A389368 a(n) = R(A007310(n)), where R(k) = (5*k+1)/(2^r*3^s), with r and s as large as possible, that is R(k) = A065330(A016861(k)).

1, 13, 1, 7, 11, 43, 1, 29, 7, 73, 13, 11, 31, 103, 1, 59, 41, 133, 23, 37, 17, 163, 7, 89, 61, 193, 11, 13, 71, 223, 19, 119, 1, 253, 43, 67, 91, 283, 1, 149, 101, 313, 53, 41, 37, 343, 29, 179, 121, 373, 7, 97, 131, 403, 17, 209, 47, 433, 73, 7

Offset

1,2

Comments

Reduced function R is applied to the odd integers not multiples of 3.

Except for 5, all odd integers which are not multiples of 3 appear in the sequence.

Conjecture: By successive iterations x->a(x) we always end up arriving at 1.

Links

Table of n, a(n) for n=1..60.

Formula

Sum_{k=1..n} a(k) ~ (25/16) * n^2. - Amiram Eldar, Mar 07 2026

Example

For n = 1 which is odd, a(1) = 1 because 5*1-3=2 to be divided by 2.
For n = 2 which is even, a(2) = 13 because 15*2-4=26 to be divided by 2.
For n = 3 which is odd, a(3) = 1 because 5*3-3=12 to be divided by 12 (2^2*3).
For n = 4 which is even, a(4) = 7 because 15*4-4=56 to be divided by 8 (2^3).
For n = 5 which is odd, a(5) = 11 because 5*5-3=22 to be divided by 2.

Mathematica

a[n_] := (#/Times @@ ({2, 3}^IntegerExponent[#, {2, 3}]))& @ If[OddQ[n], 5*n-3, 15*n-4]; Array[a, 100] (* Amiram Eldar, Feb 03 2026 *)

Prog

(PARI) a(n) = my(m = if(n%2==1, 5*n-3, 15*n-4)); m = m>>valuation(m, 2)/3^valuation(m, 3)
(Python)
from sympy import multiplicity
def A389368(n): return ((m:=5*(n+(n>>1)<<1)-4)>>(~m&m-1).bit_length())//3**multiplicity(3, m) # Chai Wah Wu, Feb 17 2026

Crossrefs

Cf. A007310, A065330, A016861.

Sequence in context: A116993 A010234 A111738 * A202134 A185409 A010235

Adjacent sequences: A389365 A389366 A389367 * A389369 A389370 A389371

Keyword

nonn,easy

Author

Alain Rocchelli, Feb 01 2026

Status

approved