OFFSET
1,2
COMMENTS
These Lucas numbers L(n) have no prime factor congruent to 3 mod 4 to an odd power.
Also, numbers n such that L(n) can be written in the form a^2 + 5*b^2.
Subsequence of A124132.
Is this A124132 without the 6? - Joerg Arndt, Sep 07 2012
Any prime factor of Lucas(n) for the prime values of n is always of the form 1 (mod 10) or 9 (mod 10).
A number n can be written in the form a^2 + 5*b^2 if and only if n is 0, or of the form 2^(2i) 5^j Product_{p==1 or 9 mod 20} p^k Product_{q==3 or 7 mod 20) q^(2m) or of the form 2^(2i+1) 5^j Product_{p==1 or 9 mod 20} p^k Product_{q==3 or 7 mod 20) q^(2m+1), for integers i,j,k,m, for primes p,q.
1501 <= a(42) <= 1531. 1531, 1651, 1747, 1849, 1951, 2053, 2413, 2449, 2467, 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 W. Weisstein, Sum of squares function
EXAMPLE
Lucas(19) = 9349 = 95^2 + 18^2.
Lucas(19) = 9349 = 23^2 + 5*42^2.
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&&i%2==1, print(i", "))) \\ a^2 + b^2 form.
(PARI) for(i=2, 500, a=factorint(fibonacci(i-1)+fibonacci(i+1))~; flag=0; flip=0; for(j=1, #a, if(((a[1, j]%20>10))&&a[2, j]%2==1, flag=1); if(((a[1, j]%20==2)||(a[1, j]%20==3)||(a[1, j]%20==7))&&a[2, j]%2==1, flip=flip+1)); if(flag==0&&flip%2==0, print(i", "))) \\ a^2 + 5*b^2 form.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
V. Raman, Aug 26 2012
EXTENSIONS
17 more terms from V. Raman, Aug 28 2012
a(38)-a(41) from Chai Wah Wu, Jul 22 2020
STATUS
approved