OFFSET
1,1
COMMENTS
Conjecture: a(n) <= 18 = a(326).
a(m) = 0 for m in A047845. - Michel Marcus, Aug 16 2022
I conjecture the opposite: a(n) is unbounded, and indeed for any k < 1 and any m there are >> x^k terms up to x with a(n) > m. At a very rough guess, there should be some n with 20-50 digits having a(n) > 18. - Charles R Greathouse IV, Oct 26 2022
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
FORMULA
a(n) is number of consecutive primes generated by (2x-1)^2+2n for x=1,2,3,4,
EXAMPLE
For n=1 we have 1^2+2*1=3 and 3^2+2*1=11 are prime but 5^2+2*1=27 is not, and thus a(1)=2.
For n=2, 1^2+2*2=5 ... 7^2+2*2=53 are prime but 9^2+2*2=85 is not, thus a(2)=4.
For n=3, 1^2+2*3=7 is prime but 3^2+2*3=15 is not thus a(3)=1.
For n=4, 1^2+2*4=9 which is not prime, thus a(4)=0.
MAPLE
f:= proc(n) local k;
for k from 1 by 2 do
if not isprime(k^2+2*n) then return (k-1)/2 fi
od
end proc:
map(f, [$1..100]); # Robert Israel, Oct 26 2023
MATHEMATICA
a[n_] := Module[{k = 1}, While[PrimeQ[k^2 + 2*n], k += 2]; (k - 1)/2]; Array[a, 100] (* Amiram Eldar, Aug 15 2022 *)
PROG
(PARI) a(n) = my(k=1); while (isprime(k^2+2*n), k+=2); (k-1)/2; \\ Michel Marcus, Aug 16 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Steven M. Altschuld, Aug 15 2022
STATUS
approved