

A185046


Smallest prime p such that 2n+1 = p^3  2q for some odd prime q, or 0 if no such prime exists.


4



5, 3, 5, 7, 13, 3, 13, 3, 5, 3, 3, 11, 0, 7, 5, 19, 37, 11, 5, 7, 5, 7, 37, 11, 5, 31, 53, 31, 13, 23, 5, 7, 5, 7, 13, 23, 13, 19, 5, 7, 421, 47, 5, 7, 5, 11, 13, 11, 5, 43, 5, 11, 61, 23, 5, 19, 5, 7, 5, 5, 53, 7, 17, 7, 13, 11, 13, 7, 113, 7, 373, 11, 17, 7
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OFFSET

1,1


COMMENTS

a(n) = 0 for n = 13, 171, 364, 1098, 2456, 3429, 6083, 7812, 9841, 12194, 14895, 17968,... and 2n+1 = 27, 343, 729,... is a class of cubes.
The corresponding primes q are in A224730.
Conjecture: The odd numbers different from a cube are of the form m = p^3  2q where p and q are prime numbers.
Remark: Its converse is false: there exists cubes m = c^3 that are in the sequence with the form c^3 = p^3  2q, where pc = 2, and q of the form x^2 +x*y+y^2 (see A007645). For example: 5^3 = 7^3  2*109.


LINKS



EXAMPLE

a(4) = 7 because, for (p, q) = (7, 167) => 2*4+1 = 9 = 7^3  2*167 = 343  334 = 9.


MAPLE

for n from 3 by 2 to 200 do:
jj:=0:
for j from 1 to 10000 while (jj=0) do:
p:=ithprime(j):q:=(p^3n)/2:
if q> 0 and type(q, prime)=true
then
jj:=1:printf(`%d, `, p):
else
fi:
od:
if jj=0 then
printf(`%d, `, 0):
else
fi:
od:


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



