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To get next term, multiply by 13, add 1 and discard any prime factors < 13.
14

%I #15 Aug 22 2023 21:26:09

%S 73,19,31,101,73,19,31,101,73,19,31,101,73,19,31,101,73,19,31,101,73,

%T 19,31,101,73,19,31,101,73,19,31,101,73,19,31,101,73,19,31,101,73,19,

%U 31,101,73,19,31,101,73,19,31,101,73,19,31,101,73,19,31,101

%N To get next term, multiply by 13, add 1 and discard any prime factors < 13.

%C This is the '13x+1' map. The 'Px+1 map': if x is divisible by any prime < P then divide out these primes one at a time starting with the smallest; otherwise multiply x by P and add 1.

%C Sequence has period 4. - _Alois P. Heinz_, Jan 19 2021

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CollatzProblem.html">Collatz problem</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,1).

%e 73 -> 13*73+1 = 950 = 2*5^2*19 -> 19, so second term is 19.

%t m13[n_]:=First[Times@@@Select[FactorInteger[13 n+1],#[[1]]>11&]]; NestList[ m13,73,80] (* or *) PadRight[{},80,{73,19,31,101}] (* _Harvey P. Dale_, Apr 16 2019 *)

%t a[n_] := a[n] = Which[n == 0, 73, n <= 4, Times @@ Power @@@ Select[ FactorInteger[13 a[n - 1] + 1], #[[1]] >= 13&], True, a[n - 4]];

%t Table[a[n], {n, 0, 59}] (* _Jean-François Alcover_, Aug 21 2023 *)

%Y Cf. A057216, A057522 (long version), A057534, A057614.

%K nonn,easy

%O 0,1

%A Murad A. AlDamen (Divisibility(AT)yahoo.com), Oct 17 2000