|
%I #55 Jul 16 2021 01:35:58
%S 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,
%T 65,42,35,512,37,1024,31,66,129,68,43,2048,257,130,39,4096,69,8192,67,
%U 74,513,16384,47,136,73,258,131,32768,85,132,71,514,1025,65536,75
%N 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).
%C 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.
%C 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).
%C 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.
%C 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.
%C For example, for n=22, n_1=21=((10)^2)(1), and s(22)=3^2*2=18.
%C 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
%H Peter J. C. Moses, <a href="/A233249/b233249.txt">Table of n, a(n) for n = 1..2500</a>
%F A059893(a(n)) = A333220(n). A124767(a(n)) = A001221(n). - _Gus Wiseman_, Apr 02 2020
%e n=10=2*5 is mapped to (1)(100)~(100)(1). Since 9 is in A114994, then a(10)=9.
%e From _Gus Wiseman_, Apr 02 2020: (Start)
%e The sequence together with the corresponding compositions begins:
%e 0: () 128: (8) 2048: (12)
%e 1: (1) 19: (3,1,1) 257: (8,1)
%e 2: (2) 34: (4,2) 130: (6,2)
%e 3: (1,1) 33: (5,1) 39: (3,1,1,1)
%e 4: (3) 256: (9) 4096: (13)
%e 5: (2,1) 23: (2,1,1,1) 69: (4,2,1)
%e 8: (4) 36: (3,3) 8192: (14)
%e 7: (1,1,1) 65: (6,1) 67: (5,1,1)
%e 10: (2,2) 42: (2,2,2) 74: (3,2,2)
%e 9: (3,1) 35: (4,1,1) 513: (9,1)
%e 16: (5) 512: (10) 16384: (15)
%e 11: (2,1,1) 37: (3,2,1) 47: (2,1,1,1,1)
%e 32: (6) 1024: (11) 136: (4,4)
%e 17: (4,1) 31: (1,1,1,1,1) 73: (3,3,1)
%e 18: (3,2) 66: (5,2) 258: (7,2)
%e 15: (1,1,1,1) 129: (7,1) 131: (6,1,1)
%e 64: (7) 68: (4,3) 32768: (16)
%e 21: (2,2,1) 43: (2,2,1,1) 85: (2,2,2,1)
%e 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.
%e (End)
%t primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
%t Table[Total[2^Accumulate[primeMS[n]]]/2,{n,100}] (* _Gus Wiseman_, Apr 02 2020 *)
%Y The sorted version is A114994.
%Y The primorials A002110 map to A246534.
%Y A partial inverse is A333219.
%Y The reversed version is A333220.
%Y Cf. A000120, A029931, A035327, A048793, A066099, A070939, A124767, A124768, A228351, A272020, A333217, A333221.
%K nonn,base,look
%O 1,3
%A _Vladimir Shevelev_, Dec 06 2013
%E More terms from _Peter J. C. Moses_, Dec 07 2013
|
