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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). 3
 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 (list; graph; refs; listen; history; text; internal format)
 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 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, 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 A163851 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

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Last modified February 27 19:23 EST 2020. Contains 332308 sequences. (Running on oeis4.)