A064989 Multiplicative with a(2^e) = 1 and a(p^e) = prevprime(p)^e for odd primes p.
1, 1, 2, 1, 3, 2, 5, 1, 4, 3, 7, 2, 11, 5, 6, 1, 13, 4, 17, 3, 10, 7, 19, 2, 9, 11, 8, 5, 23, 6, 29, 1, 14, 13, 15, 4, 31, 17, 22, 3, 37, 10, 41, 7, 12, 19, 43, 2, 25, 9, 26, 11, 47, 8, 21, 5, 34, 23, 53, 6, 59, 29, 20, 1, 33, 14, 61, 13, 38, 15, 67, 4, 71, 31, 18, 17, 35, 22, 73, 3, 16
Offset: 1
Examples
a(20) = a(2^2*5) = a(2^2)*a(5) = prevprime(5) = 3.
Links
- Harry J. Smith (first 1000 terms) & Antti Karttunen, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A000040, A008578, A027746, A032742, A049084, A052126, A055396, A061395, A112798, A249817, A249818.
Other sequences whose definition involve or are some other way related with this sequence: A105560, A108951, A118306, A122111, A156552, A163511, A200746, A241909, A243070, A243071, A243072, A243073, A244319, A245605, A245607, A246165, A246266, A246268, A246277, A246278, A246361, A246362, A246371, A246372, A246373, A246374, A246376, A246380, A246675, A246682, A249745, A250470.
Programs
-
Haskell
a064989 1 = 1 a064989 n = product $ map (a008578 . a049084) $ a027746_row n -- Reinhard Zumkeller, Apr 09 2012 (MIT/GNU Scheme, with Aubrey Jaffer's SLIB Scheme library) (require 'factor) (define (A064989 n) (if (= 1 n) n (apply * (map (lambda (k) (if (zero? k) 1 (A000040 k))) (map -1+ (map A049084 (factor n))))))) ;; Antti Karttunen, May 12 2014 (definec (A064989 n) (if (= 1 n) n (* (A008578 (A055396 n)) (A064989 (A032742 n))))) ;; One based on given recurrence and utilizing memoizing definec-macro. (definec (A064989 n) (cond ((= 1 n) n) ((even? n) (A064989 (/ n 2))) (else (A163511 (/ (- (A243071 n) 1) 2))))) ;; Corresponds to one of the alternative formulas, but is very unpractical way to compute this sequence. - Antti Karttunen, Dec 18 2014
-
Maple
q:= proc(p) prevprime(p) end proc: q(2):= 1: [seq(mul(q(f[1])^f[2], f = ifactors(n)[2]), n = 1 .. 1000)]; # Robert Israel, Dec 21 2014
-
Mathematica
Table[Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n, {n, 81}] (* Michael De Vlieger, Jan 04 2016 *)
-
PARI
{ for (n=1, 1000, f=factor(n)~; a=1; j=1; if (n>1 && f[1, 1]==2, j=2); for (i=j, length(f), a*=precprime(f[1, i] - 1)^f[2, i]); write("b064989.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 02 2009
-
PARI
a(n) = {my(f = factor(n)); for (i=1, #f~, if ((p=f[i,1]) % 2, f[i,1] = precprime(p-1), f[i,1] = 1);); factorback(f);} \\ Michel Marcus, Dec 18 2014
-
PARI
A064989(n)=factorback(Mat(apply(t->[max(precprime(t[1]-1),1),t[2]],Vec(factor(n)~))~)) \\ M. F. Hasler, Dec 29 2014
-
Python
from sympy import factorint, prevprime from operator import mul from functools import reduce def a(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]) print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 15 2017
-
Python
from math import prod from sympy import prevprime, factorint def A064989(n): return prod(prevprime(p)**e for p, e in factorint(n>>(~n&n-1).bit_length()).items()) # Chai Wah Wu, Jan 05 2023
Formula
From Antti Karttunen, Dec 18 2014: (Start)
If n = product A000040(k)^e(k) then a(n) = product A008578(k)^e(k) [where A000040(n) gives the n-th prime, and A008578(n) gives 1 for 1 and otherwise the (n-1)-th prime].
a(1) = 1; for n > 1, a(n) = A008578(A055396(n)) * a(A032742(n)). [Above formula represented as a recurrence. Cf. A252461.]
a(1) = 1; for n > 1, a(n) = A008578(A061395(n)) * a(A052126(n)). [Compare to the formula of A252462.]
This prime-shift operation is used in the definitions of many other sequences, thus it can be expressed in many alternative ways:
a(n) = A200746(n) / n.
a(n) = A122111(A122111(n)/A105560(n)) = A122111(A052126(A122111(n))). [In A112798-partition context: conjugate, remove the largest part (the largest prime factor), and conjugate again.]
a(n) = A249818(A250470(A249817(n))). [A250470 is an analogous operation for "going one step up" in the square array A083221 (A083140).]
(End)
Product_{k=1..n} a(k) = n! / A307035(n). - Vaclav Kotesovec, Mar 21 2019
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((p^2-p)/(p^2-q(p))) = 0.220703928... , where q(p) = prevprime(p) (A151799) if p > 2 and q(2) = 1. - Amiram Eldar, Nov 18 2022
Comments