A328412 -id:A328412 - cp's OEIS frontend

A328417 Numbers k such that A328412(k) sets a new record; numbers k such that (Z/mZ)* = C_2 X C_(2k) has more solutions for m than all k' < k, where (Z/mZ)* is the multiplicative group of integers modulo m.

1, 2, 6, 30, 78, 210, 690, 1050, 4830

Offset: 1

Jianing Song, Oct 14 2019

Conjecture: this sequence is infinite. That is to say, A328412 is unbounded.

It seems that a(n) == 2 (mod 4) for n > 1.

			For k = 30: (Z/mZ)* = C_2 X C_60 has 11 solutions, namely m = 143, 155, 175, 183, 225, 244, 286, 310, 350, 366, 450; for all k' < 30, (Z/mZ)* = C_2 X C_(2k') has fewer than 11 solutions. So 30 is a term.

Wikipedia, Multiplicative group of integers modulo n

Cf. A328412, A328418.

my(t=0); for(k=1, 5000, if(A328412(k)>t, print1(k, ", "); t=A328412(k))) \\ See A328412 for its program

A328418 Records in A328412.

Original entry on oeis.org

2, 4, 7, 11, 12, 13, 15, 16, 23

Offset: 1

Jianing Song, Oct 14 2019

Companion sequence of A328417.

Conjecture: this sequence is infinite. That is to say, A328412 is unbounded.

			Let (Z/mZ)* be the multiplicative group of integers modulo m. We have (Z/mZ)* = C_2 X C_60 has 11 solutions, namely m = 143, 155, 175, 183, 225, 244, 286, 310, 350, 366, 450; for all k' < 30, (Z/mZ)* = C_2 X C_(2k') has fewer than 11 solutions. So A328412(30) = 11 is a term.

Wikipedia, Multiplicative group of integers modulo n

Cf. A328412, A328417.

my(t=0); for(k=1, 5000, if(A328412(k)>t, print1(a(k), ", "); t=A328412(k)))

A328413 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 15, 16, 18, 20, 21, 22, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 39, 40, 41, 42, 44, 45, 46, 48, 50, 51, 53, 54, 55, 56, 58, 60, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 78, 81, 82, 83, 86, 87, 88, 89, 90, 95, 96, 98, 99, 102, 105, 106, 110, 111

Offset: 1

Jianing Song, Oct 14 2019

nonn

For n > 1, it is easy to see A114871(n)/2 is a term of this sequence. The smallest term here not of the form A114871(k)/2 is 24: 48 is not of the form (p-1)*p^k for any prime p, but (Z/mZ)* = C_2 X C_48 has solutions m = 119, 153, 238, 306.

			(Z/mZ)* = C_2 X C_2 has solutions m = 8, 12; (Z/mZ)* = C_2 X C_4 has solutions m = 15, 16, 20, 30; (Z/mZ)* = C_2 X C_6 has solutions m = 21, 28, 36, 42; (Z/mZ)* = C_2 X C_8 has solutions m = 32; (Z/mZ)* = C_2 X C_10 has solutions m = 33, 44, 66; (Z/mZ)* = C_2 X C_12 has solutions m = 35, 39, 45, 52, 70, 78, 90. So 1, 2, 3, 4, 5, 6 are all terms.

Cf. A328412. Complement of A328414.

Cf. also A114871.

isA328413(n) = my(r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N+1, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, return(1)); if(k==N+1, return(0)))
for(n=1, 100, if(isA328413(n), print1(n, ", ")))

A328414 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has no solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

7, 12, 13, 17, 19, 25, 28, 31, 34, 37, 38, 43, 47, 49, 52, 57, 59, 61, 62, 67, 71, 73, 76, 77, 79, 80, 84, 85, 91, 92, 93, 94, 97, 100, 101, 103, 104, 107, 108, 109, 112, 117, 118, 121, 122, 124, 127, 129, 133, 137, 139, 142, 143, 144, 148, 149, 151, 152, 157, 160, 161, 163, 164

Offset: 1

Jianing Song, Oct 14 2019

nonn

Indices of 0 in A328410, A328411 and A328412.

By definition, if there is no such m that psi(m) = 2k, psi = A002322, then m is a term of this sequence.

			12 is a term: if there exists m such that (Z/mZ)* = C_2 X C_24 = C_2 X C_8 X C_3, then m must have a factor q such that q is an odd prime power and phi(q) = 8 or phi(q) = 24, phi = A000010, which is impossible.
80 is a term: if there exists m such that (Z/mZ)* = C_2 X C_80 = C_2 X C_16 X C_5, then m must have a factor q such that q is an odd prime power and phi(q) = 80 or phi(q) = 16, which is impossible.

Wikipedia, Multiplicative group of integers modulo n

Cf. A328410, A328411, A328412. Complement of A328413.

Cf. also A000010, A002322, A005277, A079695.

isA328414(n) = my(r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N+1, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, return(0)); if(k==N+1, return(1)))
for(n=1, 200, if(isA328414(n), print1(n, ", ")))

A328415 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has exactly one solution, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

4, 16, 27, 32, 64, 256, 512, 1024, 2048, 2187, 4096, 6561, 8192, 16384, 59049, 65536, 131072, 177147, 262144, 524288, 531441, 1048576, 1594323, 2097152, 4194304, 4782969, 8388608, 14348907, 16777216, 33554432, 67108864, 134217728, 268435456, 387420489, 536870912, 1073741824

Offset: 1

Jianing Song, Oct 14 2019

nonn

Numbers k being powers of 2 or 3 such that 2*k+1 is not prime.
Proof. If m is a solution to (Z/mZ)* = C_2 X C_(2k) such that m is odd, then 2*m is also a solution, and vice versa. So if there is only one solution to (Z/mZ)* = C_2 X C_(2k), m must be a multiple of 4. If 8 divides m and m has odd prime factors, or if m has at least two distinct odd prime factors, then A046072(m) >= 3, a contradiction. So m = 2^e, e >= 3 or m = 4*p^e, p odd prime and e >= 1. If m = 4*p^e and p >= 5, then (Z/(3*p^e)Z)* = (Z/mZ)*. So we have m = 2^e, e >= 3 or m = 4*3^e, e >= 1, then (Z/mZ)* = C_2 X C_(2*2^(e-3)) or (Z/mZ)* = C_2 X C_(2*3^(e-1)).
If k = 2^(e-3) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(2^e)Z)*; if k = 3^(e-1) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(4*3^e)Z)*; on the other hand, if k is a power of 2 or a power of 3 such that 2*k+1 is not prime, then (Z/mZ)* = C_2 X C_(2k) indeed has only one solution.

			The only solution to (Z/mZ)* = C_2 X C_54 is m = 324, so 54/2 = 27 is a term.

Wikipedia, Multiplicative group of integers modulo n

Cf. A328412.

select(i->!isprime(2*i+1), upto(10^9)) \\ See A006899 for the function upto(n)

A328416 Smallest k such that (Z/mZ)* = C_2 X C_(2k) has exactly n solutions for m, or 0 if no such k exists, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

7, 4, 1, 5, 2, 10, 21, 6, 42, 90, 150, 30, 78, 210, 2730, 690, 1050

Offset: 0

Jianing Song, Oct 14 2019

Conjecture: a(n) > 0 for all n. That is to say, every number occurs in A328412.

It seems that most terms are congruent to 2 modulo 4.

			(Z/mZ)* = C_2 X C_42 has exactly 6 solutions m = 129, 147, 172, 196, 258, 294; for any k < 21, (Z/mZ)* = C_2 X C_(2k) has either fewer than or more than 6 solutions, so a(6) = 21.

Wikipedia, Multiplicative group of integers modulo n

Cf. A328412.

a(n) = my(k=1); while(A328412(k)!=n, k++); k \\ See A328412 for its program

cp's OEIS Frontend

A328417 Numbers k such that A328412(k) sets a new record; numbers k such that (Z/mZ)* = C_2 X C_(2k) has more solutions for m than all k' < k, where (Z/mZ)* is the multiplicative group of integers modulo m.

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A328418 Records in A328412.

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A328413 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

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A328414 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has no solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

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A328415 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has exactly one solution, where (Z/mZ)* is the multiplicative group of integers modulo m.

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A328416 Smallest k such that (Z/mZ)* = C_2 X C_(2k) has exactly n solutions for m, or 0 if no such k exists, where (Z/mZ)* is the multiplicative group of integers modulo m.

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