A213540 -id:A213540 - cp's OEIS frontend

A213370 a(n) = n AND 2*n, where AND is the bitwise AND operator.

0, 0, 0, 2, 0, 0, 4, 6, 0, 0, 0, 2, 8, 8, 12, 14, 0, 0, 0, 2, 0, 0, 4, 6, 16, 16, 16, 18, 24, 24, 28, 30, 0, 0, 0, 2, 0, 0, 4, 6, 0, 0, 0, 2, 8, 8, 12, 14, 32, 32, 32, 34, 32, 32, 36, 38, 48, 48, 48, 50, 56, 56, 60, 62, 0, 0, 0, 2, 0, 0, 4, 6, 0, 0, 0, 2, 8, 8

Offset: 0

Alex Ratushnyak, Jun 14 2012

Antti Karttunen, Table of n, a(n) for n = 0..16384 (terms 0..1023 from T. D. Noe)
Index entries for sequences related to binary expansion of n

Cf. A080099, A213526, A048727, A048735, A269160.
Cf. A003714: indices of 0's.
Cf. A213540: indices of 2's, indices of 4's divided by 2.

Table[BitAnd[n, 2n], {n, 0, 63}] (* Alonso del Arte, Jun 19 2012 *)

a(n) = bitand(n, 2*n); \\ Michel Marcus, Mar 26 2021

for n in range(99):
    print(2*n & n, end=", ")

a(n) = 2 * A048735(n).

a(n) = (1/2)*(A048727(n) XOR A269160(n)) = (n OR 2n) XOR (n XOR 2n). - Antti Karttunen, May 16 2021

A348710 In the binary expansion of n, decrease the length of each run of 1-bits by one.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 2, 6, 7, 0, 0, 0, 1, 0, 0, 2, 3, 8, 4, 4, 5, 12, 6, 14, 15, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 2, 6, 7, 16, 8, 8, 9, 8, 4, 10, 11, 24, 12, 12, 13, 28, 14, 30, 31, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 2, 6, 7, 0, 0, 0

Offset: 0

Kevin Ryde, Oct 30 2021

Equivalently, change bits 01 -> 0, including a 0 reckoned above the most significant 1-bit of n so change there.
A single 1-bit run decreases to nothing.  The Fibbinary numbers (A003714) are those n with only single 1-bits so that a(n) = 0 iff n is in A003714.
a(n) = 1 iff n is in A213540 since those values end with bits 011 (which become 01) and otherwise have only single 1-bits, as do the Fibbinary numbers.
Decreasing each run is the inverse of the increase A175048 so that a(A175048(k)) = k.  This n = A175048(k) is the smallest n with a(n) = k and then other occurrences of k are by inserting single 1-bits into this n, including anywhere above the most significant bit.

			n    = 14551 = binary 111 000 11 0 1 0 111
a(n) =   787 = binary  11 000  1 0   0  11

Cf. A007088 (binary), A175048 (increase 1-bits), A090077 (decrease to single 1-bits).
Cf. A003714 (indices of 0's), A213540 (indices of 1's).
Cf. A106151 (decrease 0-bits), A318921 (decrease each run).

Table[FromDigits[Flatten[Split@IntegerDigits[n,2]/. {1,a___}:>{a}],2],{n,0,82}] (* Giorgos Kalogeropoulos, Nov 01 2021 *)

a(n) = my(v=binary(n),t=0); for(i=2,#v, if(v[i-1]||!v[i], v[t++]=v[i])); fromdigits(v[1..t],2);

def a(n): return int(bin(n).replace("b", "").replace("01", "0"), 2)
print([a(n) for n in range(83)]) # Michael S. Branicky, Oct 31 2021

cp's OEIS Frontend

A213370 a(n) = n AND 2*n, where AND is the bitwise AND operator.

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