OFFSET
1,2
COMMENTS
Conjecture: Let c be any integer with |c| > 1, and let D(c,n) be the least positive integer m such that the n numbers c^(k*(k-1)/2) (k = 1..n) are pairwise incongruent modulo m.
(i) We have lim D(c,n)/n = 4 as n tends to the infinity.
(ii) If neither c nor -c is congruent to 3 modulo 8, then D(c,n) is prime for every integer n > |c| + 1.
(iii) If c is congruent to 3 or -3 modulo 8, then the set {D(c,n): n > 0} contains infinitely many powers of 2.
(iv) Let E(c) = {n > 0: D(c,n) is not prime}. Then E(2) = E(-2) = E(4) = E(6) = E(-6) = {1,3}, E(-4) = {1,5}, E(7) = {1,2}, E(-7) = E(8) = E(-8) = E(9) = E(-9) = {1}, E(10) = {1,3,4,5} and E(-10) = {1,3,4}.
See also A392775 for a supplement.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..5790 from Zhi-Wei Sun)
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory, 133 (2013), no. 8, 2794-2812.
EXAMPLE
a(4) = 5 since 2^(1*0/2) = 1, 2^(2*1/2) = 2, 2^(3*2/2) = 8 and 2^(4*3/2) = 64 are incongruent modulo 5.
MATHEMATICA
f[k_]:=f[k]=2^(k(k-1)/2);
tab={}; m=1; Do[Label[bb]; If[Length[Union[Table[Mod[f[k], m], {k, 1, n}]]]==n, tab=Append[tab, m]; Goto[aa]]; m=m+1; Goto[bb]; Label[aa], {n, 1, 60}]; Print[tab]
PROG
(Python)
from itertools import count
def A392732(n): return next(m for m in count(n) if len({pow(2, k*(k-1)>>1, m) for k in range(1, n+1)})==n) # Chai Wah Wu, Jan 21 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 21 2026
STATUS
approved
