OFFSET
1,3
COMMENTS
It is conjectured that this algorithm will always terminate at 1 or 5.
Jim Nastos verified the conjecture for n <= 354999.
The conjecture holds for all n <= 10^9. - Jon E. Schoenfield, Oct 20 2019
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..1000, May 08 2025
FORMULA
For n = 11: 11+15 = 26; 26/2 = 13; 13+15 = 28; 28/2 = 14; 14/2 = 7; 7+10 = 17; 17+21 = 38; 38/2 = 19; 19+21 = 40; 40/2 = 20; 20/2 = 10; 10/2 = 5; taking 12 steps to reach 5, so a(11) = 12.
PROG
(PARI) M326825=Map([1, 0; 5, 0]); apply( {A326825(n)=if(mapisdefined(M326825, n, &n), n, mapput(M326825, n, 1+n=A326825(if(n%2, n+binomial((sqrtint(8*n+8)+3)\2, 2), n\2))); 1+n)}, [1..77]) \\ M. F. Hasler, May 08 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Ali Sada, Oct 20 2019
EXTENSIONS
Edited by Jon E. Schoenfield and N. J. A. Sloane, Oct 20 2019
STATUS
approved
