OFFSET
1,1
COMMENTS
These Lucas numbers L(n) have no prime factor congruent to 3 mod 4 to an odd power.
Also prime numbers n such that the Lucas number L(n) can be written in the form a^2 + 5*b^2.
Any prime factor of Lucas(n) for n prime 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 (see A020669) if and only if n is 0,
or of the form 2^(2i) 5^j Prod_{p==1 or 9 mod 20} p^k Prod_{q==3 or 7 mod 20) q^(2m)
or of the form 2^(2i+1) 5^j Prod_{p==1 or 9 mod 20} p^k Prod_{q==3 or 7 mod 20) q^(2m+1),
for integers i,j,k,m, for primes p,q.
1607 <= a(34) <= 1747. 1747, 1951, 2053, 2467, 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) forprime(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", "))) \\ a^2+b^2 form.
(PARI) forprime(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
AUTHOR
V. Raman, Aug 23 2012
EXTENSIONS
a(30)-a(33) from Chai Wah Wu, Jul 22 2020
STATUS
approved