5^2.
125 is a term since floor(125^2*phi) = 159^2.
PROG
(PARI) print1("0, 1, "); my(cofr=A331692_vector_bits(1000), conv=matrix(2, #cofr)); conv[, 1]=[1, 1]~; conv[, 2]=[4, 3]~; for(n=3, #cofr, conv[, n]=cofr[n]*conv[, n-1]+conv[, n-2]; if(n%2 == 1 && (conv[1, n]^2+1)^2 - (conv[1, n]^2+1)*(conv[2, n]^2) - (conv[2, n]^2)^2 > 0, print1(conv[2, n], ", ")))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Aug 18 2022
STATUS
approved