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A354513
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The numbers whose square's position in the Wythoff array is immediately followed by another square in the next column.
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3
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11, 386, 2441, 25748423, 637519684, 2799936925, 3934324789543, 127501370029150, 21274660147684109, 644571595359295797, 15845190736671957299, 995980378496501932493, 47375682236837399943653, 213688560255016550712685, 28372206851301867342910959, 3120729065082950391169492805
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OFFSET
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1,1
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COMMENTS
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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)
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LINKS
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EXAMPLE
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11 is 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, ...
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PROG
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(PARI)
phi=quadgen(5);
nextcolumn(x) = ((x+1)*phi-1)\1; \\ A026274(x+1)
for(i=1, 10000000000, if ( issquare( nextcolumn (i^2)), print1(i, ", ")));
(PARI) 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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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