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A252463
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Hybrid shift: a(1) = 1, a(2n) = n, a(2n+1) = A064989(2n+1); shift the even numbers one bit right, shift the prime factorization of odd numbers one step towards smaller primes.
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82
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1, 1, 2, 2, 3, 3, 5, 4, 4, 5, 7, 6, 11, 7, 6, 8, 13, 9, 17, 10, 10, 11, 19, 12, 9, 13, 8, 14, 23, 15, 29, 16, 14, 17, 15, 18, 31, 19, 22, 20, 37, 21, 41, 22, 12, 23, 43, 24, 25, 25, 26, 26, 47, 27, 21, 28, 34, 29, 53, 30, 59, 31, 20, 32, 33, 33, 61, 34, 38, 35, 67, 36, 71, 37, 18, 38, 35, 39, 73, 40, 16
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OFFSET
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1,3
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COMMENTS
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For any node n >= 2 in binary trees A005940 and A163511, a(n) gives the parent node of n. (Here we assume that their initial root 1 is its own parent).
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LINKS
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FORMULA
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a(1) = 1, a(2n) = n, a(2n+1) = A064989(2n+1).
Other identities. For all n >= 1:
Sum_{k=1..n} a(k) ~ c * n^2, where c = 1/8 + (1/2) * Product_{p prime > 2} ((p^2-p)/(p^2-q(p))) = 0.2905279467..., where q(p) = prevprime(p) (A151799). - Amiram Eldar, Jan 21 2023
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MATHEMATICA
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Table[Which[n == 1, 1, EvenQ@ n, n/2, True, Times @@ Power[
Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n], {n, 81}] (* Michael De Vlieger, Sep 16 2017 *)
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PROG
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(Scheme) (define (A252463 n) (cond ((<= n 1) n) ((even? n) (/ n 2)) (else (A064989 n))))
(Python)
from sympy import factorint, prevprime
from operator import mul
def a064989(n):
f = factorint(n)
return 1 if n==1 else reduce(mul, [1 if i==2 else prevprime(i)**f[i] for i in f])
def a(n): return 1 if n==1 else n//2 if n%2==0 else a064989(n)
(PARI) a064989(n) = factorback(Mat(apply(t->[max(precprime(t[1]-1), 1), t[2]], Vec(factor(n)~))~)); \\ A064989
a(n) = if (n==1, 1, if (n%2, a064989(n), n/2)); \\ Michel Marcus, Oct 13 2021
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CROSSREFS
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A252464 gives the number of iterations needed to reach 1 from n.
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KEYWORD
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AUTHOR
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STATUS
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approved
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