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%I #12 Oct 27 2014 08:46:24
%S 1,1,0,1,1,1,0,0,1,1,1,1,0,1,1,1,0,0,1,1,0,1,1,1,1,1,1,1,1,0,0,0,1,1,
%T 1,1,1,0,1,1,0,1,0,1,1,0,0,1,1,0,1,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,0,
%U 1,0,1,0,0,1,0,0,1,1,1,1,1,1,0,1,1,1,1
%N a(n) is the number of new prime distinct divisors of n^2+1 not already present in m^2+1 for all m < n.
%C a(n) = 0 or 1. Proof:
%C If n^2+1 is composite, it is always possible to write n^2+1 = x*y where x and y are two integers. Suppose a(n)=2 with x>n and y>n (if x<n or y<n, then x or y is divisor of (n-x)^2+1 or (n-y)^2+1)). So, x*y >n^2+1, a contradiction.
%H Michel Lagneau, <a href="/A249066/b249066.txt">Table of n, a(n) for n = 1..10000</a>
%F a(A002313(n)) = 0.
%F a(A005574(n)) = 1.
%e a(3)=0 because A002522(3)= 2*5 and the prime divisors 2 and 5 exist already with A002522(1)= 2 and A002522(2)= 5.
%e a(4)= 1 because A002522(4)=17 is prime.
%e a(5)= 1 because A002522(5)=2*13 and the prime divisor 13 appears for the first time.
%p with(numtheory): nn:=600:lst:={0}:for n from 1 to 100 do:x:=factorset(n^2+1):lst1:=lst intersect x:n0:=nops(x minus lst1): printf(`%d, `,n0):lst:=lst union x:od:
%Y Cf. A002313, A002522, A005574.
%K nonn
%O 1,1
%A _Michel Lagneau_, Oct 20 2014
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