A225204 -id:A225204 - cp's OEIS frontend

A331692 Continued fraction for square root of golden ratio.

1, 3, 1, 2, 11, 3, 5, 1, 10, 1, 2, 5, 3, 5, 1, 2, 4, 1, 1, 2, 4, 51, 1, 4, 2, 2, 31, 1, 3, 1, 1, 5, 1, 1, 1, 14, 1, 1, 4, 2, 2, 8, 1, 1, 3, 23, 1, 1, 4, 16, 2, 1, 2, 13, 2, 1, 1, 1, 3, 1, 1, 4, 3, 2, 1, 1, 36, 1, 2, 1, 1, 1, 2, 3, 2, 1, 3, 3, 31, 2, 1, 2, 2, 2

Offset: 0

Jordan Paschke, Mar 03 2020

			1 + 1/(3 + 1/(1 + 1/(2 + 1/(11 + ...)))) = sqrt(phi).

Kevin Ryde, Table of n, a(n) for n = 0..10000
Kevin Ryde, PARI/GP Code

Cf. A001622, A139339 (decimal expansion).

Cf. A225204/A225205 (convergents).

ContinuedFraction[Sqrt[(1 + Sqrt[5])/2], 1000]

contfrac(sqrt((1+sqrt(5))/2)) \\ Michel Marcus, Mar 04 2020

PARI
```
\\ See Ryde link
```

A225205 Denominators of convergents to the square root of the golden ratio.

Original entry on oeis.org

1, 3, 4, 11, 125, 386, 2055, 2441, 26465, 28906, 84277, 450291, 1435150, 7626041, 9061191, 25748423, 112054883, 137803306, 249858189, 637519684, 2799936925, 143434302859, 146234239784, 728371261995, 1602976763774, 3934324789543, 123567045239607

Offset: 0

Eric W. Weisstein, May 01 2013

			1, 4/3, 5/4, 14/11, 159/125, 491/386, 2614/2055, 3105/2441, 33664/26465, ... = A225204/A225205

I. J. Good, Complex Fibonacci and Lucas Numbers, Continued Fractions, and the Square Root of the Golden Ratio (Condensed Version), Journal of the Operational Research Society, 43 (1992), 837-842.
I. J. Good, Complex Fibonacci and Lucas Numbers, Continued Fractions, and the Square Root of the Golden Ratio, The Fibonacci Quarterly 31.1 (1993):7-20.

Cf. A001622, A139339, A331692, A225204 (numerators).

Denominator[Convergents[Sqrt[GoldenRatio], 20]]

a(n) = A331692(n)*a(n-1) + a(n-2) for n >= 2. - Jianing Song, Aug 18 2022

A354513 The numbers whose square's position in the Wythoff array is immediately followed by another square in the next column.

Original entry on oeis.org

11, 386, 2441, 25748423, 637519684, 2799936925, 3934324789543, 127501370029150, 21274660147684109, 644571595359295797, 15845190736671957299, 995980378496501932493, 47375682236837399943653, 213688560255016550712685, 28372206851301867342910959, 3120729065082950391169492805

Offset: 1

Chittaranjan Pardeshi, Aug 16 2022

nonn

From Jianing Song, Aug 21 2022: (Start)
Numbers k > 0 such that floor((k^2+1)*phi) - 1 is a square, phi = A001622.
Suppose that k is a term and that floor((k^2+1)*phi) = m^2+1, then (m^2+1)/(k^2+1) < phi < (m^2+2)/(k^2+1), so |sqrt(phi) - m/k| < max{m/k - sqrt((m^2+1)/(k^2+1)), sqrt((m^2+2)/(k^2+1)) - m/k} = m/k - sqrt((m^2+1)/(k^2+1)) <= sqrt((k^2+1)*phi-1)/k - sqrt(phi) < 1/(2*sqrt(phi)*k^2). According to the Mathematics Stack Exchange link, m/k is a convergent to sqrt(phi), so this is a subsequence of A225205. The terms are b(3), b(5), b(11), b(15), b(19), b(20), ... for b = A225205.
For k = A225205(r), m = A225204(r), we have |sqrt(phi) - m/k| < 1/(k*A225205(r+1)) (by Theorem 5 of the Wikipedia link), so k = A225205(r) is a term if 1/(k*A225205(r+1)) < min{m/k - sqrt((m^2+1)/(k^2+1)), sqrt((m^2+2)/(k^2+1)) - m/k} = sqrt((m^2+2)/(k^2+1)) - m/k, or A225205(r+1) > (k*sqrt((m^2+2)/(k^2+1)) - m)^(-1).
If k = A225205(r) is a term with even r, then k is also in A354549, since m^2 < k^2*phi < k^2*(m^2+2)/(k^2+1) < m^2+phi^(-2) for m = A225204(r), so floor(k^2*phi) = m^2. Furthermore we have {k^2*phi} < phi^(-2), where {} denotes the fractional part. Conversely, if k is in A354549 and {k^2*phi} < phi^(-2), then k is in this sequence since floor((k^2+1)*phi) = floor(k^2*phi)+1 in this case. (End)

			11 is a term since 11^2 = 121 has another square, 196 = 14^2, immediately to its right in the Wythoff array. Array row: 46, 75, 121, 196, ...

Jianing Song, Table of n, a(n) for n = 1..127
Mathematics Stack Exchange, If |x - p/q| < 1/(2*q^2) then p/q is necessarily one of the convergents
Wikipedia, Continued fraction

Cf. A035513, A026274, A352538, A001622, A225204, A225205, A354549.

phi=quadgen(5);
nextcolumn(x) = ((x+1)*phi-1)\1; \\ A026274(x+1)
for(i=1, 10000000000, if ( issquare( nextcolumn (i^2)), print1(i, ", ")));

A000201(n) = (n+sqrtint(5*n^2))\2;
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(A000201(conv[2, n]^2+1) == conv[1, n]^2+1, print1(conv[2, n], ", "))) \\ Jianing Song, Aug 21 2022, modified on Aug 28 2022 according to Kevin Ryde's program for A331692

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

Original entry on oeis.org

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

Offset: 1

Jianing Song, Aug 18 2022

nonn

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.

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

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

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

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

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

A356591 Numbers k such that A225205(k) is in A354513.

Original entry on oeis.org

3, 5, 7, 15, 19, 20, 25, 27, 34, 37, 40, 44, 47, 48, 52, 57, 65, 77, 89, 91, 92, 100, 105, 107, 111, 121, 123, 126, 127, 129, 138, 141, 153, 163, 165, 167, 171, 173, 179, 182, 183, 185, 189, 193, 195, 202, 205, 209, 211, 213, 215, 222, 224, 226, 230, 232, 234, 236, 238

Offset: 1

Jianing Song, Aug 21 2022

nonn

Numbers k such that floor((A225205(k)^2+1)*phi) = A225204(k)^2+1, phi = A001622.
Numbers k such that (A225204(k)^2+1)/(A225205(k)^2+1) < phi < (A225204(k)^2+2)/(A225205(k)^2+1).
Conjecture: the odd numbers (numbers k such that A225204(k)/A225205(k) > sqrt(phi)) have relative density phi^(-1), and the even numbers (number k such that A225204(k)/A225205(k) < sqrt(phi)) have relative density phi^(-2). It is conjectured so because we have lim_{k->+oo} (m/k - sqrt((m^2+1)/(k^2+1)))/(sqrt((m^2+2)/(k^2+1)) - m/k) = phi if m/k -> sqrt(phi).
Even k is a term if and only floor(A225205(k)^2*phi) = A225204(k)^2 (k is in A356664) and {A225205(k)^2*phi} < phi^(-2), where {} denotes the fractional part; see the comments in A354513.

			3 is a term because A225204(3) = 14 and A225205(3) = 11, and floor((11^2+1)*phi) = 14^2+1.

Jianing Song, Table of n, a(n) for n = 1..283

Cf. A001622, A225204, A225205, A354513, A356664.

A000201(n) = (n+sqrtint(5*n^2))\2;
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(A000201(conv[2, n]^2+1) == conv[1, n]^2+1, print1(n-1, ", ")))
\\ Here conv[1, n] = A225204(n-1), conv[2, n] = A225205(n-1)
\\ Modified by Jianing Song, Aug 28 2022 according to Kevin Ryde's program for A331692

A354513(n) = A225205(a(n)).

A356664 Numbers k such that A225205(k) is in A354549.

Original entry on oeis.org

0, 2, 4, 10, 12, 14, 18, 20, 22, 30, 32, 34, 38, 40, 44, 48, 52, 60, 62, 72, 76, 78, 80, 82, 92, 94, 100, 104, 116, 120, 126, 130, 132, 134, 138, 140, 142, 144, 146, 148, 152, 154, 156, 158, 160, 168, 176, 180, 182, 186, 188, 192, 194, 202, 210, 222, 224, 226, 228, 230, 232

Offset: 1

Jianing Song, Aug 21 2022

nonn

Numbers k such that floor(A225205(k)^2*phi) = A225204(k)^2, phi = A001622.
Even numbers k such that (A225204(k)^2+1)/A225205(k)^2 > phi.
Even k is a term in A356591 if and only k is in this sequence and {A225205(k)^2*phi} < phi^(-2), where {} denotes the fractional part; see the comments in A354513.
Conjecture: this and A356591 have the same natural density.

			4 is a term because A225204(4) = 125 and A225205(4) = 159, and floor(125^2*phi) = 159^2.

Jianing Song, Table of n, a(n) for n = 1..293

Cf. A001622, A225204, A225205, A354549, A354513, A356591.

print1("0, "); 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(n-1, ", ")))
\\ 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

A354549(n+1) = A225205(a(n)).

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A331692 Continued fraction for square root of golden ratio.

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A225205 Denominators of convergents to the square root of the golden ratio.

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A354513 The numbers whose square's position in the Wythoff array is immediately followed by another square in the next column.

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

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A356591 Numbers k such that A225205(k) is in A354513.

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A356664 Numbers k such that A225205(k) is in A354549.

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