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 A335133 Binary interpretation of the left diagonal of the EQ-triangle 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS For any nonnegative number n, the EQ-triangle 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 self-inverse permutation of the nonnegative numbers. LINKS Rémy Sigrist, Table of n, a(n) for n = 0..8192 (n = 0..2^13) 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^k-1) = 2^k-1 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 EQ-triangle 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=#b-1, 0, -1,         if (b, v+=2^x);         b=vector(#b-1, 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

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Last modified August 12 11:16 EDT 2020. Contains 336438 sequences. (Running on oeis4.)