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Smallest b > 1 such that b^(p-1) == 1 (mod p^7) for p = prime(n).
8

%I #20 May 18 2022 07:55:26

%S 129,2186,32318,82681,758546,6826318,21444846,44702922,178042767,

%T 393747520,1548729003,4741156070,2203471139,3242334565,16609835418,

%U 114175761515,30338830655,20115543070,114457309347,370162324382,57877856575,12692933349,280646695286,127762186531

%N Smallest b > 1 such that b^(p-1) == 1 (mod p^7) for p = prime(n).

%o (PARI) a(n) = my(p=prime(n)); for(b=2, oo, if(Mod(b, p^7)^(p-1)==1, return(b)))

%o (Python)

%o from sympy import prime

%o from sympy.ntheory.residue_ntheory import nthroot_mod

%o def A353940(n): return 2**7+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**7,True)[1]) # _Chai Wah Wu_, May 17 2022

%Y Row k = 7 of A257833.

%Y Cf. similar sequences for p^k: A039678 (k=2), A249275 (k=3), A353937 (k=4), A353938 (k=5), A353939 (k=6), A353941 (k=8), A353942 (k=9), A353943 (k=10).

%K nonn

%O 1,1

%A _Felix Fröhlich_, May 12 2022

%E a(9)-a(11) from _Amiram Eldar_, May 12 2022

%E More terms from _Jinyuan Wang_, May 17 2022