OFFSET
1,1
COMMENTS
From Chai Wah Wu, Nov 21 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 2 and f(1) = a(2) + a(3) = 15.
Then f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) = Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first order recurrence relation with the initial condition f(1) = 15 shows that f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
(End)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.
FORMULA
a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. - Chai Wah Wu, Nov 21 2018
EXAMPLE
6 in binary is 110. Modify each run by appending the opposite digit to get 11001, which is 25 in decimal. So a(6) = 25.
PROG
(Python)
from re import split
def A320037(n):
return int(''.join(d+'0' if '1' in d else d+'1' for d in split('(0+)|(1+)', bin(n)[2:]) if d != '' and d != None), 2)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Chai Wah Wu, Oct 04 2018
STATUS
approved