cp's OEIS Frontend

A252606 Numbers j such that j + 2 divides 2^j + 2.

%I A252606 #59 Jan 04 2024 15:10:48
%S A252606 3,4,16,196,2836,4551,5956,25936,46775,65536,82503,540736,598816,
%T A252606 797476,1151536,3704416,4290771,4492203,4976427,8095984,11272276,
%U A252606 13362420,21235696,21537831,21549347,29640832,31084096,42913396,49960912,51127259,55137316,56786087,60296571,70254724,70836676
%N A252606 Numbers j such that j + 2 divides 2^j + 2.
%C A252606 Numbers j such that (2^j + 2)/(j + 2) is an integer. Numbers j such that (2^j - j)/(j + 2) is an integer.
%C A252606 From _Robert Israel_, Apr 09 2015: (Start)
%C A252606 The even members of this sequence (4, 16, 196, 2836, ...) are the numbers 2*k-2 where k>=3 is odd and 4^k == -8 (mod k).
%C A252606 The odd members of this sequence (3, 4551, 46775, 82503, ...) are the numbers k-2 where k>=3 is odd and 2^k == -8 (mod k). (End)
%C A252606 2^m is in this sequence for m = (2, 4, 16, 36, 120, 256, 456, 1296, 2556, ...), with the subsequence m = 2^k, k = (1, 2, 4, 8, 16, ...). - _M. F. Hasler_, Apr 09 2015
%H A252606 Chai Wah Wu, <a href="/A252606/b252606.txt">Table of n, a(n) for n = 1..78</a>
%e A252606 3 is in this sequence because (2^3 + 2)/(3 + 2) = 2.
%p A252606 select(t -> 2 &^t + 2 mod (t + 2) = 0, [$1..10^6]); # _Robert Israel_, Apr 09 2015
%t A252606 Select[Range[10^6],IntegerQ[(2^#+2)/(#+2)]&] (* _Ivan N. Ianakiev_, Apr 17 2015 *)
%o A252606 (Magma) [n: n in [0..1200000] | Denominator((2^n+2)/(n+2)) eq 1];
%o A252606 (PARI) for(n=1,10^5,if((2^n+2)%(n+2)==0,print1(n,", "))) \\ _Derek Orr_, Apr 05 2015
%o A252606 (PARI) is(n)=Mod(2,n+2)^n==-2 \\ _M. F. Hasler_, Apr 09 2015
%o A252606 (Python)
%o A252606 A252606_list = [n for n in range(10**4) if pow(2, n, n+2) == n] # _Chai Wah Wu_, Apr 16 2015
%Y A252606 Cf. A001477, A004273, A004275, A081765, A213382, A251603.
%K A252606 nonn,hard
%O A252606 1,1
%A A252606 _Juri-Stepan Gerasimov_, Mar 03 2015
%E A252606 a(17)-a(22) from _Tom Edgar_, Mar 03 2015
%E A252606 More terms from _Chai Wah Wu_, Apr 16 2015