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A064216
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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.
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116
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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
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1,2
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COMMENTS
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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 [re-expressed by Peter Munn, May 23 2020]
Not multiplicative in usual sense - but letting m=2n-1=product_j (p_j)^(e_j) then a(n)=a((m+1)/2)=product_j (p_(j-1))^(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)
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LINKS
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FORMULA
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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
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EXAMPLE
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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
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MATHEMATICA
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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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PROG
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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
(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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CROSSREFS
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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 base-2 representation of a(n), respectively).
Here obviously the variant 2, A151799(n) = A007917(n-1), of the prevprime function is used.
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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