OFFSET
1,3
COMMENTS
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.
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
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
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
LINKS
Blair Kelly, Fibonacci and Lucas factorizations
Eric Weisstein's World of Mathematics, Sum of squares function
Wikipedia, Cornacchia's algorithm
EXAMPLE
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.
MATHEMATICA
Select[Range[0, 200], SquaresR[2, LucasL[#]] > 0&] (* T. D. Noe, Aug 24 2012 *)
PROG
(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
(Python)
from itertools import count, islice
from sympy import factorint, lucas
def A124130_gen(): # generator of terms
return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(lucas(n)).items()), count(0))
CROSSREFS
KEYWORD
nonn
AUTHOR
Melvin J. Knight (melknightdr(AT)verizon.net), Nov 30 2006
EXTENSIONS
a(1)=0 and a(26)-a(45) from V. Raman, Sep 06 2012
a(46)-a(50) from Chai Wah Wu, Jul 22 2020
STATUS
approved