

A064216


Replace each p^e with prevprime(p)^e in the prime factorization of odd numbers; inverse of sequence A048673 considered as a permutation of the natural numbers.


116



1, 2, 3, 5, 4, 7, 11, 6, 13, 17, 10, 19, 9, 8, 23, 29, 14, 15, 31, 22, 37, 41, 12, 43, 25, 26, 47, 21, 34, 53, 59, 20, 33, 61, 38, 67, 71, 18, 35, 73, 16, 79, 39, 46, 83, 55, 58, 51, 89, 28, 97, 101, 30, 103, 107, 62, 109, 57, 44, 65, 49, 74, 27, 113, 82, 127, 85, 24, 131
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OFFSET

1,2


COMMENTS

a((A003961(n) + 1) / 2) = n and A003961(a(n)) = 2*n  1 for all n. If the sequence is indexed by odd numbers only, it becomes multiplicative. In this variant sequence, denoted b, even indices don't exist, and we get b(1) = a(1) = 1, b(3) = a(2) = 2, b(5) = 3, b(7) = 5, b(9) = 4 = b(3) * b(3), ... , b(15) = 6 = b(3) * b(5), and so on. This property can also be stated as: a(x) * a(y) = a(((2x  1) * (2y  1) + 1) / 2) for x, y > 0.  Reinhard Zumkeller [reexpressed by Peter Munn, May 23 2020]
Not multiplicative in usual sense  but letting m=2n1=product_j (p_j)^(e_j) then a(n)=a((m+1)/2)=product_j (p_(j1))^(e_j).  Henry Bottomley, Apr 15 2005
Several permutations that use prime shift operation A064989 in their definition yield a permutation obtained from their odd bisection when composed with this permutation from the right. For example, we have:
(End)


LINKS



FORMULA

Sum_{k=1..n} a(k) ~ c * n^2, where c = Product_{p prime > 2} ((p^2p)/(p^2q(p))) = 0.6621117868..., where q(p) = prevprime(p) (A151799).  Amiram Eldar, Jan 21 2023


EXAMPLE

For n=11, the 11th odd number is 2*11  1 = 21 = 3^1 * 7^1. Replacing the primes 3 and 7 with the previous primes 2 and 5 gives 2^1 * 5^1 = 10, so a(11) = 10.  Michael B. Porter, Jul 25 2016


MATHEMATICA

Table[Times @@ Power[If[# == 1, 1, NextPrime[#, 1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger[2 n  1], {n, 69}] (* Michael De Vlieger, Dec 18 2014, revised Mar 17 2016 *)


PROG

(PARI) a(n) = {my(f = factor(2*n1)); for (k=1, #f~, f[k, 1] = precprime(f[k, 1]1)); factorback(f); } \\ Michel Marcus, Mar 17 2016
(Python)
from sympy import factorint, prevprime
from operator import mul
def a(n):
f=factorint(2*n  1)
return 1 if n==1 else reduce(mul, [prevprime(i)**f[i] for i in f]) # Indranil Ghosh, May 13 2017


CROSSREFS

Cf. A246361 (numbers n such that a(n) <= n.)
Cf. A246362 (numbers n such that a(n) > n.)
Cf. A246371 (numbers n such that a(n) < n.)
Cf. A246372 (numbers n such that a(n) >= n.)
Cf. A246373 (primes p such that a(p) >= p.)
Cf. A246374 (primes p such that a(p) < p.)
Cf. A246343 (iterates starting from n=12.)
Cf. A246345 (iterates starting from n=16.)
Cf. A245448 (this permutation "squared", a(a(n)).)
Cf. A253894, A254044, A254045 (binary width, weight and the number of nonleading zeros in base2 representation of a(n), respectively).
Here obviously the variant 2, A151799(n) = A007917(n1), of the prevprime function is used.


KEYWORD

easy,nonn


AUTHOR



EXTENSIONS



STATUS

approved



