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.
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
Offset: 1
Examples
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
Links
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)).)
Programs
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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 *)
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PARI
a(n) = {my(f = factor(2*n-1)); for (k=1, #f~, f[k,1] = precprime(f[k,1]-1)); factorback(f);} \\ Michel Marcus, Mar 17 2016
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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
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Scheme
(define (A064216 n) (A064989 (- (+ n n) 1))) ;; Antti Karttunen, May 12 2014
Formula
a(n) = A064989(2n - 1). - Antti Karttunen, May 12 2014
Sum_{k=1..n} a(k) ~ c * n^2, where c = Product_{p prime > 2} ((p^2-p)/(p^2-q(p))) = 0.6621117868..., where q(p) = prevprime(p) (A151799). - Amiram Eldar, Jan 21 2023
Extensions
More terms from Reinhard Zumkeller, Sep 26 2001
Additional description added by Antti Karttunen, May 12 2014
Comments