A048730 Differences between A008589 (multiples of 7) and A048727, a(n) = ((n*7)-Xmult(n,7)).

0, 0, 0, 12, 0, 8, 24, 28, 0, 0, 16, 28, 48, 56, 56, 60, 0, 0, 0, 12, 32, 40, 56, 60, 96, 96, 112, 124, 112, 120, 120, 124, 0, 0, 0, 12, 0, 8, 24, 28, 64, 64, 80, 92, 112, 120, 120, 124, 192, 192, 192, 204, 224, 232, 248

Offset: 0

Author

Antti Karttunen, Apr 26 1999

Keywords

• nonn
• easy
• look

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

Links

• Paolo Xausa, Table of n, a(n) for n = 0..8191

Crossrefs

Positions of zeros are given by A048715. Cf. A048733, A342697.

Diagonal 7 of A061858.

Programs

Mathematica

A048730[n_] := 7*n - BitXor[n, 2*n, 4*n];
Array[A048730,100, 0] (* Paolo Xausa, Aug 06 2025 *)

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