A064097 A quasi-logarithm defined inductively by a(1) = 0 and a(p) = 1 + a(p-1) if p is prime and a(n*m) = a(n) + a(m) if m,n > 1.

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 6, 7, 6, 7, 5, 7, 6, 7, 6, 7, 7, 7, 6, 7, 7, 8, 7, 7, 8, 9, 6, 8, 7, 7, 7, 8, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 6, 8, 8, 9, 7, 9, 8, 9, 7, 8, 8, 8, 8, 9, 8, 9, 7, 8, 8, 9, 8, 8, 9, 9, 8, 9, 8, 9, 9, 9, 10, 9, 7, 8, 9, 9, 8, 9, 8, 9, 8

Offset: 1

Author

Thomas Schulze (jazariel(AT)tiscalenet.it), Sep 16 2001

Keywords

• nonn

Comments

Note that this is the logarithm of a completely multiplicative function. - Michael Somos

Number of iterations of r -> r - (largest divisor d < r) needed to reach 1 starting at r = n. a(n) = a(n - A032742(n)) + 1 for n >= 2. - Jaroslav Krizek, Jan 28 2010

From Antti Karttunen, Apr 04 2020: (Start)

Krizek's comment above stems from the fact that n - n/p = (p-1)*(n/p), where p is the least prime dividing n [= A020639(n), thus n/p = A032742(n)] and because this is fully additive sequence we can write a(n) = a(p) + a(n/p) = (1+a(p-1)) + a(n/p) = 1 + a((p-1)*(n/p)) = 1 + a(n - n/p), for any composite n.

Note that in above formula p can be any prime factor of n, not only the smallest, which proves Robert G. Wilson v's comment in A333123 that all such iteration paths from n to 1 are of the same length, regardless of the route taken.

(End)

From Antti Karttunen, May 11 2020: (Start)

Moreover, those paths form the chains of a graded poset, which is also a lattice. See the Mathematics Stack Exchange link.

Keeping the formula otherwise same, but changing it for primes either as a(p) = 1 + a(A064989(p)), a(p) = 1 + a(floor(p/2)) or a(p) = 1 + a(A048673(p)) gives sequences A056239, A064415 and A334200 respectively.

(End)

a(n) is the number of iterations r->A060681(r) to reach 1 starting at r=n. - R. J. Mathar, Nov 06 2023

Examples

			a(19) = 6: 19 - 1 = 18; 18 - 9 = 9; 9 - 3 = 6; 6 - 3 = 3; 3 - 1 = 2; 2 - 1 = 1. That is a total of 6 iterations. - _Jaroslav Krizek_, Jan 28 2010
From _Antti Karttunen_, Apr 04 2020: (Start)
We can follow also alternative routes, where we do not always select the largest proper divisor to subtract, for example, from 19 to 1, we could go as:
19-1 = 18; 18-(18/3) = 12; 12-(12/2) = 6; 6-(6/3) = 4; 4-(4/2) = 2; 2-(2/2) = 1, or as
19-1 = 18; 18-(18/3) = 12; 12-(12/3) = 8; 8-(8/2) = 4; 4-(4/2) = 2; 2-(2/2) = 1,
both of which also have exactly 6 iterations.
(End)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Passawan Noppakaew and Prapanpong Pongsriiam, Product of Some Polynomials and Arithmetic Functions, J. Int. Seq. (2023) Vol. 26, Art. 23.9.1.
• Hugo Pfoertner, Addition chains
• Mathematics Stack Exchange, Does a graded poset on the positive integers generated from subtracting factors define a lattice?
• Wikipedia, Addition chain

Crossrefs

Similar to A061373 which uses the same recurrence relation but a(1) = 1.

Cf. A003313, A005245, A020639, A032742, A052126, A054725, A060681, A064415, A073933, A076142, A076091, A175125, A256653, A307742, A323076, A323077, A329697, A333123, A334200, A334202, A334203, and array A334111.

Cf. A000079 (position of last occurrence), A105017 (position of records), A334197 (positions of record jumps upward).

Partial sums of A334090.

Cf. also A056239.

Programs

Haskell

import Data.List (genericIndex)
a064097 n = genericIndex a064097_list (n-1)
a064097_list = 0 : f 2 where
   f x | x == spf  = 1 + a064097 (spf - 1) : f (x + 1)
       | otherwise = a064097 spf + a064097 (x `div` spf) : f (x + 1)
       where spf = a020639 x
-- Reinhard Zumkeller, Mar 08 2013

Maple

a:= proc(n) option remember;
      add((1+a(i[1]-1))*i[2], i=ifactors(n)[2])
    end:
seq(a(n), n=1..120);  # Alois P. Heinz, Apr 26 2019
# alternative which can be even used outside this entry
A064097 := proc(n)
        option remember ;
        add((1+procname(i[1]-1))*i[2], i=ifactors(n)[2]) ;
end proc:
seq(A064097(n),n=1..100) ; # R. J. Mathar, Aug 07 2022

Mathematica

quasiLog := (Length@NestWhileList[# - Divisors[#][[-2]] &, #, # > 1 &] - 1) &;
quasiLog /@ Range[1024]
(* Terentyev Oleg, Jul 17 2011 *)
fi[n_] := Flatten[ Table[#[[1]], {#[[2]]}] & /@ FactorInteger@ n]; a[1] = 0; a[n_] := If[ PrimeQ@ n, a[n - 1] + 1, Plus @@ (a@# & /@ fi@ n)]; Array[a, 105] (* Robert G. Wilson v, Jul 17 2013 *)
a[n_] := Length@ NestWhileList[# - #/FactorInteger[#][[1, 1]] &, n, # != 1 &] - 1; Array[a, 100] (* or *)
a[n_] := a[n - n/FactorInteger[n][[1, 1]]] +1; a[1] = 0; Array[a, 100]  (* Robert G. Wilson v, Mar 03 2020 *)

PARI

NN=200; an=vector(NN);
a(n)=an[n];
for(n=2,NN,an[n]=if(isprime(n),1+a(n-1), sumdiv(n,p, if(isprime(p),a(p)*valuation(n,p)))));
for(n=1,100,print1(a(n)", "))

PARI

a(n)=if(isprime(n), return(a(n-1)+1)); if(n==1, return(0)); my(f=factor(n)); apply(a,f[,1])~ * f[,2] \\ Charles R Greathouse IV, May 10 2016

Scheme

(define (A064097 n) (if (= 1 n) 0 (+ 1 (A064097 (A060681 n))))) ;; After Jaroslav Krizek's Jan 28 2010 formula.
(define (A060681 n) (- n (A032742 n))) ;; See also code under A032742.
;; Antti Karttunen, Aug 23 2017

Formula

Conjectures: for n>1, log(n) < a(n) < (5/2)*log(n); lim n ->infinity sum(k=1, n, a(k))/(n*log(n)-n) = C = 1.8(4)... - Benoit Cloitre, Oct 30 2002

Conjecture: for n>1, floor(log_2(n)) <= a(n) < (5/2)*log(n). - Robert G. Wilson v, Aug 10 2013

a(n) = Sum_{k=1..n} a(p_k)*e_k if n is composite with factorization p_1^e_1 * ... * p_k^e_k. - Orson R. L. Peters, May 10 2016

From Antti Karttunen, Aug 23 2017: (Start)

a(1) = 0; for n > 1, a(n) = 1 + a(A060681(n)). [From Jaroslav Krizek's Jan 28 2010 formula in comments.]

a(n) = A073933(n) - 1. (End)

a(n) = A064415(n) + A329697(n) [= A054725(n) + A329697(n), for n > 1]. - Antti Karttunen, Apr 16 2020

a(n) = A323077(n) + A334202(n) = a(A052126(n)) + a(A006530(n)). - Antti Karttunen, May 12 2020

Extensions

More terms from Michael Somos, Sep 25 2001