OFFSET
2,1
COMMENTS
a(n) is the unique number k in [1, 2^n] and congruent to 3 mod 4 such that k^2 - 17 is divisible by 2^(n+1).
LINKS
Jianing Song, Table of n, a(n) for n = 2..1000
FORMULA
EXAMPLE
The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 - 17 is divisible by 8 is 3, so a(2) = 3.
a(2)^2 - 17 = -8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 7.
a(3)^2 - 17 = 32 which is divisible by 32, so a(4) = a(3) = 7.
a(4)^2 - 17 = 32 which is not divisible by 64, so a(5) = a(4) + 2^4 = 23.
a(5)^2 - 17 = 512 which is divisible by 128, so a(6) = a(5) = 23.
...
MATHEMATICA
Table[First@Select[PowerModList[17, 1/2, 2^(k+1)], Mod[#, 4]==3&], {k, 2, 35}] (* Giorgos Kalogeropoulos, Oct 22 2022 *)
PROG
(PARI) a(n) = if(n==2, 3, truncate(-sqrt(17+O(2^(n+1)))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Feb 13 2021
STATUS
approved