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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. 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

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

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 *)

PROG

(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))

(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

CROSSREFS

Cf. A105801.

Cf. A006370, A000045.

Sequence in context: A061093 A067178 A346460 * A224173 A223864 A223987

Adjacent sequences:  A181714 A181715 A181716 * A181718 A181719 A181720

KEYWORD

nonn

AUTHOR

Ralf Stephan, Nov 17 2010

STATUS

approved

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Last modified January 20 15:42 EST 2022. Contains 350472 sequences. (Running on oeis4.)