A069797 -id:A069797 - cp's OEIS frontend

A090556 Beginning with 1, a(n) = least number m > a(n-1) such that phi(a(n-1)) divides phi(m).

1, 2, 3, 4, 5, 8, 10, 12, 13, 21, 26, 28, 35, 39, 45, 52, 56, 65, 97, 119, 153, 193, 221, 291, 357, 386, 388, 416, 442, 448, 476, 485, 579, 595, 663, 765, 769, 965, 1105, 1455, 1538, 1544, 1552, 1649, 1799, 2307, 2313, 2895, 3076, 3088, 3104, 3281, 3341, 3845, 4947

Offset: 1

Amarnath Murthy, Dec 10 2003

nonn

Ivan Neretin, Table of n, a(n) for n = 1..308

Cf. A014573, A090557.

NestList[Function[n, SelectFirst[Range[n + 1, 10^4], Divisible[EulerPhi@ #, EulerPhi@ n] &]], 1, 54] (* Michael De Vlieger, May 01 2016, Version 10 *)

a(1)=1, then a(n+1) = A069797(a(n)). - Ivan Neretin, May 01 2016

Corrected and extended by Sam Handler (sam_5_5_5_0(AT)yahoo.com), Nov 03 2004

A353691 a(n) is the least number k > n such that h(k)/h(n) is an integer, where h(n) is the harmonic mean of the divisors of n, or -1 if no such k exists.

Original entry on oeis.org

6, 120, 28, 234, 30, 270, 42, 29792, 252, 1120, 66, 234, 78, 840, 140, 200, 102, 2016, 114, 1170, 945, 1320, 138, 1080, 150, 1560, 756, 270, 174, 3360, 186, 1272960, 308, 2040, 210, 9720, 222, 2280, 364, 148960, 246, 1890, 258, 2574, 1260, 2760, 282, 600, 294

Offset: 1

Amiram Eldar, May 04 2022

nonn

Does a(n) exist for all n? If m is a harmonic number (A001599) and gcd(n, m) = 1, then a(n) exists and a(n) <= m*n, since h(m*n) = h(m)*h(n) and h(m) is an integer.

			a(2) = 120 since 120 is the least number > 2 such that h(120)/h(2) = (16/3)/(4/3) = 4 is an integer.

Amiram Eldar, Table of n, a(n) for n = 1..639

Cf. A001599, A099377, A099378.

Similar sequences: A069789, A069797, A069805, A353692.

h[n_] := DivisorSigma[0, n]/DivisorSigma[-1, n]; a[n_] := Module[{k = n + 1, hn = h[n]}, While[!IntegerQ[h[k]/hn], k++]; k]; Array[a, 30]

from math import prod, gcd
from sympy import factorint
def A353691_helper(n):
    f = factorint(n).items()
    return prod(p**e*(p-1)*(e+1) for p, e in f), prod(p**(e+1)-1 for p, e in f)
def A353691(n):
    Hnp, Hnq = A353691_helper(n)
    g = gcd(Hnp, Hnq)
    Hnp //= g
    Hnq //= g
    k = n+1
    Hkp, Hkq = A353691_helper(k)
    while (Hkp*Hnq) % (Hkq*Hnp):
        k += 1
        Hkp, Hkq = A353691_helper(k)
    return k # Chai Wah Wu, May 07 2022

a(p) = 6*p for a prime p > 3.

A353692 a(n) is the least number k > n such that uh(k)/uh(n) is an integer, where uh(n) is the harmonic mean of the unitary divisors of n, or -1 if no such k exists.

Original entry on oeis.org

6, 20, 45, 72, 30, 60, 42, 272, 756, 120, 66, 18, 78, 140, 1890, 720, 102, 180, 114, 24, 315, 220, 138, 360, 150, 260, 3321, 504, 174, 7560, 186, 1440, 495, 340, 210, 52416, 222, 380, 585, 1360, 246, 420, 258, 792, 1512, 460, 282, 720, 294, 600, 765, 936, 318

Offset: 1

Amiram Eldar, May 04 2022

nonn

			a(2) = 20 since 20 is the least number > 2 such that uh(20)/uh(2) = (8/3)/(4/3) = 2 is an integer.

Amiram Eldar, Table of n, a(n) for n = 1..399

Cf. A103339, A103340.

Similar sequences: A069789, A069797, A069805, A353691.

uh[n_] := Module[{f = FactorInteger[n]}, n*2^Length[f]/Times @@ (1 + Power @@@ f)]; a[n_] := Module[{k = n + 1, uhn = uh[n]}, While[!IntegerQ[uh[k]/uhn], k++]; k]; Array[a, 30]

a(p) = 6*p for a prime p > 3.

A282016 Least k > 0 such that phi(n) divides phi(n+k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 2, 2, 4, 2, 11, 1, 8, 4, 1, 1, 15, 1, 8, 4, 5, 3, 23, 6, 8, 2, 10, 7, 29, 2, 30, 2, 8, 6, 4, 1, 20, 16, 6, 8, 14, 3, 6, 6, 7, 21, 47, 3, 37, 5, 13, 4, 53, 3, 20, 9, 6, 29, 59, 4, 16, 15, 10, 4, 32, 9, 67, 12, 20, 2, 71, 1, 18, 2, 7, 15, 16, 6, 78, 5, 28, 6, 83, 6, 43, 12, 26, 12

Offset: 1

Altug Alkan, Feb 09 2017

nonn

See A171935 for least positive k such that phi(n) = phi(n+k), or 0 if no such k exists.

See also logarithmic scatterplot of this sequence. - Altug Alkan, Feb 09 2017

			a(5) = 3 because phi(5) = 4 divides phi(5 + 3) = 4 and 3 is the least positive number with this property.

Altug Alkan, Table of n, a(n) for n = 1..10000

Cf. A000010, A069797, A171935.

f[n_] := Block[{k = 1}, While[ Mod[ EulerPhi[n + k], EulerPhi[ n]] > 0, k++]; k]; Array[f, 88] (* Robert G. Wilson v, Feb 09 2017 *)

a(n) = my(k = 1); while (eulerphi(n+k) % eulerphi(n) != 0, k++); k;

a(n) << n^5 as a consequence of Xylouris' form of Linnik's theorem: phi(n) is at most n-1, and a(n) is at most the least prime which is 1 mod phi(n). - Charles R Greathouse IV, Feb 09 2017

a(n) = A069797(n) - n. - Altug Alkan, Feb 10 2017

cp's OEIS Frontend

A090556 Beginning with 1, a(n) = least number m > a(n-1) such that phi(a(n-1)) divides phi(m).

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A353691 a(n) is the least number k > n such that h(k)/h(n) is an integer, where h(n) is the harmonic mean of the divisors of n, or -1 if no such k exists.

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A353692 a(n) is the least number k > n such that uh(k)/uh(n) is an integer, where uh(n) is the harmonic mean of the unitary divisors of n, or -1 if no such k exists.

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A282016 Least k > 0 such that phi(n) divides phi(n+k).

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