|
%I #15 May 31 2017 10:39:27
%S 1,1,1,2,1,3,1,1,2,1,3,1,1,2,5,4,2,1,2,1,3,4,2,2,5,4,1,1,2,3,5,3,1,2,
%T 6,3,1,5,4,5,4,1,2,2,1,4,1,2,2,3,3,2,5,7,1,3,3,1,2,1,4,1,1,4,1,4,1,2,
%U 2,5,3,3,1,2,1,5,4,1,5,1,3,2,10,2,1,3,6,1,2,1,4,1,5,10,3,3,2,10,7
%N Least k>0 such that k^2 + (2n+1-k)^2 is prime, or 0 if no such k exists.
%C It appears that any odd number N = 2n+1 > 1 (and also N = 2, hence all primes, see A260869) can be written as the sum of two positive integers such that the sum of their squares is prime. For an even number > 2 this is obviously not possible since k and 2n-k have the same parity and therefore the sum of their squares is even.
%C The record values 1, 2, 3, 5, 6, 7, 10, 13, 16, 29, 30, 37, 40, 41, 49, 55, 64, 67, 68, 74, 85, 88, 106, 128, ... occur for indices n (half of the odd numbers 2n+1) 1, 4, 6, 15, 35, 54, 83, 121, 172, 281, 936, 1093, 1150, 1240, 3121, 4126, 5116, 6793, 11935, 12556, 18238, 32710, 33343, 57256, ...
%H Harvey P. Dale, <a href="/A260870/b260870.txt">Table of n, a(n) for n = 1..1000</a>
%t lk[n_]:=Module[{k=1},While[!PrimeQ[k^2+(2n+1-k)^2],k++];k]; Array[lk,100] (* _Harvey P. Dale_, May 31 2017 *)
%o (PARI) A260870(n)=for(k=1,(n=2*n+1)\2,isprime(k^2+(n-k)^2)&&return(k))
%K nonn
%O 1,4
%A _M. F. Hasler_, Aug 09 2015
|
