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%I #40 Mar 02 2017 15:09:10
%S 0,1,1,1,2,1,2,1,2,2,3,2,3,1,4,3,4,1,4,4,3,2,4,1,8,4,4,3,6,3,5,3,4,4,
%T 9,3,8,4,6,6,9,2,7,4,7,5,7,3,5,7,7,6,9,4,14,4,8,4,9,4,11,7,7,6,17,5,
%U 11,6,10,8,9,5,11,6,9,7,8,3,13,9,9,5,15,5,20,8,11,8,14,7,13,9,8,6,18,7,14,10,10,8
%N Number of partitions n = x + y with y >= x > 0 such that x^2 + y^2 or (x^2 + y^2)/2 is prime.
%C Conjecture: a(n) > 0 for all n > 1.
%C We have a(n) <= phi(n)/2 for n <> 2, because must be gcd(x,y) = 1.
%C Numbers n such that a(n) = phi(n)/2 are 3, 4, 5, 6, 10, 12, 15, and 20.
%C Record values of a(n) are for n = 1, 2, 5, 11, 15, 25, 35, 55, 65, 85, 125, 145, 185, 205, 215, 235, 265, 295, 325, 365, 415, ... cf. A001750.
%H Charles R Greathouse IV, <a href="/A281543/b281543.txt">Table of n, a(n) for n = 1..10000</a>
%H Altug Alkan, <a href="/A281543/a281543.png">Alternative Scatterplot of A281543</a>
%F a(2m+1) = A036468(m) for m > 0.
%F a(2m) = A069004(m) for m > 1.
%F a(n) = O(n/log(n)).
%e a(5) = 2 because 5 = 1 + 4 and 5 = 2 + 3 are only options; 1^2 + 4^2 = 17 and 2^2 + 3^2 = 13 are primes.
%e a(6) = 1 because 6 = 1 + 5 is only option; (1^2 + 5^2)/2 = 13 is prime.
%e a(7) = 2 because 7 = 1 + 6, 7 = 2 + 5 and 7 = 3 + 4, but 3^2 + 4^2 = 5^2.
%o (PARI) a(n) = if(n==2, 1, if(n%2==0, sum(k=1, n/2-1, isprime(n^2/4+k^2)), sum(k=1, (n-1)/2, isprime(k^2+(n-k)^2))));
%Y Cf. A000010, A002313, A036468, A069004.
%K nonn
%O 1,5
%A _Thomas Ordowski_ and _Altug Alkan_, Mar 01 2017
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