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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.
1
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, 20, 54, 10, 74, 6, 84, 4, 90, 16, 94, 6, 110, 4, 116, 4, 120, 2, 124, 4, 126, 4, 130, 16, 130, 4, 146, 6, 150, 4, 156, 10, 160, 6, 170, 4, 176, 4, 180, 20, 184
OFFSET
1,3
COMMENTS
Conjecture: for n>1, A005574(n) = a+b where a and b are integers in A005574.
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
LINKS
EXAMPLE
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.
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
MAPLE
with(numtheory):lst:={}:nn:=10000:
for i from 1 to nn do:
if type(i^2+1, prime)=true
then
lst:=lst union {i}:
else
fi:
od:
n1:=nops(lst):
for n from 1 to 50 do:
m:=lst[n]:ii:=0:
for u from 1 to n1 while(ii=0)do:
for v from u to n1 while(ii=0)do:
a:=lst[u]:b:=lst[v]:
if m=a+b
then
ii:=1: printf(`%d, `, a): printf(`%d, `, b):
else
fi:
od:
od:
od:
CROSSREFS
Sequence in context: A237520 A268241 A134318 * A104295 A172068 A289195
KEYWORD
nonn
AUTHOR
Michel Lagneau, Aug 26 2014
STATUS
approved