A261393 Additive terms of the rational Collatz tree.

0, 1, 2, 4, 4, 5, 8, 13, 8, 7, 10, 14, 16, 17, 26, 40, 16, 11, 14, 16, 20, 19, 28, 41, 32, 25, 34, 44, 52, 53, 80, 121, 32, 19, 22, 20, 28, 23, 32, 43, 40, 29, 38, 46, 56, 55, 82, 122, 64, 41, 50, 52, 68, 61, 88, 125, 104, 79, 106, 134

Offset

0,3

Comments

A full binary tree can be constructed by inverting the Collatz function and removing the even and odd branch conditions. The rational numbers formed by this tree have the form c = ((2^x + a(n))/(3^y)) where (y = A000120(n)) and (x >= A029837(c+1)-y).

Links

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Formula

a(0) = 0, a(2*n) = 2*a(n), and a(2*n+1) = a(n) + 3^A000120(n).

Example

a(0) = 0.
a(1) = a(2*0+1) = a(0) + 3^A000120(0) = 0 + 3^0 = 1.
a(2) = a(2*1) = 2*a(1) = 2*1 = 2.
a(3) = a(2*1+1) = a(1) + 3^A000120(1) = 1 + 3^1 = 4.
a(4) = a(2*2) = 2*a(2) = 2*2 = 4.
a(5) = a(2*2+1) = a(2) + 3^A000120(2) = 2 + 3^1 = 5.
a(6) = a(2*3) = 2*a(3) = 2*4 = 8.
a(7) = a(2*3+1) = a(3) + 3^A000120(3) = 4 + 3^2 = 13.
a(8) = a(2*4) = 2*a(4) = 2*4 = 8.

Prog

(Python)
def h(n):
  if(n == 0):     return 0
  elif(n%2 == 0): return h(n/2)
  else:           return h((n-1)/2) + 1
def a(n):
  if(n == 0):     return 0
  elif(n%2 == 0): return 2*a(n/2)
  else:           return a((n-1)/2) + 3**h((n-1)/2)
(PARI) h(n)=if(n, n>>=valuation(n, 2); h(n\2)+1, 0)
a(n)=if(n, my(k=valuation(n, 2)); n>>=k+1; (a(n)+3^h(n))<<k, 0) \\ Charles R Greathouse IV, Aug 18 2015

Crossrefs

Cf. A000120, A029837.

Sequence in context: A344710 A184615 A151969 * A366541 A327629 A121528

Adjacent sequences: A261390 A261391 A261392 * A261394 A261395 A261396

Keyword

nonn,easy

Author

Jared Deckard, Aug 17 2015

Status

approved