

A335133


Binary interpretation of the left diagonal of the EQtriangle with first row generated from the binary expansion of n, with most significant bit given by first row.


2



0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 10, 13, 12, 14, 15, 16, 17, 18, 19, 22, 23, 20, 21, 26, 27, 24, 25, 28, 29, 30, 31, 32, 33, 35, 34, 36, 37, 39, 38, 44, 45, 47, 46, 40, 41, 43, 42, 53, 52, 54, 55, 49, 48, 50, 51, 57, 56, 58, 59, 61, 60, 62, 63, 64, 65, 66, 67
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OFFSET

0,3


COMMENTS

For any nonnegative number n, the EQtriangle for n is built by taking as first row the binary expansion of n (without leading zeros), having each entry in the subsequent rows be the EQ of the two values above it (a "1" indicates that these two values are equal, a "0" indicates that these values are different).
This sequence is a selfinverse permutation of the nonnegative numbers.


LINKS

Rémy Sigrist, Table of n, a(n) for n = 0..8192 (n = 0..2^13)
Index entries for sequences related to binary expansion of n
Index entries for sequences that are permutations of the natural numbers


FORMULA

a(floor(n/2)) = floor(a(n)/2).
abs(a(2*n+1)  a(2*n)) = 1.
a(2^k) = 2^k for any k >= 0.
a(2^k+1) = 2^k+1 for any k >= 0.
a(2^k1) = 2^k1 for any k >= 0.
Apparently, a(n) + A334727(n) = A055010(A070939(n)) for any n > 0.


EXAMPLE

For n = 42:
 the binary representation of 42 is "101010",
 the corresponding EQtriangle is:
1 0 1 0 1 0
0 0 0 0 0
1 1 1 1
1 1 1
1 1
1
 the bits on the left diagonal are: 1, 0, 1, 1, 1, 1,
 so a(42) = 2^5 + 2^3 + 2^2 + 2^1 + 2^0 = 47.


PROG

(PARI) a(n) = {
my (b=binary(n), v=0);
forstep (x=#b1, 0, 1,
if (b[1], v+=2^x);
b=vector(#b1, k, b[k]==b[k+1])
);
return (v)
}


CROSSREFS

Cf. A055010, A070939, A279645, A334727 (XOR variant).
Sequence in context: A239088 A162344 A283962 * A183083 A113220 A113218
Adjacent sequences: A335130 A335131 A335132 * A335134 A335135 A335136


KEYWORD

nonn,look,base


AUTHOR

Rémy Sigrist, May 24 2020


STATUS

approved



