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A294912 Numbers n such that 2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*ceiling((3/4)*n)-2), and (2^((n+1)/2) + floor((1/4)*n)*2^(((n+1)/2)+1)) are all congruent to 1 (mod n).

%I A294912 #57 Feb 13 2020 02:44:50
%S A294912 3,11,19,43,59,67,83,107,131,139,163,179,211,227,251,283,307,331,347,
%T A294912 379,419,443,467,491,499,523,547,563,571,587,619,643,659,683,691,739,
%U A294912 787,811,827,859,883,907,947,971,1019,1051,1091,1123,1163,1171,1187,1259
%N A294912 Numbers n such that 2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*ceiling((3/4)*n)-2), and (2^((n+1)/2) + floor((1/4)*n)*2^(((n+1)/2)+1)) are all congruent to 1 (mod n).
%C A294912 It appears that A007520 is a subsequence.
%C A294912 The first composite term is a(9969) = 476971 = 11*131*331. - _Alois P. Heinz_, Nov 10 2017
%C A294912 From _Hilko Koning_, Dec 03 2019: (Start)
%C A294912 The next composite terms < 1999979 are
%C A294912 a(17428) = 877099  = 307*2857
%C A294912 a(25090) = 1302451 = 571*2281
%C A294912 a(25518) = 1325843 = 499*2657
%C A294912 a(26785) = 1397419 = 67*20857
%C A294912 a(27549) = 1441091 = 347*4153
%C A294912 a(28715) = 1507963 = 971*1553
%C A294912 a(29117) = 1530787 = 619*2473
%C A294912 a(35635) = 1907851 = 11*251*691
%C A294912 (End)
%C A294912 From _Hilko Koning_, Dec 05 2019: (Start)
%C A294912 The next composite terms < 24999971 are
%C A294912 a(37344) = 2004403 = 307*6529
%C A294912 a(55773) = 3090091 = 1163*2657
%C A294912 a(56189) = 3116107 = 883*3529
%C A294912 a(91332) = 5256091 = 811*6481
%C A294912 a(102027) = 5919187 = 1777*3331
%C A294912 a(133230) = 7883731 = 811*9721
%C A294912 a(156407) = 9371251 = 1531*6121
%C A294912 a(182911) = 11081459 = 227*48817
%C A294912 a(189922) = 11541307 = 1699*6793
%C A294912 a(201043) = 12263131 = 811*15121
%C A294912 a(213203) = 13057787 = 467*27961
%C A294912 a(217484) = 13338371 = 3163*4217
%C A294912 a(257526) = 15976747 = 3739*4273
%C A294912 a(274961) = 17134043 = 1097*15619
%C A294912 a(299096) = 18740971 = 1531*12241
%C A294912 a(308928) = 19404139 = 2011*9649
%C A294912 a(321676) = 20261251 = 2251*9001
%C A294912 a(341902) = 21623659 = 1163*18593
%C A294912 a(348622) = 22075579 = 163*135433
%C A294912 a(380162) = 24214051 = 281*86171
%C A294912 The composite terms < 25*10^6 match the terms of A244628.
%C A294912 (End)
%C A294912 It appears that composites of the form 2k+1 such that 3*(2k+1) divides 2^k+1 are the composite terms of this sequence. - _Hilko Koning_, Dec 09 2019
%H A294912 Jonas Kaiser, <a href="https://arxiv.org/abs/1608.00862">On the relationship between the Collatz conjecture and Mersenne prime numbers</a>, arXiv:1608.00862 [math.GM], 2016.
%t A294912 okQ[n_] := AllTrue[{2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*Ceiling@((3/4)*n) - 2), (2^((n+1)/2) + Floor@(n/4)*2^(((n+1)/2)+1))}, Mod[#, n] == 1&];
%t A294912 Select[Range[1300], okQ] (* _Jean-François Alcover_, Feb 18 2019 *)
%o A294912 (PARI)  isok(n) = (n%2) && lift((Mod(2, n)^(n-1))==1)&&lift((Mod((2*n-1), n)*Mod(2, n)^((n-1)/2)) == 1)&&lift((Mod(((4*ceil((3/4)*n)-2)), n) )== 1)&&lift((Mod(2, n)^((n+1)/2) +Mod(floor((1/4)*n),n)*Mod(2, n)^(((n+1)/2)+1 ))== 1)
%Y A294912 Cf. A244626, A294717, A293394, A070179, A007520.
%K A294912 nonn
%O A294912 1,1
%A A294912 _Jonas Kaiser_, Nov 10 2017
%E A294912 More terms from _Alois P. Heinz_, Nov 10 2017