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

%I #14 Jun 29 2022 10:27:45

%S 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,

%T 112,124,112,120,120,124,0,0,0,12,0,8,24,28,64,64,80,92,112,120,120,

%U 124,192,192,192,204,224,232,248

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

%C 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

%o (PARI) a(n)=7*n - bitxor(n, bitxor(2*n, 4*n)) \\ _Charles R Greathouse IV_, Oct 03 2016

%o (Python)

%o def A048730(n): return 7*n-(n^n<<1^n<<2) # _Chai Wah Wu_, Jun 29 2022

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

%Y Diagonal 7 of A061858.

%K nonn,easy

%O 0,4

%A _Antti Karttunen_, Apr 26 1999