cp's OEIS Frontend

A352406 The number of terms before reaching zero when starting at n and iterating: f(n) = n, f(n+1) = n+1; f(n+k) = (f(n+k-2) + f(n+k-1)) (mod (n+k)), where k>=2.

%I A352406 #14 Mar 19 2022 09:43:03
%S A352406 5,2,7,3,8,14,5,7,4,12,46,34,18,21,21,16,5,10,8,25,128,237,79,25,266,
%T A352406 25,10,74,34,27,6,11,22,23,72,75,26,267,16,893,28,40,8,113,27,16,163,
%U A352406 41,13,27,169,48,837,7,88,436,23,144,59,36,77,71,466,96,14,226,371,72,231,463,377,29
%N A352406 The number of terms before reaching zero when starting at n and iterating: f(n) = n, f(n+1) = n+1; f(n+k) = (f(n+k-2) + f(n+k-1)) (mod (n+k)), where k>=2.
%C A352406 The first term corresponds to the number of terms before zero in A079777. This sequence starts at n and counts the number of terms before zero using the same iterative formula.
%C A352406 The sequence shows large variations in its value, e.g., a(496) = 222, a(497) = 1087851. In the first 50000 terms the largest value is a(21897) = 1248976431. In the same range the smallest number greater than 1 not to have appeared is 414, although it is likely all numbers eventually appear.
%H A352406 Scott R. Shannon, <a href="/A352406/a352406.png">Line graph of the first 50000 terms</a>.
%e A352406 a(0) = 5 as starting at 0 and 1 gives 0+1 % 2 = 1, 1+1 % 3 = 2, 1+2 % 4 = 3, 2+3 % 5 = 0, with five terms before reaching zero. See A079777.
%e A352406 a(1) = 2 as starting at 1 and 2 gives 1+2 % 3 = 0, with two terms before reaching zero. This is the smallest possible value and the only term to equal 2.
%e A352406 a(2) = 7 as starting at 2 and 3 gives 2+3 % 4 = 1, 3+1 % 5 = 4, 1+4 % 6 = 5, 4+5 % 7 = 2, 5+2 % 8 = 7, 2+7 % 9 = 0, with seven terms before reaching zero.
%Y A352406 Cf. A079777, A352407 (multiplication).
%K A352406 nonn
%O A352406 0,1
%A A352406 _Scott R. Shannon_, Mar 15 2022