A179738 a(n) = length of (eventual) period of the sequence defined by s(0) = 1, s(n+1) = odd_part(2n-1 + (s(n) if n odd else s(n)*3)), where odd_part = A000265.

1, 2, 4, 1, 4, 8, 2, 4, 8, 4, 2, 2, 1, 4, 12, 4, 4, 26, 2, 12, 6, 2, 4, 4, 4, 16, 4, 8, 10, 6, 4, 8, 4, 2, 6, 8, 12, 12, 28, 1, 22, 16, 2, 16, 12, 12, 16, 6, 4, 40, 24, 4, 32, 6, 26, 16, 6, 8, 4, 8, 12, 2, 36, 6, 46, 12, 2, 36, 60, 6, 16, 8, 4, 18, 20, 4, 60, 36

Offset: 2

Author

Vladimir Shevelev, Jul 25 2010

Keywords

• nonn

Comments

Original definition (edited): Let x, y be odd numbers and the operation x <+> y := A000265(x+y). Consider sequence s(0) = x <+> y, s(2*k+1) = x <+> 3*s(2*k), s(2*k+2) = x <+> s(2*k+1); a(n) is the smallest period in case x = 2*n-1, y = 1.

The operation x <+> y = A000265(x+y) = odd part of x+y is also considered in A179382.

Record values are: a(2) = 1, a(3) = 2, a(4) = 4, a(7) = 8, a(16) = 12, a(19) = 26, a(40) = 28, a(51) = 40, a(66) = 46, a(70) = 60, a(111) = 64, a(126) = 70, a(147) = 80, a(162) = 96, a(225) = 120, a(379) = 170, a(619) = 184, a(640) = 228, a(727) = 248, a(759) = 256, a(916) = 348, ... - M. F. Hasler, Feb 16 2025

Examples

			For n = 4, 2*n-1 = 7, we get: 7 <+> 1 = 1, 7 <+> 3*1 = 5, 7 <+> 5 = 3, 7 <+> 3*3 = 1, and from here on it starts over with 7 <+> 1 = 1, etc., so the period is [1, 5, 3, 1], of length 4, whence a(4) = 4.
For n = 6, 2*n-1 = 11, we get:
  11 <+> 1 = 3, 11 <+> 3*3 =  5, 11 <+>  5  = 1, 11 <+> 3*1 = 7,
  11 <+> 7 = 9, 11 <+> 3*9 = 19, 11 <+> 19 = 15, 11 <+> 3*15 = 7, 11 <+> 7 = 9, ...
Thus we have an eventually periodic sequence with the smallest period 4 (with elements 7, 9, 19, 15). Thus a(6) = 4.
		

Links

• Jason Yuen, Table of n, a(n) for n = 2..10000 (terms 2..1000 from M. F. Hasler)

Crossrefs

Cf. A179382, A179480, A179686.

Programs

PARI

apply( {A179738(n, y=1, T(y, x=2*n-1)=(x+y)>>valuation(x+y,2))=my(s=[], P); until(, s=concat(s, y=T(3^(#s%2)*y)); for(L=1, #s\3, P=[vecextract(s,Str(-L-t,"..-",1+t)) | t<-[0,L,2*L]]; P[1]==P[2] && P[1]==P[3] && return(#P[1])))}, [2..90])

Python

def A179738(n):
    s = [1]; x = 2*n-1; odd = lambda z: all(z&1 or(z:=z>>1)for _ in range(z))and z
    while not(p := next((p for p in range(1, len(s)//3+1) if
        s[-p:]==s[-2*p:-p]==s[-3*p:-2*p]), 0)): s.append(odd(x+3**(len(s)&1)*s[-1]))
    return p
print([A179738(n)for n in range(2,99)]) # M. F. Hasler, Feb 16 2025

Extensions

Name edited and corrections proposed by Jason Yuen, Feb 09 2025

Edited, a(4) and a(18) corrected, and extended by M. F. Hasler, Feb 15 2025