A078440 Numbers n with property that n is not a power of 2 and the finite sequence n, f(n), f(f(n)), ...., 1 in the Collatz (or 3x + 1) problem contains exactly one prime. (The earliest "1" is meant.)

21, 42, 84, 85, 168, 170, 336, 340, 341, 453, 672, 680, 682, 906, 909, 1344, 1360, 1364, 1365, 1812, 1813, 1818, 2688, 2720, 2728, 2730, 3624, 3626, 3636, 5376, 5440, 5456, 5460, 5461, 7248, 7252, 7272, 7281, 9669

Offset: 1

Author

Joseph L. Pe, Dec 31 2002

Keywords

• nonn

Comments

f(n) = n/2 if n is even, = 3n + 1 if n is odd. Powers 2^n trivially have exactly one prime in n, f(n), f(f(n)), ..., 2, 1, namely 2 and so are excluded from the sequence.

A055509(a(n)) = 0; A078350(a(n)) <= 1.

Examples

			n, f(n), f(f(n)), .... for n = 21 is: 21, 64, 32, 16, 8, 4, 2, 1, which has exactly one prime, that is, 2. Hence 21 belongs to the sequence.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Collatz Problem
• Wikipedia, Collatz conjecture
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

A006370; subsequence of A196871 (with binary powers).

Programs

Haskell

a078440 n = a078440_list !! (n-1)
a078440_list = filter notbp a196871_list where
   notbp x = m > 0 && x > 1 || m == 0 && notbp x' where
      (x',m) = divMod x 2
-- Reinhard Zumkeller, Oct 08 2011

Mathematica

f[n_] := n/2 /; Mod[n, 2] == 0 f[n_] := 3 n + 1 /; Mod[n, 2] == 1 g[n_] := Module[{i, p}, i = n; p = 0; While[i > 1, If[PrimeQ[i], p = p + 1]; i = f[i]]; p]; Select[Range[10^4], g[ # ] == 1 && ! IntegerQ[Log[2, # ]] &]
pQ[n_]:=Count[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n, #>1&], ?PrimeQ] == 1; With[ {nn=10000},Complement[Select[Range[nn],pQ],2^Range[Floor[ Log[ 2,nn]]]]] (* _Harvey P. Dale, Oct 19 2011 *)