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A335133
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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.
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2
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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
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0,3
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COMMENTS
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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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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PROG
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(PARI) a(n) = {
my (b=binary(n), v=0);
forstep (x=#b-1, 0, -1,
if (b[1], v+=2^x);
b=vector(#b-1, k, b[k]==b[k+1])
);
return (v)
}
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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