OFFSET
1,7
COMMENTS
Conjecture: a(n) <= 2, empirically checked for the first 10^6 primes.
The conjecture is true, because by the uniqueness part of Fermat's two-squares theorem, at most one duplicate of a^2 + b^4 can exist. Namely, if a is a square, say a = B^2, then a^2 + b^4 = A^2 + B^4 where A = b^2. - Jonathan Sondow, Oct 03 2015
Friedlander and Iwaniec proved that a(n) > 0 infinitely often. - Jonathan Sondow, Oct 05 2015
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Art of Problem Solving, Fermat's Two Squares Theorem
John Friedlander and Henryk Iwaniec, Using a parity-sensitive sieve to count prime values of a polynomial, PNAS, vol. 94 no. 4, pp. 1054-1058.
Wikipedia, Friedlander-Iwaniec theorem
EXAMPLE
First numbers n, such that a(n) > 0:
. k | n | prime(n) | a(n)
. ----+----+-------------------------------+-----
. 1 | 1 | 2 = 1^2 + 1^4 | 1
. 2 | 3 | 5 = 2^2 + 1^4 | 1
. 3 | 7 | 17 = 1^2 + 2^4 = 4^2 + 1^4 | 2
. 4 | 12 | 37 = 6^2 + 1^4 | 1
. 5 | 13 | 41 = 5^2 + 2^4 | 1
. 6 | 25 | 97 = 4^2 + 3^4 = 9^2 + 2^4 | 2
. 7 | 33 | 101 = 10^2 + 1^4 | 1
. 8 | 42 | 181 = 10^2 + 3^4 | 1
. 9 | 45 | 197 = 14^2 + 1^4 | 1
. 10 | 53 | 241 = 15^2 + 2^4 | 1
. 11 | 55 | 257 = 1^2 + 4^4 = 16^2 + 1^4 | 2
. 12 | 59 | 277 = 14^2 + 3^4 | 1
. 13 | 60 | 281 = 5^2 + 4^4 | 1
. 14 | 68 | 337 = 9^2 + 4^4 = 16^2 + 3^4 | 2
. 15 | 79 | 401 = 20^2 + 1^4 | 1
. 16 | 88 | 457 = 21^2 + 2^4 | 1 .
PROG
(Haskell)
a256852 n = a256852_list !! (n-1)
a256852_list = f a000040_list [] $ tail a000583_list where
f ps'@(p:ps) us vs'@(v:vs)
| p > v = f ps' (v:us) vs
| otherwise = (sum $ map (a010052 . (p -)) us) : f ps us vs'
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Apr 11 2015
STATUS
approved