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, 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

Michel Lagneau, Table of n, a(n) for n = 1..5000

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

Cf. A002496, A005574.

Sequence in context: A237520 A268241 A134318 * A104295 A172068 A289195

Adjacent sequences: A246449 A246450 A246451 * A246453 A246454 A246455

Keyword

nonn

Author

Michel Lagneau, Aug 26 2014

Status

approved