A326923 a(n) is the number of iterations needed to reach 1 or 9 starting at n and using the map k -> (k/2 if k is even, otherwise k + (largest triangular number < k)). Set a(n) = -1 if the trajectory never reaches 1 or 9.

0, 1, 3, 2, 4, 4, 9, 3, 0, 5, 4, 5, 8, 10, 10, 4, 6, 1, 8, 6, 3, 5, 7, 6, 9, 9, 8, 11, 15, 11, 17, 5, 19, 7, 19, 2, 27, 9, 41, 7, 33, 4, 39, 6, 47, 8, 10, 7, 12, 10, 9, 10, 14, 9, 12, 12, 14, 16, 16, 12, 18, 18, 18, 6, 14, 20, 26, 8, 32, 20, 24, 3, 26, 28, 40, 10, 32

Offset: 1

Author

Ali Sada, Oct 21 2019

Keywords

• nonn
• look

Comments

It is conjectured that this algorithm will always terminate at 1 or 9.

Jim Nastos verified the conjecture for n <= 64*10^5.

Jim Nastos verified the conjecture for n <= 45248000.

Examples

			For n = 11: 11+10=21; 21+15=36; 36/2=18; 18/2=9; taking 4 steps to reach 9, so a(11)=4.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006577, A057944, A326823, A326825.

Programs

Maple

LT:= proc(k) local n;
   n:= ceil((sqrt(1+8*k)-1)/2);
   n*(n-1)/2
end proc:
f:= proc(k) option remember; if k::even then 1+procname(k/2) else 1+procname(k+LT(k))fi
end proc:
f(1):=0: f(9):= 0:
map(f, [$1..100]); # Robert Israel, Oct 23 2019

Mathematica

LT[k_] := Module[{n}, n = Ceiling[(Sqrt[1+8k]-1)/2]; n(n-1)/2]; f[k_] := f[k] = If[EvenQ[k], 1+f[k/2], 1+f[k+LT[k]]]; f[1] = 0; f[9] = 0;
Array[f, 100] (* Jean-François Alcover, Aug 27 2022, after Robert Israel *)