cp's OEIS Frontend

A246452 Let A be the set of all numbers m for which m^2+1 is prime (see A005574). The sequence lists the pairs of numbers (a, b) such that every m > 1 included in A can be written in the form m = a + b for a <= b both in A and a minimum.

%I A246452 #18 Feb 02 2015 03:43:22
%S A246452 1,1,2,2,2,4,4,6,4,10,2,14,4,16,4,20,2,24,10,26,4,36,14,40,2,54,10,56,
%T A246452 20,54,10,74,6,84,4,90,16,94,6,110,4,116,4,120,2,124,4,126,4,130,16,
%U A246452 130,4,146,6,150,4,156,10,160,6,170,4,176,4,180,20,184
%N A246452 Let A be the set of all numbers m for which m^2+1 is prime (see A005574). The sequence lists the pairs of numbers (a, b) such that every m > 1 included in A can be written in the form m = a + b for a <= b both in A and a minimum.
%C A246452 Conjecture: for n>1, A005574(n) = a+b where a and b are integers in A005574.
%C A246452 By definition, a(2n-1) + a(2n) = A005574(n+1), and a(2n-1) is the minimal solution where the pair (a(2n-1), a(2n)) are both in A005574. The above conjecture says such a pair always exists. - _Jens Kruse Andersen_, Aug 27 2014
%H A246452 Michel Lagneau, <a href="/A246452/b246452.txt">Table of n, a(n) for n = 1..5000</a>
%e A246452 The 3rd pair (2,4) is in the sequence because for m = 2,4 and 2+4 = 6 we obtain three primes 2^2+1 = 5, 4^2+1 = 17 and 6^2+1 = 37.
%e A246452 The 6th number above 1 in A005574 is 16 which can be written as 2+14 or 6+10 using a pair in A005574. 2 is the minimal number in a solution so the 6th pair is (a(11), a(12)) = (2, 14). - _Jens Kruse Andersen_, Aug 27 2014
%p A246452 with(numtheory):lst:={}:nn:=10000:
%p A246452 for i from 1 to nn do:
%p A246452   if type(i^2+1,prime)=true
%p A246452   then
%p A246452   lst:=lst union {i}:
%p A246452   else
%p A246452   fi:
%p A246452 od:
%p A246452 n1:=nops(lst):
%p A246452   for n from 1 to 50 do:
%p A246452   m:=lst[n]:ii:=0:
%p A246452     for u from 1 to n1 while(ii=0)do:
%p A246452       for v from u to n1 while(ii=0)do:
%p A246452         a:=lst[u]:b:=lst[v]:
%p A246452         if m=a+b
%p A246452         then
%p A246452         ii:=1: printf(`%d, `,a): printf(`%d, `,b):
%p A246452         else
%p A246452         fi:
%p A246452       od:
%p A246452     od:
%p A246452    od:
%Y A246452 Cf. A002496, A005574.
%K A246452 nonn
%O A246452 1,3
%A A246452 _Michel Lagneau_, Aug 26 2014