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 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. 7
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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 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}] PROG (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)) (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 CROSSREFS Cf. A000012, A000045, A006370, A181663. Cf. A181717. Sequence in context: A033468 A047816 A095845 * A086064 A076374 A142954 Adjacent sequences: A105798 A105799 A105800 * A105802 A105803 A105804 KEYWORD nonn AUTHOR Zak Seidov, Sep 12 2006 EXTENSIONS Incorrect formula deleted by Colin Barker, Jul 29 2013 STATUS approved

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Last modified August 13 17:56 EDT 2024. Contains 375144 sequences. (Running on oeis4.)