A326823 -id:A326823 - cp's OEIS frontend

A326825 a(n) is the number of iterations needed to reach 1 or 5 starting at n and using the map k -> (k/2 if k is even, otherwise k + (smallest triangular number > k)). Set a(n) = -1 if the trajectory never reaches 1 or 5.

0, 1, 6, 2, 0, 7, 7, 3, 5, 1, 12, 8, 10, 8, 8, 4, 6, 6, 4, 2, 15, 13, 58, 9, 56, 11, 52, 9, 36, 9, 15, 5, 28, 7, 13, 7, 20, 5, 18, 3, 18, 16, 16, 14, 59, 59, 59, 10, 14, 57, 57, 12, 55, 53, 51, 10, 22, 37, 55, 10, 24, 16, 22, 6, 35, 29, 14, 8, 27, 14, 12, 8, 10, 21, 23, 6, 27, 19

Offset: 1

Ali Sada, Oct 20 2019

nonn

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

M. F. Hasler, Table of n, a(n) for n = 1..1000, May 08 2025

Cf. A006577, A326823 (similar with squares instead of triangular numbers).

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

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.

Edited by Jon E. Schoenfield and N. J. A. Sloane, Oct 20 2019

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.

Original entry on oeis.org

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

Ali Sada, Oct 21 2019

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.

			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.

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

Cf. A006577, A057944, A326823, A326825.

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

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 *)

cp's OEIS Frontend

A326825 a(n) is the number of iterations needed to reach 1 or 5 starting at n and using the map k -> (k/2 if k is even, otherwise k + (smallest triangular number > k)). Set a(n) = -1 if the trajectory never reaches 1 or 5.

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