A273258 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.

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, 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, 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

Offset: 1

Author

Michael De Vlieger, Aug 28 2016

Keywords

• easy
• base
• nonn

Comments

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.

This sequence decodes f(n) by reversing the digits.

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.

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.

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).

Examples

			a(3) = 2 since f(3) = "10" reversed becomes "01", loses leading zeros to become "1" -> 2.
a(6) = a(12) = "11" reversed stays the same -> 2 * 3 = 6.
a(15) = "110" reversed becomes "011", loses leading zeros to become "11" -> 6.
a(42) = "1101" reversed becomes "1011" -> 70 (a(70) = 42).
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A007947, A019565, A030101, A054841 (analogous encoding algorithm), A069799, A087207, A137502, A276379, A293448 (a bijective variant of this sequence).

Programs

Mathematica

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}]

Scheme

(define (A273258 n) (A019565 (A030101 (A087207 n)))) ;; Antti Karttunen, Jun 18 2017

Formula

a(n) = A019565(A030101(A087207(n))). - Antti Karttunen, Jun 18 2017

For all n, a(A039956(n)) = A293448(A039956(n)). - Antti Karttunen, Nov 21 2017