A341335 For any number n with binary expansion (b_1, ..., b_k), the binary expansion of a(n), say (c_1, ..., c_k) satisfies c_m = Sum_{d | m} b_d mod 2 for m = 1..k.
0, 1, 3, 2, 7, 6, 5, 4, 15, 14, 13, 12, 10, 11, 8, 9, 31, 30, 29, 28, 27, 26, 25, 24, 21, 20, 23, 22, 17, 16, 19, 18, 63, 62, 61, 60, 59, 58, 57, 56, 54, 55, 52, 53, 50, 51, 48, 49, 42, 43, 40, 41, 46, 47, 44, 45, 35, 34, 33, 32, 39, 38, 37, 36, 127, 126, 125
Offset: 0
Examples
For n = 42: - the binary expansion of 42 is (1, 0, 1, 0, 1, 0), - the binary expansion of a(42) has 6 digits: - the 1st digit = 1 mod 2 = 1, - the 2nd digit = 1 + 0 mod 2 = 1, - the 3rd digit = 1 + 1 mod 2 = 0, - the 4th digit = 1 + 0 + 0 mod 2 = 1, - the 5th digit = 1 + 1 mod 2 = 0, - the 6th digit = 1 + 0 + 1 + 0 mod 2 = 0, - so the binary expansion of a(42) is "110100", - and a(42) = 52.
Links
Programs
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PARI
a(n) = { my (b=binary(n), c=vector(#b)); for (m=1, #b, fordiv (m, d, c[m]=(c[m] + b[d])%2)); fromdigits(c, 2) }
Formula
a(n) < 2^k for any n < 2^k.
a(floor(n/2)) = floor(a(n)/2).
a(2^k) = 2^(k+1) - 1 for any k >= 0.
Comments