OFFSET
1,1
COMMENTS
Conjecture: a(n) is the smallest integer k > 1 such that 2^(k-1) == 1 (mod a(0)*...*a(n-1)*k), with a(0) = 1. - Thomas Ordowski, Mar 13 2019
Either a(n) > a(n-1), or a(n) = a(n-1) is a Wieferich prime (A001220). - Max Alekseyev, Sep 29 2024
LINKS
Max Alekseyev, Table of n, a(n) for n = 1..1000
MATHEMATICA
i=1; Do[p=Prime[n]; If[Mod[2^(p-1)-1, p*i]==0, Print[p]; i=p*i], {n, 2, 78498}]
PROG
(PARI) findprime(prd) = {forprime(p=2, , if (Mod(2, p*prd)^(p-1) == 1, return (p)); ); }
lista(nn) = {my(prd = 1, na); for (n=1, nn, na = findprime(prd); print1(na, ", "); prd *= na; ); } \\ Michel Marcus, Mar 14 2019
(PARI) { a175257_first_terms(N=1000) = my(P, L, t); P=[3]; L=2; for(n=#P, N, print(n, " ", P[n]); forstep(p=P[n], oo, Mod(1, L), if(p==P[n], if(Mod(2, p^2)^(p-1)==1, error("Wieferich prime!"), next)); if(ispseudoprime(p), P=concat(P, [p]); t=Mod(2, p)^L; fordiv((p-1)\L, d, if(t^d==1, L*=d; break)); break))); P; } \\ Max Alekseyev, Sep 29 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Manuel Valdivia, Mar 15 2010
EXTENSIONS
a(17)-a(26) from Amiram Eldar, Feb 03 2019
Name corrected by Thomas Ordowski, Mar 13 2019
a(27) from Hans Havermann, Mar 29 2019
Eliminated a(0)=1 in the definition (empty products equal 1). - R. J. Mathar, Jun 19 2021
Terms a(28) onward from Max Alekseyev, Sep 29 2024
STATUS
approved