A069214 - OEIS

%I #22 Nov 23 2022 10:08:27

%S 5,4,5,6,8,8,11,10,14,12,17,14,20,16,23,18,26,20,29,22,32,24,35,26,38,

%T 28,41,30,44,32,47,34,50,36,53,38,56,40,59,42,62,44,65,46,68,48,71,50,

%U 74,52,77,54,80,56,83,58,86,60,89,62,92,64,95,66,98,68,101,70,104,72

%N Let u(n,k) be the recursion defined by u(n,1)=1, u(n,2)=2, u(n,3)=n, and u(n,k+3) = (u(n,k+2) + u(n,k+1))/u(n,k) if u(n,k) divides u(n,k+2) + u(n,k+1), u(n,k+3) = u(n,k) otherwise. Then u(n,k) is periodic and a(n) = Max(u(n,k), k >= 1).

%C Let p(n) denote the period of u(n,k) (i.e., p(n) is the smallest integer such that u(n,k) = u(n, k+p(n))). p(n) = 22,12,22,21,15,9,15,9,15,9,... for n = 1,2,3,4,5,6,.... Hence for n > 4, p(n) = 15 if n is odd; p(n) = 9 if n is even.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).

%F a(1)=5, a(2n)=2n+2, a(2n+1)=3n+2.

%F a(n) = (5*n+5-(n-3)*(-1)^n)/4 for n>=2, with a(1)=5. - _Wesley Ivan Hurt_, Sep 05 2022

%e E.g., for k=1..15, u(7, k) = 1, 2, 7, 9, 8, 7, 9, 2, 7, 1, 4, 7, 11, 4, 7; hence a(7)=11.

%t LinearRecurrence[{0, 2, 0, -1}, {5, 4, 5, 6}, 70] (* _Georg Fischer_, Nov 23 2022 *)

%o (Perl)

%o use strict; use integer;

%o for (my $n = 1; $n <= 70; $n ++) { my @u = (0, 1, 2, $n);

%o     for (my $k = 1; $k <= 64; $k ++) {

%o         my $sum = $u[$k + 1] + $u[$k + 2];

%o         $u[$k + 3] = ($sum % $u[$k] == 0) ? $sum / $u[$k] : $u[$k];

%o     }

%o     my $max = 0;

%o     for (my $k = 1; $k < scalar(@u); $k ++) {

%o         if ($max < $u[$k]) { $max = $u[$k]; }

%o     }

%o     print "$n $max\n";

%o } # _Georg Fischer_, Nov 23 2022

%K nonn

%O 1,1

%A _Benoit Cloitre_, Apr 11 2002

%E a(65), a(67), a(69) corrected by _Georg Fischer_, Nov 23 2022