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).

3, 11, 19, 43, 59, 67, 83, 107, 131, 139, 163, 179, 211, 227, 251, 283, 307, 331, 347, 379, 419, 443, 467, 491, 499, 523, 547, 563, 571, 587, 619, 643, 659, 683, 691, 739, 787, 811, 827, 859, 883, 907, 947, 971, 1019, 1051, 1091, 1123, 1163, 1171, 1187, 1259

Offset

1,1

Comments

It appears that A007520 is a subsequence.

The first composite term is a(9969) = 476971 = 11*131*331. - Alois P. Heinz, Nov 10 2017

From Hilko Koning, Dec 03 2019: (Start)

The next composite terms < 1999979 are

a(17428) = 877099 = 307*2857

a(25090) = 1302451 = 571*2281

a(25518) = 1325843 = 499*2657

a(26785) = 1397419 = 67*20857

a(27549) = 1441091 = 347*4153

a(28715) = 1507963 = 971*1553

a(29117) = 1530787 = 619*2473

a(35635) = 1907851 = 11*251*691

(End)

From Hilko Koning, Dec 05 2019: (Start)

The next composite terms < 24999971 are

a(37344) = 2004403 = 307*6529

a(55773) = 3090091 = 1163*2657

a(56189) = 3116107 = 883*3529

a(91332) = 5256091 = 811*6481

a(102027) = 5919187 = 1777*3331

a(133230) = 7883731 = 811*9721

a(156407) = 9371251 = 1531*6121

a(182911) = 11081459 = 227*48817

a(189922) = 11541307 = 1699*6793

a(201043) = 12263131 = 811*15121

a(213203) = 13057787 = 467*27961

a(217484) = 13338371 = 3163*4217

a(257526) = 15976747 = 3739*4273

a(274961) = 17134043 = 1097*15619

a(299096) = 18740971 = 1531*12241

a(308928) = 19404139 = 2011*9649

a(321676) = 20261251 = 2251*9001

a(341902) = 21623659 = 1163*18593

a(348622) = 22075579 = 163*135433

a(380162) = 24214051 = 281*86171

The composite terms < 25*10^6 match the terms of A244628.

(End)

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

Links

Table of n, a(n) for n=1..52.

Jonas Kaiser, On the relationship between the Collatz conjecture and Mersenne prime numbers, arXiv:1608.00862 [math.GM], 2016.

Mathematica

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&];
Select[Range[1300], okQ] (* Jean-François Alcover, Feb 18 2019 *)

Prog

(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)

Crossrefs

Cf. A244626, A294717, A293394, A070179, A007520.

Sequence in context: A192717 A163183 A007520 * A309027 A213891 A336790

Adjacent sequences: A294909 A294910 A294911 * A294913 A294914 A294915

Keyword

nonn

Author

Jonas Kaiser, Nov 10 2017

Extensions

More terms from Alois P. Heinz, Nov 10 2017

Status

approved