cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A066498 Numbers k such that 3 divides phi(k).

Original entry on oeis.org

7, 9, 13, 14, 18, 19, 21, 26, 27, 28, 31, 35, 36, 37, 38, 39, 42, 43, 45, 49, 52, 54, 56, 57, 61, 62, 63, 65, 67, 70, 72, 73, 74, 76, 77, 78, 79, 81, 84, 86, 90, 91, 93, 95, 97, 98, 99, 103, 104, 105, 108, 109, 111, 112, 114, 117, 119, 122, 124, 126, 127, 129, 130, 133
Offset: 1

Views

Author

Benoit Cloitre, Jan 04 2002

Keywords

Comments

Numbers k such that x^3 == 1 (mod k) has solutions 1 < x < k.
Terms are multiple of 9 or of a prime of the form 6k+1.
If k is a term of this sequence, then G = is a non-abelian group of order 3k, where 1 < r < n and r^3 == 1 (mod k). For example, G can be the subgroup of GL(2, Z_k) generated by x = {{1, 1}, {0, 1}} and y = {{r, 0}, {0, 1}}. - Jianing Song, Sep 17 2019
The asymptotic density of this sequence is 1 (Dressler, 1975). - Amiram Eldar, Mar 21 2021

Examples

			If n < 7 then x^3 = 1 (mod n) has no solution 1 < x < n, but {2,4} are solutions to x^3 == 1 (mod 7), hence a(1) = 7.

Links

Crossrefs

Complement of A088232.
Cf. A000010, A066499, A066500, A066501, A066502.
A007645 gives the primes congruent to 1 mod 3.
Column k=2 of A277915.

Programs

  • Mathematica
    Select[Range[150], Divisible[EulerPhi[#], 3]&] (* Harvey P. Dale, Jan 12 2011 *)
  • PARI
    isok(k)={ eulerphi(k)%3 == 0 } \\ Harry J. Smith, Feb 18 2010

Extensions

Simpler definition from Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 25 2003
Corrected and extended by Ray Chandler, Nov 05 2003

A319100 Number of solutions to x^6 == 1 (mod n).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 6, 4, 6, 2, 2, 4, 6, 6, 4, 4, 2, 6, 6, 4, 12, 2, 2, 8, 2, 6, 6, 12, 2, 4, 6, 4, 4, 2, 12, 12, 6, 6, 12, 8, 2, 12, 6, 4, 12, 2, 2, 8, 6, 2, 4, 12, 2, 6, 4, 24, 12, 2, 2, 8, 6, 6, 36, 4, 12, 4, 6, 4, 4, 12, 2, 24, 6, 6, 4, 12, 12, 12, 6, 8, 6, 2
Offset: 1

Views

Author

Jianing Song, Sep 10 2018

Keywords

Comments

All terms are 3-smooth. a(n) is even for n > 2. Those n such that a(n) = 2 are in A066501.

Examples

			Solutions to x^6 == 1 (mod 13): x == 1, 3, 4, 9, 10, 12 (mod 13).
Solutions to x^6 == 1 (mod 27): x == 1, 8, 10, 17, 19, 26 (mod 27) (x == 1, 8 (mod 9)).
Solutions to x^6 == 1 (mod 37): x == 1, 10, 11, 26, 27, 36 (mod 37).

Links

Crossrefs

Number of solutions to x^k == 1 (mod n): A060594 (k=2), A060839 (k=3), A073103 (k=4), A319099 (k=5), this sequence (k=6), A319101 (k=7), A247257 (k=8).
Cf. A046072, A066501, A088232, A293483, A000010.
Mobius transform gives A307381.

Programs

  • PARI
    a(n)=my(Z=znstar(n)[2]); prod(i=1, #Z, gcd(6, Z[i]))

Formula

Multiplicative with a(2) = 1, a(4) = 2, a(2^e) = 4 for e >= 3; a(3) = 2, a(3^e) = 6 if e >= 2; for other primes p, a(p^e) = 6 if p == 1 (mod 6), a(p^e) = 2 if p == 5 (mod 6).
If the multiplicative group of integers modulo n is isomorphic to C_{k_1} x C_{k_2} x ... x C_{k_m}, where k_i divides k_j for i < j; then a(n) = Product_{i=1..m} gcd(6, k_i).
a(n) = A060594(n)*A060839(n).
For n > 2, a(n) = A060839(n)*2^A046072(n).
a(n) = A060594(n) iff n is not divisible by 9 and no prime factor of n is congruent to 1 mod 6, that is, n in A088232.
a(n) = A000010(n)/A293483(n). - Jianing Song, Nov 10 2019
Sum_{k=1..n} a(k) ~ c * n * log(n)^3, where c = (1/Pi^4) * Product_{p prime == 1 (mod 6)} (1 - (12*p-4)/(p+1)^3) = 0.0075925601... (Finch et al., 2010). - Amiram Eldar, Mar 26 2021

A066502 Numbers k such that 7 divides phi(k).

Original entry on oeis.org

29, 43, 49, 58, 71, 86, 87, 98, 113, 116, 127, 129, 142, 145, 147, 172, 174, 196, 197, 203, 211, 213, 215, 226, 232, 239, 245, 254, 258, 261, 281, 284, 290, 294, 301, 319, 337, 339, 343, 344, 348, 355, 377, 379, 381, 387, 392, 394, 406, 421, 422, 426, 430
Offset: 1

Views

Author

Benoit Cloitre, Jan 04 2002

Keywords

Comments

Related to the equation x^7 == 1 (mod k): sequence gives values of k such there are solutions 1 < x < k of x^7 == 1 (mod k).
If k is a term of this sequence, then G = is a non-abelian group of order 7k, where 1 < r < n and r^7 == 1 (mod k). For example, G can be the subgroup of GL(2, Z_k) generated by x = {{1, 1}, {0, 1}} and y = {{r, 0}, {0, 1}}. - Jianing Song, Sep 17 2019
The asymptotic density of this sequence is 1 (Dressler, 1975). - Amiram Eldar, May 23 2022

Examples

			x^7 == 1 (mod k) has solutions 1 < x < k for k = 29, 43, 49, ...

Links

Crossrefs

Cf. A045465, A066498, A066499, A066500, A066501, A000010.
Column k=4 of A277915.

Programs

  • Mathematica
    Select[Range[500],Divisible[EulerPhi[#],7]&] (* Harvey P. Dale, Apr 12 2012 *)
  • PARI
    isok(k) = { eulerphi(k)%7 == 0 } \\ Harry J. Smith, Feb 18 2010

Formula

a(n) are the numbers generated by 7^2 = 49 and all primes congruent to 1 mod 7 (A045465). Hence sequence gives all k such that k == 0 (mod A045465(n)) for some n > 1 or k == 0 (mod 49).

Extensions

Simpler definition from Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 25 2003

A066500 Numbers k such that 5 divides phi(k).

Original entry on oeis.org

11, 22, 25, 31, 33, 41, 44, 50, 55, 61, 62, 66, 71, 75, 77, 82, 88, 93, 99, 100, 101, 110, 121, 122, 123, 124, 125, 131, 132, 142, 143, 150, 151, 154, 155, 164, 165, 175, 176, 181, 183, 186, 187, 191, 198, 200, 202, 205, 209, 211, 213, 217, 220, 225, 231, 241
Offset: 1

Views

Author

Benoit Cloitre, Jan 04 2002

Keywords

Comments

Related to the equation x^5 == 1 (mod k): sequence gives values of k such there are solutions 1 < x < k of x^5 == 1 (mod k).
If k is a term of this sequence, then G = is a non-abelian group of order 5k, where 1 < r < n and r^5 == 1 (mod k). For example, G can be the subgroup of GL(2, Z_k) generated by x = {{1, 1}, {0, 1}} and y = {{r, 0}, {0, 1}}. - Jianing Song, Sep 17 2019
The asymptotic density of this sequence is 1 (Dressler, 1975). - Amiram Eldar, May 23 2022

Examples

			x^5 == 1 (mod 11) has solutions 1 < x < 11, namely {3,4,5,9}.

Links

Crossrefs

Cf. A000010, A045453, A066498, A066499, A066501, A066502.
Column k=3 of A277915.

Programs

  • Mathematica
    Select[Range[250], Divisible[EulerPhi[#], 5] &] (* Amiram Eldar, May 23 2022 *)
  • PARI
    isok(k) = { eulerphi(k)%5 == 0 } \\ Harry J. Smith, Feb 18 2010

Formula

a(n) are the numbers generated by 5^2 = 25 and all primes congruent to 1 mod 5 (A045453). Hence sequence gives all k such that k == 0 (mod A045453(n)) for some n > 1 or k == 0 (mod 25).

Extensions

Simpler definition from Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 25 2003
Extended by Ray Chandler, Nov 06 2003

A066499 Numbers k such that phi(k) == 2 (mod 4).

Original entry on oeis.org

3, 4, 6, 7, 9, 11, 14, 18, 19, 22, 23, 27, 31, 38, 43, 46, 47, 49, 54, 59, 62, 67, 71, 79, 81, 83, 86, 94, 98, 103, 107, 118, 121, 127, 131, 134, 139, 142, 151, 158, 162, 163, 166, 167, 179, 191, 199, 206, 211, 214, 223, 227, 239, 242, 243, 251, 254, 262, 263, 271
Offset: 1

Views

Author

Benoit Cloitre, Jan 04 2002

Keywords

Comments

Related to the equation x^4 = 1 (mod y): sequence gives values of n such x^4 = 1 (mod n) has no solution 1 < x < n-1.
k is of the form p^m or 2*p^m where p is A002145 (with the exception of a(2)=4).
All prime numbers here belong also to A002145, prime numbers of the form 4n+3. - Enrique Pérez Herrero, Sep 07 2011

References

  • W. J. LeVeque, Fundamentals of Number Theory, pp. 57 Problem 15, Dover NY 1996.

Links

Crossrefs

Cf. A066498, A066500, A066501, A066502, A000010.
Essentially the same as A097987.
Cf. A002145.

Programs

  • Mathematica
    Select[Range[300],Mod[EulerPhi[#],4]==2&] (* Harvey P. Dale, Feb 18 2018 *)
  • PARI
    isok(k) = { eulerphi(k)%4 == 2 } \\ Harry J. Smith, Feb 18 2010

Extensions

Simpler definition from Lekraj Beedassy, Jul 21 2003
Corrected and extended by Ray Chandler, Nov 06 2003
Showing 1-5 of 5 results.

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