|
|
A230362
|
|
Least prime p with 2*p^2 - 1 and 2*(n-p)^2 -1 both prime, or 0 if such a prime p does not exist.
|
|
2
|
|
|
3, 13, 7, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 7, 2, 2, 3, 2, 2, 3, 7, 2, 3, 7, 11, 13, 7, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 7, 2, 2, 3, 2, 3, 7, 7, 2, 3, 11, 2, 3, 7, 2, 2, 2, 3, 43, 7, 7
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Conjecture: 0 < a(n) < sqrt(2n)*(log n) except for n = 1, 2, 3, 232, 1478, 6457.
By the conjecture in the comments in A230351, 0 < a(n) < n for all n > 3.
|
|
LINKS
|
|
|
EXAMPLE
|
a(12) = 2 since 2*2^2 - 1 and 2*(12-2)^2 - 1 = 199 are both prime.
|
|
MATHEMATICA
|
Do[Do[If[PrimeQ[2Prime[i]^2-1]&&PrimeQ[2(n-Prime[i])^2-1], Print[n, " ", Prime[i]]; Goto[aa]], {i, 1, Max[13, PrimePi[n-1]]}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 1, 70}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|