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%I #19 Aug 21 2023 08:23:29
%S 61,173,1471,521,4429,4183,2963,257,437,743,1579,2237,3803,2309,19627,
%T 5561,47269,14881,3833,32581,263,43,61,173,1471,521,4429,4183,2963,
%U 257,437,743,1579,2237,3803,2309,19627,5561,47269,14881,3833,32581,263,43
%N To get next term, multiply by 17, add 1 and discard any prime factors < 17.
%C This is the '17x+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 22. - _Alois P. Heinz_, Jan 15 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_22">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).
%e 61 -> 17*61+1 = 1038 = 2*3*173 -> 173, so second term is 173.
%t a[n_] := a[n] = Which[n == 0, 61, n <= 22, Times @@ Power @@@ Select[ FactorInteger[17 a[n - 1] + 1], #[[1]] >= 17&], True, a[n - 22]];
%t Table[a[n], {n, 0, 43}] (* _Jean-François Alcover_, Aug 21 2023 *)
%o (PARI) lista(nn) = {my(x=61); for (n=1, nn, print1(x, ", "); my(f=factor(17*x+1)); for (k=1, #f~, if (f[k,1] < 17, f[k,1] = 1)); x = factorback(f););} \\ _Michel Marcus_, Jan 19 2021
%Y Cf. A057446, A057522, A057534 (long version), A057614.
%K nonn,easy
%O 0,1
%A Murad A. AlDamen (Divisibility(AT)yahoo.com), Oct 17 2000
%E More terms from _James A. Sellers_ and Larry Reeves (larryr(AT)acm.org), Oct 18 2000
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