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%I #26 Nov 21 2017 21:35:19
%S 1,2,2,2,2,6,2,2,2,10,2,6,2,14,6,2,2,6,2,10,10,22,2,6,2,26,2,14,2,30,
%T 2,2,14,34,6,6,2,38,22,10,2,70,2,22,6,46,2,6,2,10,26,26,2,6,10,14,34,
%U 58,2,30,2,62,10,2,14,154,2,34,38,42,2,6,2,74,6,38,6,286,2,10,2,82,2,70,22,86
%N Write the distinct prime divisors p of n in the (PrimePi(p) - 1)-th place, ignoring multiplicity. Decode the resulting number after first reversing the code, ignoring any leading zeros.
%C Encode n with the function f(n) = noting the distinct prime divisors p of n by writing "1" in the (PrimePi(n) - 1)-th place, e.g, f(6) = f(12) = "11". This function is akin to A054841(n) except we don't note the multiplicity e of p in n, rather merely note "1" if e > 0.
%C This sequence decodes f(n) by reversing the digits.
%C If we decode f(n) without reversal, we have A007947(n), since f(n) sets any multiplicity e > 1 of prime divisor p of n to 1.
%C All terms except a(1) are of the form 2x with x odd. a(1) = 1, since f(1) = "0" and stands unaffected in reversal and decoding, and any zeros to the right of all 1's are lost in reversal. Thus f(15) = "110" reversed becomes "011" -> "11" decoded equals 2 * 3 = 6. Because we lose leading zeros, we always have 1 in position 1, which decoded is interpreted as the factor 2.
%C a(p) for p prime = 2, since primes are written via f(p) as 1 in the (PrimePi(p)-1)-th place. There is only one 1 in this number (similar to a perfect power of ten decimally) and when it is reversed, the number loses all leading zeros to become "1" -> 2. This also applies to prime powers p^e, since e is rendered as 1 by f(p^e), i.e., f(p^e) = f(p).
%H Michael De Vlieger, <a href="/A273258/b273258.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F a(n) = A019565(A030101(A087207(n))). - _Antti Karttunen_, Jun 18 2017
%F For all n, a(A039956(n)) = A293448(A039956(n)). - _Antti Karttunen_, Nov 21 2017
%e a(3) = 2 since f(3) = "10" reversed becomes "01", loses leading zeros to become "1" -> 2.
%e a(6) = a(12) = "11" reversed stays the same -> 2 * 3 = 6.
%e a(15) = "110" reversed becomes "011", loses leading zeros to become "11" -> 6.
%e a(42) = "1101" reversed becomes "1011" -> 70 (a(70) = 42).
%t Table[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ If[# == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> 1 &, f]]@ FactorInteger@ #] &@ n, {n, 86}]
%o (Scheme) (define (A273258 n) (A019565 (A030101 (A087207 n)))) ;; _Antti Karttunen_, Jun 18 2017
%Y Cf. A007947, A019565, A030101, A054841 (analogous encoding algorithm), A069799, A087207, A137502, A276379, A293448 (a bijective variant of this sequence).
%K easy,base,nonn
%O 1,2
%A _Michael De Vlieger_, Aug 28 2016
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