|
|
A306261
|
|
Least k > 0 such that 2n - p is prime where p is some prime divisor of 4n^2 - (2k-1)^2 for n >= 4.
|
|
3
|
|
|
1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 1, 2, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 4, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
4,3
|
|
COMMENTS
|
Conjecture: a(n) exists for n >= 4.
The conjecture holds up to 10^6. Records: a(4) = 1, a(6) = 2, a(34) = 3, a(75) = 4, a(154) = 9, a(1027) = 10, a(1097) = 11, a(1477) = 14, a(1552) = 17, a(5179) = 18, a(10684) = 29, a(70201) = 32, a(79861) = 43, a(519632) = 45, a(1018804) = 46, a(1713031) = 47, .... - Charles R Greathouse IV, Feb 17 2019
|
|
LINKS
|
|
|
EXAMPLE
|
a(4) = 1 because 4*4^2 - (2*1-1)^2 = 63 = 3^2*7 and 2*4 - 3 = 5 is prime;
a(5) = 1 because 4*5^2 - (2*1-1)^2 = 99 = 3^2*11 and 2*5 - 3 = 7 is prime;
a(6) = 2 because 4*6^2 - (2*1-1)^2 = 143 = 11*13 and 2*6 - 11 = 1 is not a prime, 2*6 - 13 = -1 is not a prime, but 4*6^2 -(2*2-1)^2 = 135 = 3^3*5 and 2*6 - 5 = 7 is prime.
|
|
PROG
|
(PARI) a(n)=for(k=1, 2*n, my(f=factor(4*n^2-(2*k-1)^2)[, 1]); for(i=1, #f, if(isprime(2*n-f[i]), return(k)))); "does not exist" \\ Charles R Greathouse IV, Feb 17 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|