A105801 Fibonacci-Collatz sequence: a(1)=1, a(2)=2; for n > 2, let fib = a(n-1) + a(n-2); if fib is odd then a(n) = 3*fib + 1 else a(n) = fib/2.

1, 2, 10, 6, 8, 7, 46, 160, 103, 790, 2680, 1735, 13246, 44944, 29095, 222118, 753640, 487879, 3724558, 12637312, 8180935, 62454742, 211907032, 137180887, 1047263758, 3553333936, 2300298847, 17560898350, 59583591592, 38572244971

Offset: 1

Author

Zak Seidov, Sep 12 2006

Keywords

• nonn

Comments

Taking a(1)=1, a(2)=1 leads to the all-ones sequence 1,1,1,1,1,1,... (A000012); similarly a(1)=a(2)=b gives "all b's" sequence b,b,b,b,b,....

Apparently, for n > 2, the sequence is periodic modulo 2 with period 3. However, this regularity is disrupted when starting at a(34)=4937737952464 a run of 6 even terms appears.

It is easy to prove that all the terms a(n) with n >= 10 are congruent to 7 mod 9.

Conjecture: for every k > 0 there is an index m such that all the a(n) with n > m have the same residue mod 3^k. - Giovanni Resta, Nov 17 2010

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..3000

Crossrefs

Cf. A000012, A000045, A006370, A181663.

Cf. A181717.

Programs

Haskell

a105801 n = a105801_list !! (n-1)
a105801_list = 1 : 2 : fc 2 1 where
   fc x x' = y : fc y x where y = a006370 (x + x')
-- Reinhard Zumkeller, Oct 09 2011

Maple

M:=3000: a:=1: b:=2: lprint(1,1): lprint(2,2): for n from 3 to M do c:=a+b: if (c mod 2) = 0 then d:=c/2 else d:=3*c+1: fi: lprint(n,d): a:=b: b:=d: od: # N. J. A. Sloane, Nov 20 2010

Mathematica

a[1]=1; a[2]=2; a[n_]:=a[n]=(fib=a[n-1]+a[n-2]; col=If[OddQ[fib],3*fib+1,fib/2]); Table[a[n],{n,30}]

PARI

A105801(n)=if(n<3,if(n<2,1,2),f=A105801(n-1)+A105801(n-2); if(f%2,3*f+1,f/2))

Extensions

Incorrect formula deleted by Colin Barker, Jul 29 2013