A028476 -id:A028476 - cp's OEIS frontend

A015126 Least k such that phi(k) = phi(n).

1, 1, 3, 3, 5, 3, 7, 5, 7, 5, 11, 5, 13, 7, 15, 15, 17, 7, 19, 15, 13, 11, 23, 15, 25, 13, 19, 13, 29, 15, 31, 17, 25, 17, 35, 13, 37, 19, 35, 17, 41, 13, 43, 25, 35, 23, 47, 17, 43, 25, 51, 35, 53, 19, 41, 35, 37, 29, 59, 17, 61, 31, 37, 51, 65, 25, 67, 51, 69, 35, 71, 35, 73

Offset: 1

Vladeta Jovovic, Jan 12 2002

From Jianing Song, Nov 11 2022: (Start)
The first even term is a(33817088) = 16842752 (see A002181 and A143510).
Conjecture: a(n) is always odd for odd n. (End)

Antti Karttunen, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A000010, A028476, A066412, A066659.

Cf. A002181, A143510.

a(n) = {my(en = eulerphi(n)); k = 1; while (eulerphi(k) != en, k++); return (k);} \\ Michel Marcus, Jun 17 2013

a(n) = vecmin(select(x -> x<=n, invphi(eulerphi(n)))); \\ Amiram Eldar, Nov 14 2024, using Max Alekseyev's invphi.gp

A043343 Numbers m such that there is no k > m such that phi(k) = phi(m), where phi is Euler's totient function.

Original entry on oeis.org

2, 6, 12, 18, 22, 30, 42, 46, 54, 58, 60, 62, 66, 90, 94, 98, 106, 118, 120, 126, 134, 138, 142, 150, 158, 162, 166, 174, 198, 206, 210, 214, 240, 242, 250, 254, 262, 270, 274, 276, 278, 282, 294, 298, 302, 318, 330, 334, 346, 348, 354, 358, 378, 382, 394, 398

Offset: 1

Vladeta Jovovic, Jan 12 2002

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..980 from Paul Tek)
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A000010, A028476, A066659.

Flatten@ Position[#, 0] &@ Table[k = n + 1; While[And[k <= 2 n + 1, EulerPhi@ k != EulerPhi@ n], k++]; Boole[k < 2 n + 1] k, {n, 400}] (* Michael De Vlieger, Dec 31 2016 *)

is(k) = #select(x -> x>k, invphi(eulerphi(k))) == 0; \\ Amiram Eldar, Nov 12 2024, using Max Alekseyev's invphi.gp

A066659(a(n)) = 0.

A028476(a(n)) = a(n).

Offset corrected by Paul Tek, Sep 25 2015

Definition clarified by Alonso del Arte, Dec 31 2016

A317993 Number of k such that (Z/kZ)* is isomorphic to (Z/nZ), where (Z/nZ) is the multiplicative group of integers modulo n.

Original entry on oeis.org

2, 2, 3, 3, 2, 3, 4, 2, 4, 2, 2, 2, 2, 4, 4, 4, 2, 4, 4, 4, 4, 2, 2, 1, 2, 2, 4, 4, 2, 4, 2, 1, 3, 2, 7, 4, 2, 4, 7, 3, 2, 4, 4, 3, 7, 2, 2, 3, 4, 2, 4, 7, 2, 4, 5, 3, 4, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 4, 3, 7, 2, 3, 2, 2, 5, 4, 7, 7, 2, 1, 2, 2, 2, 3, 2, 4, 3

Offset: 1

Jianing Song, Oct 03 2018

nonn

To find solutions for k to (Z/kZ)* = (Z/nZ)*, it's sufficient to check for A015126(n) <= k <= A028476(n).
It seems that this sequence is unbounded. For example, there are 59 solutions to (Z/nZ)* = C_2 X C_6 X C_1260.
Conjecture: Every number occurs in this sequence.

			The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 4.
The solutions to (Z/kZ)* = C_2 X C_20 are k = 55, 75, 100, 110 and 150, so a(55) = a(75) = a(100) = a(110) = a(150) = 5.
The solutions to (Z/kZ)* = C_2 X C_12 are k = 35, 39, 45, 52, 70, 78 and 90, so a(35) = a(39) = a(45) = a(52) = a(70) = a(78) = a(90) = 7.

Jianing Song, Table of n, a(n) for n = 1..20000
Wikipedia, Multiplicative group of integers modulo n

Cf. A015126, A028476, A057635.

Earliest occurrence of m is A303712(m).

a(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, search_max = A057635(eulerphi(n))); for(j=eulerphi(n)+1, search_max, if(znstar(j)[2]==znstar(n)[2], i++)); i) \\ search_max is the largest k such that phi(k) = phi(n). See A057635 for its program

A319928 Numbers k such that there is no other m such that (Z/mZ)* is isomorphic to (Z/kZ), where (Z/kZ) is the multiplicative group of integers modulo k.

Original entry on oeis.org

24, 32, 80, 96, 120, 128, 160, 168, 240, 252, 256, 264, 324, 384, 400, 408, 416, 456, 480, 504, 512, 544, 552, 640, 648, 672, 696, 768, 840, 928, 1040, 1088, 1128, 1272, 1280, 1312, 1320, 1360, 1408, 1416, 1504, 1536, 1632, 1696, 1704, 1840, 1848, 1896, 1920, 1992

Offset: 1

Jianing Song, Oct 03 2018

nonn

Numbers such that A317993(k) = 1.
To find such k, it's sufficient to check for A015126(k) <= m <= A028476(k).
This is a subsequence of A296233. As a result, all members in this sequence should not satisfy any congruence mentioned there. Specially, all terms here are divisible by 4.
There are only 218 terms <= 10000 and 396 terms <= 20000.

			(Z/24Z)* = C_2 X C_2 X C_2, and there is no other m such that (Z/mZ)* = C_2 X C_2 X C_2, so 24 is a term.
(Z/96Z)* = C_2 X C_2 X C_8, and there is no other m such that (Z/mZ)* = C_2 X C_2 X C_8, so 24 is a term.

Jianing Song, Table of n, a(n) for n = 1..396 (all terms <= 20000)
Wikipedia, Multiplicative group of integers modulo n

Cf. A015126, A028476, A057635, A296233, A317993.

b(n) = my(i=0, search_max = A057635(eulerphi(n))); for(j=eulerphi(n)+1, search_max, if(znstar(j)[2]==znstar(n)[2], i++)); i \\ search_max is the largest k such that phi(k) = phi(n). See A057635 for its program
isA319928(n) = if(n>2, b(n)==1, 0)

A320046 Largest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

Original entry on oeis.org

2, 2, 6, 6, 10, 6, 18, 12, 18, 10, 22, 12, 26, 18, 30, 30, 34, 18, 54, 30, 42, 22, 46, 24, 50, 26, 54, 42, 58, 30, 62, 32, 66, 34, 90, 42, 74, 54, 90, 60, 82, 42, 98, 66, 90, 46, 94, 60, 98, 50, 102, 90, 106, 54, 150, 84, 114, 58, 118, 60, 122, 62, 126, 102, 130, 66, 134, 102, 138, 90

Offset: 1

Jianing Song, Oct 04 2018

nonn

All terms are even because (Z/kZ)* is isomorphic to (Z/(2k)Z)* for odd k. Most terms are congruent to 2 modulo 4. Among the first 10000 terms there are 8980 ones congruent to 2 modulo 4.

			The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 18.
The solutions to (Z/kZ)* = C_2 X C_20 are k = 55, 75, 100, 110 and 150, so a(55) = a(75) = a(100) = a(110) = a(150) = 150.
The solutions to (Z/kZ)* = C_2 X C_12 are k = 35, 39, 45, 52, 70, 78 and 90, so a(35) = a(39) = a(45) = a(52) = a(70) = a(78) = a(90) = 90.

Jianing Song, Table of n, a(n) for n = 1..10000
Wikipedia, Multiplicative group of integers modulo n

Cf. A028476, A057635, A320045.

a(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, search_max = A057635(eulerphi(n))); for(j=eulerphi(n)+1, search_max, if(znstar(j)[2]==znstar(n)[2], i=j)); i) \\ search_max is the largest k such that phi(k) = phi(n). See A057635 for its program

n <= a(n) <= A028476(n).

A066705 Greatest k < n such that phi(k) = phi(n) if such k exists, otherwise 0.

Original entry on oeis.org

0, 1, 0, 3, 0, 4, 0, 5, 7, 8, 0, 10, 0, 9, 0, 15, 0, 14, 0, 16, 13, 11, 0, 20, 0, 21, 19, 26, 0, 24, 0, 17, 25, 32, 0, 28, 0, 27, 35, 34, 0, 36, 0, 33, 39, 23, 0, 40, 43, 44, 0, 45, 0, 38, 41, 52, 37, 29, 0, 48, 0, 31, 57, 51, 0, 50, 0, 64, 0, 56, 0, 70, 0, 63, 55, 74, 61, 72, 0, 68, 0

Offset: 1

Vladeta Jovovic, Jan 14 2002

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A000010, A015126, A028476, A043343, A066659.

a(n) = {my(v = select(x -> xAmiram Eldar, Nov 14 2024, using Max Alekseyev's invphi.gp

cp's OEIS Frontend

A015126 Least k such that phi(k) = phi(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

PARI

A043343 Numbers m such that there is no k > m such that phi(k) = phi(m), where phi is Euler's totient function.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A317993 Number of k such that (Z/kZ)* is isomorphic to (Z/nZ)*, where (Z/nZ)* is the multiplicative group of integers modulo n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A319928 Numbers k such that there is no other m such that (Z/mZ)* is isomorphic to (Z/kZ)*, where (Z/kZ)* is the multiplicative group of integers modulo k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A320046 Largest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A066705 Greatest k < n such that phi(k) = phi(n) if such k exists, otherwise 0.

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

A317993 Number of k such that (Z/kZ)* is isomorphic to (Z/nZ), where (Z/nZ) is the multiplicative group of integers modulo n.

A319928 Numbers k such that there is no other m such that (Z/mZ)* is isomorphic to (Z/kZ), where (Z/kZ) is the multiplicative group of integers modulo k.