

A233249


a(1)=0; let, for k>=1, prime(k) map to 10...0 with k1 zeros and prime(k)*prime(m) map to concatenation in binary of 2^(k1) and 2^(m1). Let, for n>=2, prime power factorization of n is mapped to r(n). a(n) is the term in A114994 which is cequivalent to r(n) (see there our comment).


7



0, 1, 2, 3, 4, 5, 8, 7, 10, 9, 16, 11, 32, 17, 18, 15, 64, 21, 128, 19, 34, 33, 256, 23, 36, 65, 42, 35, 512, 37, 1024, 31, 66, 129, 68, 43, 2048, 257, 130, 39, 4096, 69, 8192, 67, 74, 513, 16384, 47, 136, 73, 258, 131, 32768, 85, 132, 71, 514, 1025, 65536, 75
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OFFSET

1,3


COMMENTS

Let (10...0)_i (i>=0) denote 2^i in binary. Under (10...0)_i^k we understand a concatenation of (10...0)_i k times.
If n=prod {i=1,...,m}p_i^t_i is prime power factorization of n, then in the name r(n)=concatenation{i=1,...,m} ((10...0_(i1)^t_i).
Numbers q and s are called cequivalent, if their binary expansions contain the same set of parts of the form 10...0. For example, 14=(1)(1)(10)~(10)(1)(1)=11.
Conversely, if n~n_1, such that n_1 is in A114994 and has cfactorization: n_1= concatenation{i=m,...,0} ((10...0)_i^t_i), one can consider "converse" sequence {s(n)}, where s(n)=prod {i=m,...,0}p_(i+1)^t_i.
For example, for n=22, n_1=21=((10)^2)(1), and s(22)=3^2*2=18.


LINKS

Peter J. C. Moses, Table of n, a(n) for n = 1..2500


EXAMPLE

n=10=2*5 is mapped to (1)(100)~(100)(1). Since 9 is in A114994, then a(10)=9.


CROSSREFS

Cf. A114994.
Sequence in context: A245822 A069797 A158979 * A330573 A309369 A091893
Adjacent sequences: A233246 A233247 A233248 * A233250 A233251 A233252


KEYWORD

nonn,base,look


AUTHOR

Vladimir Shevelev, Dec 06 2013


EXTENSIONS

More terms from Peter J. C. Moses, Dec 07 2013


STATUS

approved



