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Numbers k having at least one prime factor p such that p^2 divides 2^(k-1) - 1.
2

%I #31 Sep 08 2022 08:46:19

%S 1093,3511,398945,796797,1194649,1592501,1990353,2388205,2786057,

%T 3183909,3581761,3979613,4377465,4775317,5173169,5571021,5968873,

%U 6165316,6366725,6764577,7162429,7560281,7958133,8355985,8753837,9151689,9549541,9947393,10345245

%N Numbers k having at least one prime factor p such that p^2 divides 2^(k-1) - 1.

%C Another version of A001220.

%C Sequence is infinite since if k is a term then also k^m is a term, for every m >= 2.

%C What is the smallest number in this sequence which is not of the form 13*n + 1?

%C Complete factorizations of the first 15 terms:

%C a(1) = 1093

%C a(2) = 3511

%C a(3) = 5 * 73 * 1093

%C a(4) = 3^6 * 1093

%C a(5) = 1093^2

%C a(6) = 31 * 47 * 1093

%C a(7) = 3 * 607 * 1093

%C a(8) = 5 * 19 * 23 * 1093

%C a(9) = 1093 * 2549

%C a(10) = 3 * 971 * 1093

%C a(11) = 29 * 113 * 1093

%C a(12) = 11 * 331 * 1093

%C a(13) = 3^2 * 5 * 89 * 1093

%C a(14) = 17 * 257 * 1093

%C a(15) = 1093 * 4733

%C These are the numbers k for which gcd(k^2, 2^(k-1)-1) is not squarefree. However, numbers k such that gcd(k^2, 2^(k-1)-1) > k are a proper subset of them. Are there infinitely many such numbers? See A331021. - _Amiram Eldar_ and _Thomas Ordowski_, Jan 06 2020

%H Giovanni Resta, <a href="/A291194/b291194.txt">Table of n, a(n) for n = 1..10000</a>

%o (Magma) lst:=[]; for n in [2..10345245] do f:=Factorization(n); if not IsNull([x: x in [1..#f] | Modexp(2, n-1, f[x][1]^2) eq 1]) then Append(~lst, n); end if; end for; lst;

%Y Cf. A190991, A270833. A001220 gives the primes.

%K nonn

%O 1,1

%A _Arkadiusz Wesolowski_, Aug 20 2017