A181717 Fibonacci-Collatz sequence: a(1)=0, a(2)=1; 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.

0, 1, 4, 16, 10, 13, 70, 250, 160, 205, 1096, 3904, 2500, 3202, 2851, 18160, 63034, 40597, 310894, 1054474, 682684, 868579, 4653790, 16567108, 10610449, 81532672, 276429364, 178981018, 227705191, 1220058628, 4343291458, 2781675043, 21374899504, 72469723642

Offset: 1

Author

Ralf Stephan, Nov 17 2010

Keywords

• nonn

Comments

It is easy to prove that all the terms a(n) with n>=7 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

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000045, A006370, A105801.

Programs

Haskell

a181717 n = a181717_list !! (n-1)
a181717_list = 0 : 1 : fc 1 0 where
   fc x x' = y : fc y x where y = a006370 (x + x')
-- Reinhard Zumkeller, Oct 09 2011

Maple

a:= proc(n) option remember; local f;
      if n<3 then return n-1 fi;
      f:= a(n-1) +a(n-2);
      `if`(irem(f, 2)=0, f/2, 3*f+1)
    end:
seq(a(n), n=1..50); # Alois P. Heinz, Oct 09 2011

Mathematica

nxt[{a_,b_}]:=Module[{fib=a+b},If[OddQ[fib],{b,3fib+1},{b,fib/2}]]; Transpose[NestList[nxt,{0,1},40]][[1]] (* Harvey P. Dale, Mar 21 2012 *)

PARI

v=vector(60,n,0); v[2]=1; for(n=3,60,f=v[n-1]+v[n-2]; v[n]=if(f%2,3*f+1,f/2))

SageMath

@CachedFunction
def a(n):
    if n<3: return n-1
    elif (a(n-1)+a(n-2))%2==1: return 3*(a(n-1)+a(n-2))+1
    else: return (a(n-1)+a(n-2))/2
[a(n) for n in range(1,51)] # G. C. Greubel, Mar 25 2024