

A124130


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


4



0, 1, 3, 6, 7, 13, 19, 30, 31, 37, 43, 49, 61, 67, 73, 78, 79, 91, 111, 127, 150, 163, 169, 183, 199, 223, 307, 313, 349, 361, 390, 397, 433, 511, 523, 541, 606, 613, 619, 709, 750, 823, 907, 1087, 1123
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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 + adbc^2 = acbd^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


LINKS

Table of n, a(n) for n=1..45.
Blair Kelly, Fibonacci and Lucas factorizations
Eric W. Weisstein, 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(i1)+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


CROSSREFS

Intersection of A000032 or A000204 = Lucas numbers and A001481.
Cf. A215809, A215906, A215907.
Sequence in context: A107850 A216514 A051218 * A124132 A064291 A245394
Adjacent sequences: A124127 A124128 A124129 * A124131 A124132 A124133


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


STATUS

approved



