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A233249
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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).
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27
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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
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1,3
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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n=10=2*5 is mapped to (1)(100)~(100)(1). Since 9 is in A114994, then a(10)=9.
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)
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MATHEMATICA
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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 *)
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CROSSREFS
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Cf. A000120, A029931, A035327, A048793, A066099, A070939, A124767, A124768, A228351, A272020, A333217, A333221.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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