A015126 -id:A015126 - cp's OEIS frontend

A028476 Greatest k such that phi(k) = phi(n), where phi is Euler's totient function.

2, 2, 6, 6, 12, 6, 18, 12, 18, 12, 22, 12, 42, 18, 30, 30, 60, 18, 54, 30, 42, 22, 46, 30, 66, 42, 54, 42, 58, 30, 62, 60, 66, 60, 90, 42, 126, 54, 90, 60, 150, 42, 98, 66, 90, 46, 94, 60, 98, 66, 120, 90, 106, 54, 150, 90, 126, 58, 118, 60, 198, 62, 126, 120, 210, 66, 134

Offset: 1

Vladeta Jovovic, Jan 12 2002

nonn

Every number in this sequence occurs at least twice. For all n > 6, a(n) > phi(n)^2 is impossible. - Alonso del Arte, Dec 31 2016

			phi(1) = 1 and phi(2) = 1 also. There is no greater k such that phi(k) = 1, so therefore a(1) = a(2) = 2.
phi(3) = phi(4) = phi(6) = 2, and there is no greater k such that phi(k) = 6, hence a(3) = a(4) = a(6) = 6.

Antti Karttunen, Table of n, a(n) for n = 1..10000 (computed from the b-file of A057826 provided by T. D. Noe)
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A000010, A015126, A057826, A066412, A066659, A071181.

Table[Module[{k = (2 Boole[n <= 6]) + #^2}, While[EulerPhi@ k != #, k--]; k] &@ EulerPhi@ n, {n, 120}] (* Michael De Vlieger, Dec 31 2016 *)

a(n) = invphiMax(eulerphi(n)); \\ Amiram Eldar, Nov 14 2024, using Max Alekseyev's invphi.gp

a(1) = a(2) = 2, for n > 2, a(n) = A057826(A000010(n)/2). - Antti Karttunen, Aug 07 2017

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)

A289828 a(n) is the least k such that phi(k) = n*phi(n).

Original entry on oeis.org

1, 3, 7, 15, 25, 13, 43, 51, 81, 41, 121, 65, 157, 129, 143, 255, 289, 109, 361, 187, 301, 253, 529, 193, 625, 313, 487, 337, 841, 241, 961, 771, 661, 685, 899, 433, 1369, 1083, 937, 641, 1681, 551, 1849, 881, 1147, 1013, 2209, 769, 2401, 1111, 1751, 1249, 2809, 1141, 2323, 1469, 2053, 1711

Offset: 1

Altug Alkan, Aug 13 2017

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A000010, A002618, A015126.

With[{s = EulerPhi /@ Range[10^4]}, Table[First@ FirstPosition[s, n EulerPhi@ n], {n, 58}]] (* Michael De Vlieger, Aug 14 2017 *)

a(n) = my(k=1); while(1, if(eulerphi(k)==n*eulerphi(n), return(k)); k++) \\ Felix Fröhlich, Aug 14 2017

a(n) = invphiMin(n * eulerphi(n)); \\ Amiram Eldar, Nov 15 2024, using Max Alekseyev's invphi.gp

a(n) = A015126(n^2).

n^2/log log n << a(n) <= n^2. More specifically (on the lower bound), a(n) > n^2(e^-gamma + o(1))/log log n. - Charles R Greathouse IV, Aug 14 2017

A320045 Smallest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

Original entry on oeis.org

1, 1, 3, 3, 5, 3, 7, 8, 7, 5, 11, 8, 13, 7, 15, 15, 17, 7, 19, 15, 21, 11, 23, 24, 25, 13, 19, 21, 29, 15, 31, 32, 33, 17, 35, 21, 37, 19, 35, 40, 41, 21, 43, 33, 35, 23, 47, 40, 43, 25, 51, 35, 53, 19, 55, 56, 57, 29, 59, 40, 61, 31, 63, 51, 65, 33, 67, 51, 69, 35

Offset: 1

Jianing Song, Oct 04 2018

nonn

a(n) = n iff n is a term in A296233.
Most terms are odd. Among the first 10000 terms there are 8837 odd ones. The even terms are divisible by 4 because (Z/kZ)* is isomorphic to (Z/(2k)Z)* for odd k.
A015126(n) <= a(n) <= n.

			The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 7.
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) = 55.
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) = 35.

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

Cf. A015126, A296233, A320046.

a(n) = my(i=eulerphi(n)); while(znstar(i)[2]!=znstar(n)[2], i++); i

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

A028476 Greatest k such that phi(k) = phi(n), where phi is Euler's totient function.

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

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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.

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A289828 a(n) is the least k such that phi(k) = n*phi(n).

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A320045 Smallest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

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PARI

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

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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.