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Numbers n such that L_n = a^2 + b^2, where L_n is the n-th Lucas number with a and b integers.
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%I #41 Jun 27 2022 18:50:12

%S 0,1,3,6,7,13,19,30,31,37,43,49,61,67,73,78,79,91,111,127,150,163,169,

%T 183,199,223,307,313,349,361,390,397,433,511,523,541,606,613,619,709,

%U 750,823,907,1087,1123,1129,1147,1213,1279,1434

%N Numbers n such that L_n = a^2 + b^2, where L_n is the n-th Lucas number with a and b integers.

%C Congruence considerations eliminate many indices, but the remaining numbers were factored. These have no prime factors of the form p=4m+3 dividing them to an odd power. Joint work with Kevin O'Bryant and Dennis Eichhorn.

%C To write Lucas(n) as a^2+b^2: find the a^2+b^2 representation for the individual prime factors by Cornacchia's algorithm, and 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_, Oct 04 2012

%C Values of A000032(n) such that A000032(n) or A000032(n)/2 is a square are only 1, 2, 4, 18. So a and b must be distinct and nonzero for all values of this sequence except 0, 1, 3, 6. - _Altug Alkan_, May 04 2016

%C 1501 <= a(51) <= 1531. 1531, 1651, 1747, 1758, 1849, 1950, 1951, 2053, 2413, 2449, 2467, 3030, 4069, 5107, 5419, 5851, 7243, 7741, 8467, 13963, 14449, 14887, 15511, 15907, 35449, 51169, 193201, 344293, 387433, 574219, 901657, 1051849 are terms. - _Chai Wah Wu_, Jul 22 2020

%H Blair Kelly, <a href="http://mersennus.net/fibonacci">Fibonacci and Lucas factorizations</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SumofSquaresFunction.html">Sum of squares function</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Cornacchia%27s_algorithm">Cornacchia's algorithm</a>

%e a(5) = 13 because the first five Lucas numbers that are the sum of two squares are L_1, L_3, L_6, L_7 and L_13 = 521 = 11^2 + 20^2.

%t Select[Range[0, 200], SquaresR[2, LucasL[#]] > 0&] (* _T. D. Noe_, Aug 24 2012 *)

%o (PARI) for(i=2, 500, a=factorint(fibonacci(i-1)+fibonacci(i+1))~; has=0; for(j=1, #a, if(a[1, j]%4==3&&a[2, j]%2==1, has=1; break)); if(has==0, print(i", "))) \\ _V. Raman_, Aug 23 2012

%o (Python)

%o from itertools import count, islice

%o from sympy import factorint, lucas

%o def A124130_gen(): # generator of terms

%o return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(lucas(n)).items()),count(0))

%o A124130_list = list(islice(A124130_gen(),20)) # _Chai Wah Wu_, Jun 27 2022

%Y Intersection of A000032 or A000204 = Lucas numbers and A001481.

%Y Cf. A215809, A215906, A215907.

%K nonn

%O 1,3

%A Melvin J. Knight (melknightdr(AT)verizon.net), Nov 30 2006

%E a(1)=0 and a(26)-a(45) from _V. Raman_, Sep 06 2012

%E a(46)-a(50) from _Chai Wah Wu_, Jul 22 2020