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A243589 Numbers returned when each binary digit of n is replaced by the sum modulo 2 of the digits to its (wrapped) left and (wrapped) right. 1
0, 0, 0, 3, 5, 6, 0, 5, 15, 0, 10, 15, 5, 10, 0, 9, 27, 12, 30, 3, 17, 6, 20, 29, 15, 24, 10, 23, 5, 18, 0, 17, 51, 20, 54, 27, 57, 30, 60, 5, 39, 0, 34, 15, 45, 10, 40, 57, 27, 60, 30, 51, 17, 54, 20, 45, 15, 40, 10, 39, 5, 34, 0, 33, 99, 36, 102, 43, 105, 46 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

a(n) = ror(n) XOR rol(n), where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 24 2016

Numbers returned by the following function: take the t binary digits of n, d(1)..d(t), and replace each with the sum d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) maps to t and (i+1 > t) maps to 1.

LINKS

Anthony Sand, Table of n, a(n) for n = 1..1000

FORMULA

for digits d(1)..d(t), d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) -> t, (i+1 > t) -> 1.

EXAMPLE

For 1, the function returns d(1) = (d(1) + d(1)) mod 2 = (1 + 1) mod 2 = 0.

For 5, the initial digits are (1,0,1).

d(1) = (d(3) + d(2)) mod 2 = (1 + 0) mod 2 = 1; d(2) = (d(1) + d(3)) mod 2 = (1 + 1) mod 2 = 0; d(3) = (d(2) + d(1)) mod 2 = (0 + 1) mod 2 = 1.

The function returns (1,0,1) = 101 = 5 in base 10.

PROG

(Python)

for n in range(1, 100):

BL = len(bin(n))-2

x = (n>>1) + ((n&1) << (BL-1)) # A038572(n)

x^= (n*2) - (1<<BL) + 1 # A006257(n) for n>0

print str(x)+', ',

CROSSREFS

Cf. A006257, A035327, A038572, A055120.

Sequence in context: A186969 A111950 A274792 * A178255 A154467 A152713

Adjacent sequences: A243586 A243587 A243588 * A243590 A243591 A243592

KEYWORD

nonn,easy,base,look

AUTHOR

Anthony Sand, Jun 07 2014

STATUS

approved

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Last modified December 1 22:12 EST 2022. Contains 358484 sequences. (Running on oeis4.)