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%I #45 May 05 2020 06:01:12
%S 0,2,4,386,20136,59140,373164544
%N Numbers k such that k^2 + 1 divides 2^k + 1.
%C a(8) > 2*10^9. - _Hiroaki Yamanouchi_, Nov 29 2014
%C a(8) > 2*10^10. - _Chai Wah Wu_, Dec 07 2014
%C All terms of the sequence are even. a(5), a(6) and a(7) are of the form 2*p + 2 where p is a prime and mod(p, 14) = 1. - _Farideh Firoozbakht_, Dec 07 2014
%C From _Jianing Song_, Jan 13 2019: (Start)
%C Among the known terms only a(3) and a(4) are of the form 2*p where p is a prime.
%C a(n)^2 + 1 is prime for 2 <= n <= 7. Among these primes, the multiplicative order of 2 modulo a(n)^2 + 1 is 2*a(n) except for n = 5, in which case it is 2*a(n)/3. (End)
%C a(8) > 10^12. - _Giovanni Resta_, May 05 2020
%e 0 is in this sequence because 0^2 + 1 = 1 divides 2^0 + 1 = 2.
%o (PARI) for(n=0,10^5,if(Mod(2,n^2+1)^n==-1,print1(n,", "))); \\ _Joerg Arndt_, Nov 30 2014
%o (Python)
%o from gmpy2 import powmod
%o A247220_list = [i for i in range(10**7) if powmod(2,i,i*i+1) == i*i]
%o # _Chai Wah Wu_, Dec 03 2014
%K nonn,more
%O 1,2
%A _Juri-Stepan Gerasimov_, Nov 26 2014
%E a(7) from _Hiroaki Yamanouchi_, Nov 29 2014
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