A233249 a(1)=0; for k >= 1, let prime(k) map to 10...0 with k-1 zeros and let prime(k)*prime(m) map to the concatenation in binary of 2^(k-1) and 2^(m-1). For n >= 2, let the prime power factorization of n be mapped to r(n). a(n) is the term in A114994 which is c-equivalent to r(n) (see there our comment).

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

Offset: 1

Author

Vladimir Shevelev, Dec 06 2013

Keywords

• nonn
• base
• look

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=Product_{i=1..m} p_i^t_i is the prime power factorization of n, then in the name r(n)=concatenation{i=1..m} ((10...0_(i-1)^t_i).

Numbers q and s are called c-equivalent 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 c-factorization: n_1 = concatenation{i=m,...,0} ((10...0)i^t_i), one can consider "converse" sequence {s(n)}, where s(n) = Product{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.

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary binary expansion of k, prepending 0, taking first differences, and reversing again. Then a(n) is the number k such that the k-th composition in standard order consists of the prime indices of n in weakly decreasing order (the partition with Heinz number n). - Gus Wiseman, Apr 02 2020

Examples

			n=10=2*5 is mapped to (1)(100)~(100)(1). Since 9 is in A114994, then a(10)=9.
From _Gus Wiseman_, Apr 02 2020: (Start)
The sequence together with the corresponding compositions begins:
   0: ()             128: (8)             2048: (12)
   1: (1)             19: (3,1,1)          257: (8,1)
   2: (2)             34: (4,2)            130: (6,2)
   3: (1,1)           33: (5,1)             39: (3,1,1,1)
   4: (3)            256: (9)             4096: (13)
   5: (2,1)           23: (2,1,1,1)         69: (4,2,1)
   8: (4)             36: (3,3)           8192: (14)
   7: (1,1,1)         65: (6,1)             67: (5,1,1)
  10: (2,2)           42: (2,2,2)           74: (3,2,2)
   9: (3,1)           35: (4,1,1)          513: (9,1)
  16: (5)            512: (10)           16384: (15)
  11: (2,1,1)         37: (3,2,1)           47: (2,1,1,1,1)
  32: (6)           1024: (11)             136: (4,4)
  17: (4,1)           31: (1,1,1,1,1)       73: (3,3,1)
  18: (3,2)           66: (5,2)            258: (7,2)
  15: (1,1,1,1)      129: (7,1)            131: (6,1,1)
  64: (7)             68: (4,3)          32768: (16)
  21: (2,2,1)         43: (2,2,1,1)         85: (2,2,2,1)
For example, the Heinz number of (2,2,1) is 18, and the 21st composition in standard order is (2,2,1), so a(18) = 21.
(End)
		

Links

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

Crossrefs

The sorted version is A114994.

The primorials A002110 map to A246534.

A partial inverse is A333219.

The reversed version is A333220.

Cf. A000120, A029931, A035327, A048793, A066099, A070939, A124767, A124768, A228351, A272020, A333217, A333221.

Programs

Mathematica

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Total[2^Accumulate[primeMS[n]]]/2,{n,100}] (* Gus Wiseman, Apr 02 2020 *)

Formula

A059893(a(n)) = A333220(n). A124767(a(n)) = A001221(n). - Gus Wiseman, Apr 02 2020

Extensions

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