A354549 Numbers k such that floor(k^2*phi) is a square, where phi = A001622 is the golden ratio.

0, 1, 4, 125, 84277, 1435150, 9061191, 249858189, 2799936925, 146234239784, 1139643680683264, 7471434609455791, 21274660147684109, 2911209509190673141, 15845190736671957299, 995980378496501932493, 213688560255016550712685, 28372206851301867342910959

Offset: 1

Author

Jianing Song, Aug 18 2022

Keywords

• nonn

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.

Examples

			4 is a term since floor(4^2*phi) = 5^2.
125 is a term since floor(125^2*phi) = 159^2.
		

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

Crossrefs

Cf. A001622, A000201, A003622, A035513, A225204, A225205.

Programs

PARI

A000201(n) = (n+sqrtint(5*n^2))\2; isok(n) = issquare(A000201(n^2))

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], ", ")))
\\ Here conv[1,n] = A225204(n-1), conv[2,n] = A225205(n-1), n odd implies conv[1,n]/conv[2,n] < sqrt((1+sqrt(5))/2); let A = conv[1,n]^2+1, B = conv[2,n]^2, then A^2 - A*B - B^2 > 0 implies A/B > (1+sqrt(5))/2
\\ Modified by Jianing Song, Aug 28 2022 according to Kevin Ryde's program for A331692