A288162 -id:A288162 - cp's OEIS frontend

A109395 Denominator of phi(n)/n = Product_{p|n} (1 - 1/p); phi(n)=A000010(n), the Euler totient function.

1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 3, 13, 7, 15, 2, 17, 3, 19, 5, 7, 11, 23, 3, 5, 13, 3, 7, 29, 15, 31, 2, 33, 17, 35, 3, 37, 19, 13, 5, 41, 7, 43, 11, 15, 23, 47, 3, 7, 5, 51, 13, 53, 3, 11, 7, 19, 29, 59, 15, 61, 31, 7, 2, 65, 33, 67, 17, 69, 35, 71, 3, 73, 37, 15, 19, 77, 13, 79, 5, 3

Offset: 1

Franz Vrabec, Aug 26 2005

a(n)=2 iff n=2^k (k>0); otherwise a(n) is odd. If p is prime, a(p)=p; the converse is false, e.g.: a(15)=15. It is remarkable that this sequence often coincides with A006530, the largest prime P dividing n. Theorem: a(n)=P if and only if for every prime p < P in n there is some prime q in n with p|(q-1). - Franz Vrabec, Aug 30 2005

			a(10) = 10/gcd(10,phi(10)) = 10/gcd(10,4) = 10/2 = 5.

Antti Karttunen, Table of n, a(n) for n = 1..16384 (terms 1..1000 from T. D. Noe)
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Cf. A076512 for the numerator.
Cf. A000010, A009195, A054741, A059956, A082695, A318304, A318305, A323170.
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).

Table[Denominator[EulerPhi[n]/n], {n, 81}] (* Alonso del Arte, Sep 03 2011 *)

a(n)=n/gcd(n,eulerphi(n)) \\ Charles R Greathouse IV, Feb 20 2013

A007947(n) = factorback(factorint(n)[, 1]); \\ From A007947
A173557(n) = my(f=factor(n)[, 1]); prod(k=1, #f, f[k]-1); \\ From A173557
A109395(n) = denominator(A173557(n)/A007947(n)); \\ Antti Karttunen, Feb 09 2019

a(n) = n/gcd(n, phi(n)) = n/A009195(n).
From Antti Karttunen, Feb 09 2019: (Start)
a(n) = denominator of A173557(n)/A007947(n).
a(2^n) = 2 for all n >= 1.
(End)
From Amiram Eldar, Jul 31 2020: (Start)
Asymptotic mean of phi(n)/n: lim_{m->oo} (1/m) * Sum_{n=1..m} A076512(n)/a(n) = 6/Pi^2 (A059956).
Asymptotic mean of n/phi(n): lim_{m->oo} (1/m) * Sum_{n=1..m} a(n)/A076512(n) = zeta(2)*zeta(3)/zeta(6) (A082695). (End)

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).

A343300 a(n) is p1^1 + p2^2 + ... + pk^k where {p1,p2,...,pk} are the distinct prime factors in ascending order in the prime factorization of n.

Original entry on oeis.org

0, 2, 3, 2, 5, 11, 7, 2, 3, 27, 11, 11, 13, 51, 28, 2, 17, 11, 19, 27, 52, 123, 23, 11, 5, 171, 3, 51, 29, 136, 31, 2, 124, 291, 54, 11, 37, 363, 172, 27, 41, 354, 43, 123, 28, 531, 47, 11, 7, 27, 292, 171, 53, 11, 126, 51, 364, 843, 59, 136, 61, 963, 52, 2, 174, 1342, 67, 291, 532, 370, 71, 11, 73

Offset: 1

Giorgos Kalogeropoulos, Apr 11 2021

nonn

From Bernard Schott, May 07 2021: (Start)
a(n) depends only on prime factors of n (see formulas).
Primes are fixed points of this sequence.
Terms are in increasing order in A344023. (End)

			a(60) = 136 because the distinct prime factors of 60 are {2, 3, 5} and 2^1 + 3^2 + 5^3 = 136.

Michel Marcus, Table of n, a(n) for n = 1..10000

Cf. A027748, A344023 (terms ordered).

A339794 a(n) is the least integer k satisfying rad(k)^2 < sigma(k) and whose prime factors set is the same as the prime factors set of A005117(n+1).

Original entry on oeis.org

4, 9, 25, 18, 49, 80, 121, 169, 112, 135, 289, 361, 441, 352, 529, 416, 841, 360, 961, 891, 1088, 875, 1369, 1216, 1053, 1681, 672, 1849, 1472, 2209, 2601, 2809, 3025, 3249, 1856, 3481, 3721, 1984, 4225, 1584, 4489, 4761, 1960, 5041, 5329, 4736, 5929, 2496, 6241

Offset: 1

Michel Marcus, Dec 17 2020

nonn

Equivalently, subsequence of terms of A339744 excluding terms whose prime factor set has already been encountered.
a(n) = A005117(n + 1)^2 when A005117(n + 1) is prime. Proof: if A005117(n + 1) is a prime p then rad(A005117(n + 1))^2 = rad(p)^2 = p^2 and so integers whose prime factors set is the same as the prime factors set of A005117(n + 1) = p are p^m where m >= 1. p^2 > sigma(p^1) = p + 1 but p^2 < sigma(p^2) = p^2 + p + 1. Q.E.D. - David A. Corneth, Dec 19 2020
From Bernard Schott, Jan 19 2021: (Start)
Indeed, a(n) satisfies the double inequality A005117(n+1) < a(n) <= A005117(n+1)^2.
It is also possible that a(n) = A005117(n+1)^2, even when A005117(n+1) is not prime; the smallest such example is for a(13) = 441 = 21^2 = A005117(14)^2. (End)

			   n  a(n) prime factor set
   1    4  [2]           A000079
   2    9  [3]           A000244
   3   25  [5]           A000351
   4   18  [2, 3]        A033845
   5   49  [7]           A000420
   6   80  [2, 5]        A033846
   7  121  [11]          A001020
   8  169  [13]          A001022
   9  112  [2, 7]        A033847
  10  135  [3, 5]        A033849
  11  289  [17]          A001026
  12  361  [19]          A001029
  13  441  [3, 7]        A033850
  14  352  [2, 11]       A033848
  15  529  [23]          A009967
  16  416  [2, 13]       A288162
  17  841  [29]          A009973
  18  360  [2, 3, 5]     A143207

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Michel Marcus)

Cf. A000203 (sigma), A007947 (rad).
Cf. A005117 (squarefree numbers), A027748, A265668, A339744.
Subsequence: A001248 (squares of primes).

u(n) = {my(fn=factor(n)[,1]); for (k = n, n^2, my(fk = factor(k)); if (fk[,1] == fn, if (factorback(fk[,1])^2 < sigma(fk), return (k));););}
lista(nn) = {for (n=2, nn, if (issquarefree(n), print1(u(n), ", ");););}

a(n) <= A005117(n+1)^2. - David A. Corneth, Dec 19 2020

cp's OEIS Frontend

A109395 Denominator of phi(n)/n = Product_{p|n} (1 - 1/p); phi(n)=A000010(n), the Euler totient function.

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

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A343300 a(n) is p1^1 + p2^2 + ... + pk^k where {p1,p2,...,pk} are the distinct prime factors in ascending order in the prime factorization of n.

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A339794 a(n) is the least integer k satisfying rad(k)^2 < sigma(k) and whose prime factors set is the same as the prime factors set of A005117(n+1).

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