|
|
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
|
|
|
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:
|
|
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')
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,changed
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|