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A124132
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Positive integers n such that Fibonacci(2*n) is the sum of two squares.
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6
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1, 3, 6, 7, 13, 19, 31, 37, 43, 49, 61, 67, 73, 79, 91, 111, 127, 163, 169, 183, 199, 223, 307, 313, 349, 361, 397, 433, 511, 523, 541, 613, 619, 709, 823, 907, 1087, 1123, 1129, 1147, 1213, 1279
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OFFSET
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1,2
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COMMENTS
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This sequence excludes Fibonacci numbers with odd indices, all of which are sums of two squares.
Fibonacci numbers with even indices factor as F_2n = F_n L_n, L_n being n-th Lucas number. Thus F_2n factors further as F_2n = F_m L_m L_2m L_4m...L_n, with m odd. Since F_m is a sum of two squares and the pairs of Lucas numbers all have GCD dividing 2, the conclusion for F_2n depends on each Lucas number being a sum of two squares. Joint work with Kevin O'Bryant and Dennis Eichhorn.
To write Fibonacci(n) as a^2+b^2: find the a^2+b^2 representation for the individual prime factors, by using Cornacchia's algorithm, and then merge them by using the formulas (a^2+b^2)(c^2+d^2) = |ac+bd|^2 + |ad-bc|^2 = |ac-bd|^2 + |ad+bc|^2. - V. Raman, Aug 29 2012
All corresponding Fibonacci(2*n) values are the sum of two nonzero distinct squares except n = 1, 3, 6. - Altug Alkan, May 04 2016
1501 <= a(43) <= 1651. 1651, 1849, 2449 are terms. - Chai Wah Wu, Jul 22 2020
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LINKS
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EXAMPLE
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a(4) = 7 because the first four Fibonacci numbers with even indices that are the sum of two squares are F_2, F_6, F_12 and F_14, 14 being 2*a(4) and F_14 = 377 = 11^2+16^2.
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MATHEMATICA
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Select[Range[100], Length[FindInstance[x^2 + y^2 == Fibonacci[2 #], {x, y}, Integers]] > 0 &] (* T. D. Noe, Aug 27 2012 *)
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PROG
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(PARI) for(i=2, 500, a=factorint(fibonacci(i))~; has=0; for(j=1, #a, if(a[1, j]%4==3&&a[2, j]%2==1, has=1; break)); if(has==0&&i%2==0, print((i/2)", "))) \\ V. Raman, Aug 27 2012
(Python)
from itertools import count, islice
from sympy import factorint, fibonacci
def A124132_gen(): # generator of terms
return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(fibonacci(2*n)).items()), count(1))
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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Melvin J. Knight (melknightdr(AT)verizon.net), Nov 30 2006
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EXTENSIONS
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STATUS
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approved
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