A116640 a(n) = A116623(A059893(n)).

1, 5, 7, 19, 11, 23, 29, 65, 19, 31, 37, 73, 49, 85, 103, 211, 35, 47, 53, 89, 65, 101, 119, 227, 89, 125, 143, 251, 179, 287, 341, 665, 67, 79, 85, 121, 97, 133, 151, 259, 121, 157, 175, 283, 211, 319, 373, 697, 169, 205, 223, 331, 259, 367, 421, 745, 331

Offset: 0

Author

Antti Karttunen, Feb 20 2006. Proposed by Pierre Lamothe (plamothe(AT)aei.ca), May 21 2004

Keywords

• nonn
• base
• tabf

Comments

Viewed as a binary tree, this is (1); 5; 7,19; 11,23,29,65; ... Cf. A116623.

If we treat (2n+1) as a binary number with the nonzero bits numbered (highest bit first) from 0..k and the regular binary place value of each nonzero bit numbered from b(0) to b(k) then a(n) = 3^0 * b(0) + 3^1 * b(1) + .. + 3^k. For instance, if n=6 then 2n+1 = 13, which is equal to 8+4+1 or 1101 base(2); and a(n)=29 which is 8*1 + 4*3 + 1*9. - Joe Slater, Jan 23 2016

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A000079, A000225, A001047, A062709.

Programs

Maple

a:= proc(n) option remember; piecewise(
    n mod 4 = 0, 3*procname(n/2) - 2*procname(n/4),
  n mod 4 = 1, 6*procname((n-1)/4) - procname((n-1)/2),
  n mod 4 = 2, procname(n/2) + 2*procname((n-2)/4),
  5*procname((n-1)/2) - 6*procname((n-3)/4))
end proc:
a(0):= 1:
map(a, [$0..100]); # Robert Israel, Jan 19 2016

Mathematica

a[n_] := a[n] = Switch[Mod[n, 4], 0, 3a[Floor[n/2]] - 2a[Floor[n/4]], 1, 6a[Floor[n/4]] - a[Floor[n/2]], 2, a[Floor[n/2]] + 2a[Floor[n/4]], 3, 5a[Floor[n/2]] - 6a[Floor[n/4]]]; a[0]=1; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 28 2016 *)

PARI

a(n) = if(n==0, return(1)); 2*a(n\2) - (-1)^n * 3^hammingweight(n) \\ Charles R Greathouse IV, Jan 21 2016

PARI

a(n) = my(p=2*n+1,v=vecextract(vector(#binary(p),j,2^(j-1)),p));sum(i=0,#v-1,3^i*v[#v-i]) \\ Joe Slater, May 09 2017

Formula

a(A000225(n)) = A001047(n+1).

For n>= 1 a(A000079(n)) = A062709(n+1).

From Joe Slater, Jan 19 2016: (Start)

a(0) = 1,

a(n) = 3*a(floor(n/2)) - 2*a(floor(n/4)) for n=0 (mod 4) and n>0,

a(n) = 6*a(floor(n/4)) - a(floor(n/2)) for n=1 (mod 4),

a(n) = a(floor(n/2)) + 2*a(floor(n/4)) for n=2 (mod 4),

a(n) = 5*a(floor(n/2)) - 6*a(floor(n/4)) for n=3 (mod 4)

(End)

a(0) = 1, a(n) = 2*a(floor(n/2)) - A033999(n) * A048883(n) for n>0. -

Joe Slater, Jan 22 2016