A007694 -id:A007694 - cp's OEIS frontend

A049237 Quotient n/phi(n) for n in A007694.

1, 2, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2

Offset: 1

Labos Elemer

nonn

Here phi(n) denotes Euler's totient function A000010.
As n increases, the proportion of 3's seems to approach 100 percent (it is 40 percent for the first 10 results; 82 percent for 100 results; 87.5 percent for 200 results while up to 200 million, for the first 235 results, is 88.51 percent). - Zoltan Galantai, Jul 28 2019
From Bernard Schott, Jul 30 2019: (Start)
According to [Ecker and Beslin], the quotients n/phi(n) when phi(n) divides n can take only 3 distinct values:
n/phi(n) = 1 iff n = 1,
n/phi(n) = 2 iff n = 2^w, w >= 1,
n/phi(n) = 3 iff n = 2^w * 3^u, w >= 1, u >= 1.
The previous comment follows because between 2^k and 2^(k+1) there are two consecutive integers for which n/phi(n) = 2, and there are floor(k*(log(2)/log(3))) integers of the form 2^b*3^c (b and c>=1) for which n/phi(n) = 3. (End)

			For powers of 2 the quotient is 2.
a(95) = 124416/phi(124416) = 124416/41472 = 3.

Sárközy A. and Suranyi J., Number Theory Problem Book (in Hungarian), Tankonyvkiado, Budapest, 1972.

Jinyuan Wang, Table of n, a(n) for n = 1..10000
Michael W. Eckert and Scott J. Beslin, Find all positive integers n such that phi(n) divides n, AMM, Vol. 93, No 8, Oct. 86, 656-657, E 3037.

Cf. A000010, A007694, A062356.

v:=[m:m in [1..150000]|m mod EulerPhi(m) eq 0];[v[k]/EulerPhi(v[k]):k in [1..#v]]; // Marius A. Burtea, Jul 28 2019

Select[#/EulerPhi@ # & /@ Range[10^6], IntegerQ] (* Michael De Vlieger, Jul 02 2016 *)

lista(NN) = for(n=1,NN,if(n%eulerphi(n)==0,print1(n/eulerphi(n),", "))); \\ Jinyuan Wang, Jul 31 2019

n/phi(n) is an integer iff n = 1 or n = 2^w*3^u for w = 1, 2, ... and u = 0, 1, 2, ...

Edited by M. F. Hasler, Jul 02 2016

A174030 Partial sums of A007694.

Original entry on oeis.org

1, 3, 7, 13, 21, 33, 49, 67, 91, 123, 159, 207, 261, 325, 397, 493, 601, 729, 873, 1035, 1227, 1443, 1699, 1987, 2311, 2695, 3127, 3613, 4125, 4701, 5349, 6117, 6981, 7953, 8977, 10129, 11425, 12883, 14419, 16147, 18091, 20139, 22443, 25035, 27951, 31023

Offset: 1

Jonathan Vos Post, Mar 06 2010

nonn

Partial sums of numbers k such that phi(k) divides k.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A000010, A007694, A259823.

Accumulate[Select[Range[5000], Divisible[#, EulerPhi[#]] &]] (* Amiram Eldar, Nov 05 2024 *)

a(n) = Sum_{i=1..n} A007694(i).

a(n) = 2*A259823(n-1) + 1. - Amiram Eldar, Nov 05 2024

A285483 a(n) is the smallest positive integer that makes a(n)*A007694(n)-1 a prime number, while a(n) and A007694(n) are coprimes.

Original entry on oeis.org

7, 1, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 17, 1, 5, 1, 1, 5, 5, 1, 7, 5, 5, 5, 1, 1, 13, 7, 5, 1, 11, 1, 1, 5, 1, 17, 19, 17, 19, 5, 25, 5, 1, 7, 5, 13, 11, 5, 1, 5, 5, 7, 1, 1, 19, 1, 1, 17, 5, 7, 29, 1, 5, 1, 5, 7, 7, 17, 1, 7, 7, 1, 7, 49, 5, 13, 13, 25, 5, 5, 23

Offset: 2

Lei Zhou, Apr 19 2017

All terms are elements of A007310, which are free of prime factor 2 and 3, since if a(n) has a factor of 2, 2*A007694(n) is also an element of A007694. Ditto for a(n) is divisible by 3 cases.

a(1) is not defined since any odd number greater than 3 minus 1 is an even nonprime number.

			For n = 2, A007694(2) = 2, testing k*2-1 for k in set {1, 5, 7, 11, 13, 17, 19... }, we find that 7*2 - 1 = 13 is the first prime number found.  So a(2) = 7;
In the similar way, 1*A007694(3) - 1 = 1*4 - 1 = 3 is the first prime number found for n = 3, so a(3) = 1.
For n = 7, A007694(7) = 16, 5*16 - 1 = 89 is the smallest prime found, so a(7) = 5.

Lei Zhou, Table of n, a(n) for n = 2..10001

Cf. A003586, A007310, A007694.

b = 2; a = {b}; sm = {}; r = a; While[Length[sm] < 81, f = 0;
While[f++; (fc = FactorInteger[f];
    MemberQ[{2, 3}, fc[[1, 1]]]) || (! PrimeQ[f*a[[Length[a]]] - 1])];
  AppendTo[sm, f]; c = r*2; d = r*3; e = Sort[Union[c, d]]; i = 1;
While[e[[i]] <= a[[Length[a]]], i++]; AppendTo[a, e[[i]]];
AppendTo[r, e[[i]]];
While[(3*r[[1]]) < r[[Length[r]]], r = Delete[r, 1]]]; sm

A285486 a(n) is the smallest positive integer that makes a(n)*A007694(n)+1 a prime number, while a(n) and A007694(n) are coprimes.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 1, 13, 11, 1, 5, 5, 7, 1, 1, 1, 5, 7, 1, 1, 11, 1, 7, 5, 7, 1, 1, 23, 1, 11, 1, 19, 5, 13, 1, 1, 1, 5, 5, 5, 29, 19, 1, 1, 5, 1, 1, 37, 5, 5, 17, 11, 13, 25, 13, 5, 41, 7, 1, 5, 1, 25, 7, 5, 7, 1, 1, 5, 11, 11, 5, 5, 5, 5, 11, 5, 1, 11, 7

Offset: 1

Lei Zhou, Apr 19 2017

All terms are elements of A007310, which are free of prime factor 2 and 3, since if a(n) has a factor of 2, 2*A007694(n) is also an element of A007694. Ditto for a(n) is divisible by 3 cases.

			For n = 1, A007694(1) = 1, testing k*2-1 for k in set {1, 5, 7, 11, 13, 17, 19... }, we find that 1*1 + 1 = 2 is the first prime number found.  So a(1) = 1;
In the similar way, 1*A007694(2) + 1 = 1*2 + 1 = 3 is the first prime number found for n = 2, so a(2) = 1.
...
For n = 5, A007694(5) = 8, 5*8 + 1 = 41 is the smallest prime found, so a(5) = 5.

Lei Zhou, Table of n, a(n) for n = 1..10000

Cf. A003586, A007310, A007694, A285483.

b = 2; a = {1, b}; sp = {1}; r = {b}; While[Length[sp] < 79, f = 0;
While[f++; (fc = FactorInteger[f];
    MemberQ[{2, 3}, fc[[1, 1]]]) || (! PrimeQ[f*a[[Length[a]]] + 1])];
  AppendTo[sp, f]; c = r*2; d = r*3; e = Sort[Union[c, d]]; i = 1;
While[e[[i]] <= a[[Length[a]]], i++]; AppendTo[a, e[[i]]];
AppendTo[r, e[[i]]];
While[(3*r[[1]]) < r[[Length[r]]], r = Delete[r, 1]]]; sp

A379851 Numbers k such that phi(k) does not divide k. Complement of A007694.

Original entry on oeis.org

3, 5, 7, 9, 10, 11, 13, 14, 15, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89

Offset: 1

Franz Vrabec, Jan 04 2025

nonn

Let PHI(n) the set of all numbers x such there is a k-fold iteration of Euler's totient function phi = A000010 on x resulting in n. The numbers a(n) are exactly the numbers for which PHI(a(n)) is a finite set (possibly empty).

Contains A007617.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000010, A007617. Complement of A007694.

filter:= n -> n mod numtheory:-phi(n) <> 0:
select(filter, [$1..100]); # Robert Israel, Feb 04 2025

Select[Range[100], ! Divisible[#, EulerPhi[#]] &] (* Amiram Eldar, Jan 08 2025 *)

A003586 3-smooth numbers: numbers of the form 2^i*3^j with i, j >= 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256, 288, 324, 384, 432, 486, 512, 576, 648, 729, 768, 864, 972, 1024, 1152, 1296, 1458, 1536, 1728, 1944, 2048, 2187, 2304, 2592, 2916, 3072, 3456, 3888

Offset: 1

Paul Zimmermann, Dec 11 1996

This sequence is easily confused with A033845, which gives numbers of the form 2^i*3^j with i, j >= 1. Don't simply say "numbers of the form 2^i*3^j", but specify which sequence you mean. - N. J. A. Sloane, May 26 2024
These numbers were once called "harmonic numbers", see Lenstra links. - N. J. A. Sloane, Jul 03 2015
Successive numbers k such that phi(6k) = 2k. - Artur Jasinski, Nov 05 2008
Where record values greater than 1 occur in A088468: A160519(n) = A088468(a(n)). - Reinhard Zumkeller, May 16 2009
Also numbers that are divisible by neither 6k - 1 nor 6k + 1, for all k > 0. - Robert G. Wilson v, Oct 26 2010
Also numbers m such that the rooted tree with Matula-Goebel number m has m antichains. The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T. The vertices of a rooted tree can be regarded as a partially ordered set, where u<=v holds for two vertices u and v if and only if u lies on the unique path between v and the root. An antichain is a nonempty set of mutually incomparable vertices. Example: m=4 is in the sequence because the corresponding rooted tree is \/=ARB (R is the root) having 4 antichains (A, R, B, AB). - Emeric Deutsch, Jan 30 2012
A204455(3*a(n)) = 3, and only for these numbers. - Wolfdieter Lang, Feb 04 2012
The number of terms less than or equal to n is Sum_{i=0..floor(log_2(n))} floor(log_3(n/2^i) + 1), or Sum_{i=0..floor(log_3(n))} floor(log_2(n/3^i) + 1), which requires fewer terms to compute. - Robert G. Wilson v, Aug 17 2012
Named 3-friables in French. - Michel Marcus, Jul 17 2013
In the 14th century Levi Ben Gerson proved that the only pairs of terms which differ by 1 are (1,2), (2,3), (3,4), and (8,9); see A235365, A235366, A236210. - Jonathan Sondow, Jan 20 2014
Range of values of A000005(n) (and also A181819(n)) for cubefree numbers n. - Matthew Vandermast, May 14 2014
A036561 is a permutation of this sequence. - L. Edson Jeffery, Sep 22 2014
Also the sorted union of A000244 and A007694. - Lei Zhou, Apr 19 2017
The sum of the reciprocals of the 3-smooth numbers is equal to 3. Brief proof: 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + ... = (Sum_{k>=0} 1/2^k) * (Sum_{m>=0} 1/3^m) = (1/(1-1/2)) * (1/(1-1/3)) = (2/(2-1)) * (3/(3-1)) = 3. - Bernard Schott, Feb 19 2019
Also those integers k for which, for every prime p > 3, p^(2k) - 1 == 0 (mod 24k). - Federico Provvedi, May 23 2022
For n>1, the exponents’ parity {parity(i), parity(j)} of one out of four consecutive terms is {odd, odd}. Therefore, for n>1, at least one out of every four consecutive terms is a Zumkeller number (A083207). If for the term whose parity is {even, odd}, even also means nonzero, then this term is also a Zumkeller number (as is the case with the last of the four consecutive terms 1296, 1458, 1536, 1728). - Ivan N. Ianakiev, Jul 10 2022
Except the initial terms 2, 3, 4, 8, 9 and 16, these are numbers k such that k^6 divides 6^k. Except the initial terms 2, 3, 4, 6, 8, 9, 16, 18 and 27, these are numbers k such that k^12 divides 12^k. - Mohammed Yaseen, Jul 21 2022
In music theory, a comma is a ratio, close to 1 (typically less than 1.04), between two natural numbers divisible by only small primes (typically single digit). In this sequence, a(131) / a(130) = 531441 / 524288 ~ 1.013643 is the Pythagorean comma (A221363), the difference between 12 perfect fifths and 7 octaves. - Hal M. Switkay, Mar 23 2025

J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 654 pp. 85, 287-8, Ellipses Paris 2004.
S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962, p. xxiv.
R. Tijdeman, Some applications of Diophantine approximation, pp. 261-284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.

Lei Zhou, Table of n, a(n) for n = 1..10000 (first 501 terms from Franklin T. Adams-Watters)
R. Blecksmith, M. McCallum and J. L. Selfridge, 3-smooth representations of integers, Amer. Math. Monthly, 105 (1998), 529-543.
Thierry Bousch, La Tour de Stockmeyer, Séminaire Lotharingien de Combinatoire 77 (2017), Article B77d.
Benoit Cloitre, a(n)/((1/sqrt(6))*exp(sqrt(2*log(2)*log(3)*n))) for 0

Natalia da Silva, Serban Raianu, and Hector Salgado, Differences of Harmonic Numbers and the abc-Conjecture, arXiv:1708.00620 [math.NT], 2017.
Emeric Deutsch, Rooted tree statistics from Matula numbers, arXiv:1111.4288 [math.CO], 2011.
David Eppstein, Making Change in 2048, arXiv:1804.07396 [cs.DM], 2018.
F. Goebel, On a 1-1-correspondence between rooted trees and natural numbers, J. Combin. Theory, B 29 (1980), 141-143.
I. Gutman and A. Ivic, On Matula numbers, Discrete Math., 150, 1996, 131-142.
I. Gutman and Yeong-Nan Yeh, Deducing properties of trees from their Matula numbers, Publ. Inst. Math., 53 (67), 1993, 17-22.
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 252. Book's website
H. W. Lenstra Jr., Harmonic Numbers
H. W. Lenstra, Jr., Harmonic Numbers and the ABC-conjecture, Abstract of talk, May 30, 2001 [Annotated scanned copy]
D. Matula, A natural rooted tree enumeration by prime factorization, SIAM Rev. 10 (1968) 273.
D. J. Mintz, 2,3 sequence as a binary mixture, Fib. Quarterly, Vol. 19, No 4, Oct 1981, pp. 351-360.
I. Peterson, Medieval Harmony
Raphael Schumacher, The Formulas for the Distribution of the 3-Smooth, 5-Smooth, 7-Smooth and all other Smooth Numbers, arXiv preprint arXiv:1608.06928 [math.NT], 2016.
Eric Weisstein's World of Mathematics, Smooth Number

Cf. A051037, A002473, A051038, A080197, A080681, A080682, A117221, A105420, A062051, A117222, A117220, A090184, A131096, A131097, A186711, A186712, A186771, A088468, A061987, A080683 (p-smooth numbers with other values of p), A025613 (a subsequence).
Cf. also A235365, A235366, A236210, A036561.
Cf. also A000244, A007694. - Lei Zhou, Apr 19 2017
Cf. A191475 (successive values of i), A191476 (successive values of j), A022330 (indices of the pure terms 2^i), A022331 (indices of the pure terms 3^j). - N. J. A. Sloane, May 26 2024
Cf. A221363.

import Data.Set (Set, singleton, insert, deleteFindMin)
smooth :: Set Integer -> [Integer]
smooth s = x : smooth (insert (3*x) $ insert (2*x) s')
  where (x, s') = deleteFindMin s
a003586_list = smooth (singleton 1)
a003586 n = a003586_list !! (n-1)
-- Reinhard Zumkeller, Dec 16 2010

[n: n in [1..4000] | PrimeDivisors(n) subset [2,3]]; // Bruno Berselli, Sep 24 2012

A003586 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+1 do numtheory[factorset](a) minus {2,3} ; if % = {} then return a; end if; end do: end if; end proc: # R. J. Mathar, Feb 28 2011
with(numtheory): for i from 1 to 23328 do if(i/phi(i)=3)then print(i/6) fi od; # Gary Detlefs, Jun 28 2011

a[1] = 1; j = 1; k = 1; n = 100; For[k = 2, k <= n, k++, If[2*a[k - j] < 3^j, a[k] = 2*a[k - j], {a[k] = 3^j, j++}]]; Table[a[i], {i, 1, n}] (* Hai He (hai(AT)mathteach.net) and Gilbert Traub, Dec 28 2004 *)
aa = {}; Do[If[EulerPhi[6 n] == 2 n, AppendTo[aa, n]], {n, 1, 1000}]; aa (* Artur Jasinski, Nov 05 2008 *)
fQ[n_] := Union[ MemberQ[{1, 5}, # ] & /@ Union@ Mod[ Rest@ Divisors@ n, 6]] == {False}; fQ[1] = True; Select[ Range@ 4000, fQ] (* Robert G. Wilson v, Oct 26 2010 *)
powerOfTwo = 12; Select[Nest[Union@Join[#, 2*#, 3*#] &, {1}, powerOfTwo-1], # < 2^powerOfTwo &] (* Robert G. Wilson v and T. D. Noe, Mar 03 2011 *)
fQ[n_] := n == 3 EulerPhi@ n; Select[6 Range@ 4000, fQ]/6 (* Robert G. Wilson v, Jul 08 2011 *)
mx = 4000; Sort@ Flatten@ Table[2^i*3^j, {i, 0, Log[2, mx]}, {j, 0, Log[3, mx/2^i]}] (* Robert G. Wilson v, Aug 17 2012 *)
f[n_] := Block[{p2, p3 = 3^Range[0, Floor@ Log[3, n] + 1]}, p2 = 2^Floor[Log[2, n/p3] + 1]; Min[ Select[ p2*p3, IntegerQ]]]; NestList[f, 1, 54] (* Robert G. Wilson v, Aug 22 2012 *)
Select[Range@4000, Last@Map[First, FactorInteger@#] <= 3 &] (* Vincenzo Librandi, Aug 25 2016 *)
Select[Range[4000],Max[FactorInteger[#][[All,1]]]<4&] (* Harvey P. Dale, Jan 11 2017 *)

test(n)=for(p=2,3, while(n%p==0, n/=p)); n==1;
for(n=1,4000,if(test(n),print1(n",")))

list(lim)=my(v=List(),N);for(n=0,log(lim\1+.5)\log(3),N=3^n;while(N<=lim,listput(v,N);N<<=1));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

is_A003586(n)=n<5||vecmax(factor(n,5)[, 1])<5 \\ M. F. Hasler, Jan 16 2015

list(lim)=my(v=List(), N); for(n=0, logint(lim\=1,3), N=3^n; while(N<=lim, listput(v, N); N<<=1)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018

from itertools import count, takewhile
def aupto(lim):
    pows2 = list(takewhile(lambda x: xMichael S. Branicky, Jul 08 2022

from sympy import integer_log
def A003586(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Sep 15 2024

# faster for initial segment of sequence
import heapq
from itertools import islice
def A003586gen(): # generator of terms
    v, oldv, h, psmooth_primes, = 1, 0, [1], [2, 3]
    while True:
        v = heapq.heappop(h)
        if v != oldv:
            yield v
            oldv = v
            for p in psmooth_primes:
                heapq.heappush(h, v*p)
print(list(islice(A003586gen(), 65))) # Michael S. Branicky, Sep 17 2024
(C++) // Returns A003586 <= threshold without approximations nor sorting
#include 
std::forward_list A003586(const int threshold) {
    std::forward_list sequence;
    auto start_it = sequence.before_begin();
    for (int i = 1; i <= threshold; i *= 2) {
        for (int inc = 1; std::next(start_it) != sequence.end() && inc <= i; inc *= 3)
            ++start_it;
        auto it = start_it;
        for (int j = 1; i * j <= threshold; j *= 3) {
            sequence.emplace_after(it, i * j);
            for (int inc = 1; std::next(it) != sequence.end() && inc <= i; inc *= 2)
                ++it;
        }
    }
    return sequence;
} // Eben Gino Lester, Apr 17 2025

def isA003586(n) :
    return not any(d != 2 and d != 3 for d in prime_divisors(n))
@CachedFunction
def A003586(n) :
    if n == 1 : return 1
    k = A003586(n-1) + 1
    while not isA003586(k) : k += 1
    return k
[A003586(n) for n in (1..55)] # Peter Luschny, Jul 20 2012

An asymptotic formula for a(n) is roughly a(n) ~ 1/sqrt(6)*exp(sqrt(2*log(2)*log(3)*n)). - Benoit Cloitre, Nov 20 2001
A061987(n) = a(n + 1) - a(n), a(A084791(n)) = A084789(n), a(A084791(n) + 1) = A084790(n). - Reinhard Zumkeller, Jun 03 2003
Union of powers of 2 and 3 with n such that psi(n) = 2*n, where psi(n) = n*Product_(1 + 1/p) over all prime factors p of n = A001615(n). - Lekraj Beedassy, Sep 07 2004; corrected by Franklin T. Adams-Watters, Mar 19 2009
a(n) = 2^A022328(n)*3^A022329(n). - N. J. A. Sloane, Mar 19 2009
The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} moebius(6*n)*x^n/(1 - x^n). - Paul D. Hanna, Sep 18 2011
a(n) = A007694(n+1)/2. - Lei Zhou, Apr 19 2017

Deleted claim that this sequence is union of 2^n (A000079) and 3^n (A000244) sequences -- this does not include the terms which are not pure powers. - Walter Roscello (wroscello(AT)comcast.net), Nov 16 2008

A076512 Denominator of cototient(n)/totient(n).

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 6, 1, 2, 2, 10, 1, 12, 3, 8, 1, 16, 1, 18, 2, 4, 5, 22, 1, 4, 6, 2, 3, 28, 4, 30, 1, 20, 8, 24, 1, 36, 9, 8, 2, 40, 2, 42, 5, 8, 11, 46, 1, 6, 2, 32, 6, 52, 1, 8, 3, 12, 14, 58, 4, 60, 15, 4, 1, 48, 10, 66, 8, 44, 12, 70, 1, 72, 18, 8, 9, 60, 4, 78, 2, 2, 20, 82, 2, 64, 21

Offset: 1

Reinhard Zumkeller, Oct 15 2002

a(n)=1 iff n=A007694(k) for some k.
Numerator of phi(n)/n=Prod_{p|n} (1-1/p). - Franz Vrabec, Aug 26 2005
From Wolfdieter Lang, May 12 2011: (Start)
For n>=2, a(n)/A109395(n) = sum(((-1)^r)*sigma_r,r=0..M(n)) with the elementary symmetric functions (polynomials) sigma_r of the indeterminates {1/p_1,...,1/p_M(n)} if n = prod((p_j)^e(j),j=1..M(n)) where M(n)=A001221(n) and sigma_0=1.
This follows by expanding the above given product for phi(n)/n.
The n-th member of this rational sequence 1/2, 2/3, 1/2, 4/5, 1/3, 6/7, 1/2, 2/3, 2/5,... is also (2/n^2)*sum(k,with 1<=k=2.
Therefore, this scaled sum depends only on the distinct prime factors of n.
See also A023896. Proof via PIE (principle of inclusion and exclusion). (End)
In the sequence of rationals r(n)=eulerphi(n)/n: 1, 1/2, 2/3, 1/2, 4/5, 1/3, 6/7, 1/2, 2/3, 2/5, 10/11, 1/3, ... one can observe that new values are obtained for squarefree indices (A005117); while for a nonsquarefree number n (A013929), r(n) = r(A007947(n)), where A007947(n) is the squarefree kernel of n. - Michel Marcus, Jul 04 2015

T. D. Noe, Table of n, a(n) for n=1..1000

Cf. A076511 (numerator of cototient(n)/totient(n)), A051953.

Phi(m)/m = k: A000079 \ {1} (k=1/2), A033845 (k=1/3), A000244 \ {1} (k=2/3), A033846 (k=2/5), A000351 \ {1} (k=4/5), A033847 (k=3/7), A033850 (k=4/7), A000420 \ {1} (k=6/7), A033848 (k=5/11), A001020 \ {1} (k=10/11), A288162 (k=6/13), A001022 \ {1} (12/13), A143207 (k=4/15), A033849 (k=8/15), A033851 (k=24/35).

[Numerator(EulerPhi(n)/n): n in [1..100]]; // Vincenzo Librandi, Jul 04 2015

Table[Denominator[(n - EulerPhi[n])/EulerPhi[n]], {n, 80}] (* Alonso del Arte, May 12 2011 *)

vector(80, n, numerator(eulerphi(n)/n)) \\ Michel Marcus, Jul 04 2015

a(n) = A000010(n)/A009195(n).

cp's OEIS Frontend

A049237 Quotient n/phi(n) for n in A007694.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A174030 Partial sums of A007694.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A285483 a(n) is the smallest positive integer that makes a(n)*A007694(n)-1 a prime number, while a(n) and A007694(n) are coprimes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A285486 a(n) is the smallest positive integer that makes a(n)*A007694(n)+1 a prime number, while a(n) and A007694(n) are coprimes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A379851 Numbers k such that phi(k) does not divide k. Complement of A007694.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

A003586 3-smooth numbers: numbers of the form 2^i*3^j with i, j >= 0.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

Python

Python

Python

Sage

Formula

Extensions

A076512 Denominator of cototient(n)/totient(n).