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.

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A328420 Numbers k such that A317993(k) sets a new record; numbers k such that (Z/mZ)* = (Z/kZ)* has more solutions for m than all k' < k, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

1, 3, 7, 35, 104, 143, 371, 385, 2233, 6149, 16555, 17081
Offset: 1

Views

Author

Jianing Song, Oct 14 2019

Keywords

Comments

It seems that this sequence is infinite (i.e., A317993 is unbounded), but for each n, to really construct a number k such that A317993(k) > A317993(a(n)) seems impossible.

Examples

			For k = 104: (Z/mZ)* = (Z/104Z)* has 8 solutions, namely m = 104, 105, 112, 140, 144, 156, 180, 210; for all k' < 104, (Z/mZ)* = (Z/k'Z)* has fewer than 8 solutions. So 104 is a term.

Links

Crossrefs

Cf. A317993, A328421.

Programs

  • PARI
    b(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, k=eulerphi(n), N=floor(exp(Euler)*k*log(log(k^2))+2.5*k/log(log(k^2)))); for(j=k+1, N, if(znstar(j)[2]==znstar(n)[2], i++)); i)
my(t=0); for(k=1, 20000, if(b(k)>t, print1(k, ", "); t=b(k))) \\ Warning: program runs for about 30 min

A328421 Records in A317993.

Original entry on oeis.org

2, 3, 4, 7, 8, 11, 12, 17, 30, 39, 52, 59
Offset: 1

Views

Author

Jianing Song, Oct 14 2019

Keywords

Comments

Companion sequence of A328420.
It seems that this sequence is infinite (i.e., A317993 is unbounded), but for each n, to really construct a number k such that A317993(k) > a(n) seems impossible.

Examples

			Let (Z/mZ)* be the multiplicative group of integers modulo m. We have (Z/mZ)* = (Z/104Z)* has 8 solutions, namely m = 104, 105, 112, 140, 144, 156, 180, 210; for all k' < 104, (Z/mZ)* = (Z/k'Z)* has fewer than 8 solutions. So A317993(104) = 8 is a term.

Links

Crossrefs

Cf. A317993, A328420.

Programs

  • PARI
    b(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, k=eulerphi(n), N=floor(exp(Euler)*k*log(log(k^2))+2.5*k/log(log(k^2)))); for(j=k+1, N, if(znstar(j)[2]==znstar(n)[2], i++)); i)
my(t=0); for(k=1, 20000, if(b(k)>t, print1(b(k), ", "); t=b(k))) \\ Warning: program runs for about 30 min

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

Views

Author

Jianing Song, Oct 03 2018

Keywords

Comments

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.

Examples

			(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.

Links

Crossrefs

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

Programs

  • PARI
    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)
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