OFFSET
0,4
COMMENTS
For n = binary n[k],n[k-1],...,n[0], bits a(n) = binary b[k+1],b[k],...,b[0] are b[i] = 1 when n[i-1] + n[i-2] + n[i-3] >= 2, so the majority bit 0 or 1 among the 3 bits of n below position i (with 0 bits below the radix point of n as necessary). This is since 7*n = 4*n + 2*n + n is n[i-1] + n[i-2] + n[i-3] at position i-1, and 4*n XOR 2*n XOR n is the same but no carry, so b[i] is the carry only. - Kevin Ryde, Mar 26 2021
PROG
(PARI) a(n)=7*n - bitxor(n, bitxor(2*n, 4*n)) \\ Charles R Greathouse IV, Oct 03 2016
(Python)
def A048730(n): return 7*n-(n^n<<1^n<<2) # Chai Wah Wu, Jun 29 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Apr 26 1999
STATUS
approved