1,1

Theorem: if 5^((n-1)/2) = -1 (mod n) then n = 2 or 3 (mod 5) (see Crandall and Pomerance).

Start with 2. The next number, 3, can not be written as the sum of two of the previous terms. So 3 is in. 4=2+2, 5=2+3, 6=3+3, so these are not in. But you can not obtain 7, so the next term is 7. And so on. - Fabian Rothelius (fabian.rothelius(AT)telia.com), Mar 13 2001

Primitive roots of 5. The first differences are periodic: 1,4,1,4,1,4.... - Paolo P. Lava (ppl(AT)spl.at), Feb 29 2008

Also, numbers n such that n^2 = -1 mod(5) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 20 2009]

Also, numbers n such that n^2 = -1 mod(p), with p (prime)=4k+1; example 128^2=-1 mod (5), 128^2=-1 mod (29). [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Feb 06 2009]

Also, except the first term, a(n)=5*n-a(n-1), (with a(1)=3) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 17 2009]

R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2001; see Exercise 3.24, p. 154.

a(n)=-3+(1/2)*Sum_{k=0..n}{5-3*(-1)^k}, with n>=1 - Paolo P. Lava (ppl(AT)spl.at), Feb 29 2008

P:=proc(n, m) local a, i, ok; for i from 1 by 1 to n do if (i^(m-1) mod m)=1 then a:=1; ok:=1; while a<m-1 do if (i^a mod m)=1 then ok:=0; fi; a:=a+1; od; if ok=1 then print(i); fi; fi; od; end: P(100, 5); - Paolo P. Lava (ppl(AT)spl.at), Feb 29 2008

Sequence in context: A002274 A102664 A055053 this_sequence A032967 A111101 A003307

Adjacent sequences: A047218 A047219 A047220 this_sequence A047222 A047223 A047224

nonn,easy

N. J. A. Sloane (njas(AT)research.att.com).

More terms from Larry Reeves (larryr(AT)acm.org), Apr 08 2002