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A233569
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Canonical parts power representation of n: n = concatenation((1)^k_1,(10)^k_2,...).
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6
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0, 1, 2, 3, 4, 6, 6, 7, 8, 12, 10, 14, 12, 14, 14, 15, 16, 24, 20, 28, 20, 26, 26, 30, 24, 28, 26, 30, 28, 30, 30, 31, 32, 48, 40, 56, 36, 52, 52, 60, 40, 52, 42, 58, 52, 58, 58, 62, 48, 56, 52, 60, 52, 58, 58, 62, 56, 60, 58, 62, 60, 62, 62, 63, 64, 96, 80
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OFFSET
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0,3
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COMMENTS
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Two numbers n_1 and n_2 are called c-equivalent (n_1~n_2) if in the binary they have the same parts of the form 10...0 with k>=0 zeros up to a permutation of them. For example, 6~5, 14~13~11, 12~9.
Denote by (10...0)^k the concatenation k the same consecutive parts (10...0). By agreement, (10...0)^0 denotes the absence of the corresponding part in the binary of n. Let n contains k_i parts with i-1 zeros, i=1,2,... . Then n~concatenation((1)^k_1, (10)^k_2,(100)^k_3,...). The latter number is a(n). Thus a(n_1)=a(n_2) if and only if n_1~n_2. For example, since a(19)=28 which is in binary 11100, then the canonical representation of 19 is (1)^2[*](100), where [*] means concatenation. Analogously, since a(23)=30 which in binary 11110, then the canonical representation of 23 is (1)^3[*](10).
As a natural application, consider a notion of parts power divisor of canonical representation of n. We consider parts power divisors only of the form a(m).
If the canonical representation of n is a(n)=(1)^k_1[*](10)^k_2[*](100)^k_3[*]..., then number a(m) is a parts power divisor of a(n), iff a(m)=(1)^t_1[*](10)^t_2[*](100)^t_3[*]... with all t_i<=k_i. In particular, 0 (with all t_i=0) is parts power divisor of every a(n). From this it follows that the number of primes power divisors of a(n) is (k_1+1)*(k_2+1)*... This number is an upper estimate for A124771(n).
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LINKS
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MATHEMATICA
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bitPatt[n_]:=bitPatt[n]=Split[IntegerDigits[n, 2], #1>#2||#2==0&]; Map[FromDigits[Flatten[Sort[bitPatt[#]]], 2]&, Range[0, 33]] (* Peter J. C. Moses, Dec 14 2013 *)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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