A264907 -id:A264907 - cp's OEIS frontend

A350340 a(n) is the smallest k such that k^2 is an abelian order with precisely 2^n groups.

1, 2, 35, 595, 13685, 506345, 26836285, 1702480351, 80016576497, 5681176931287, 414725915983951, 40228413850443247, 4304440281997427429, 546663915813673283483, 75986284298100586404137, 10144780646398552482233711, 1511572316313384319852822939, 252432576824335181415421430813, 49729217634394030738838021870161

Offset: 0

Jianing Song, Dec 25 2021

If m is an abelian order, then m = (p_1)^2 * (p_2)^2 * ... * (p_r)^2 * q_1 * q_2 * ... * q_s, where p_1, p_2, ... p_r, q_1, q_2, ..., q_s are distinct primes such that (p_i)^2 !== 1 (mod p_j) for i != j, (p_i)^2 !== 1 (mod q_j), q_i !== 1 (mod p_j), q_i !== 1 (mod q_j) for i != j. In this case there are 2^r groups of order m.
Note that the smallest abelian order with precisely 2^n groups must be the square of a squarefree number.
a(n) is the smallest k with n distinct prime factors such that k^2 is an abelian order.
a(n) is the smallest number of the form p_1*p_2*...*p_n where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.
a(n) exists for all n.
Except for a(1) = 2, no term can be divisible by 2 or 3. Conjecture: lpf(a(n+1)) >= lpf(a(n)) for all n, where lpf = least prime factor. - David A. Corneth and Jianing Song, Jan 03 2022

			a(2) = 35 = 5*7 since the smallest k with 2 distinct prime factors such that k^2 is an abelian order is 35.
a(3) = 595 = 5*7*17 since the smallest k with 3 distinct prime factors such that k^2 is an abelian order is 595.

David A. Corneth, Table of n, a(n) for n = 0..30
David A. Corneth, PARI program

Cf. A051532 (abelian orders), A264907, A350341.

isA051532(n) = my(f=factor(n), v=vector(#f[, 1])); for(i=1, #v, if(f[i, 2]>2, return(0), v[i]=f[i, 1]^f[i, 2])); for(i=1, #v, for(j=i+1, #v, if(v[i]%f[j, 1]==1 || v[j]%f[i, 1]==1, return(0)))); 1 \\ Charles R Greathouse IV's program for A051532
a(n) = for(k=1, oo, if(issquarefree(k) && omega(k)==n && isA051532(k^2), return(k)))

A350341(n) = a(n)^2.

a(7)-a(18) from David A. Corneth, Jan 02 2022

A350341 a(n) is the smallest abelian order with precisely 2^n groups.

Original entry on oeis.org

1, 4, 1225, 354025, 187279225, 256385259025, 720186192601225, 2898439345541083201, 6402652514300252791009, 32275771324587574319476369, 171997585388727183548489570401, 1618325280922534070007738367903009, 18528206141282092567518596574121550041, 298841436852738871021507444144006480611289

Offset: 0

Jianing Song, Dec 25 2021

If m is an abelian order, then m = (p_1)^2 * (p_2)^2 * ... * (p_r)^2 * q_1 * q_2 * ... * q_s, where p_1, p_2, ... p_r, q_1, q_2, ..., q_s are distinct primes such that (p_i)^2 !== 1 (mod p_j) for i != j, (p_i)^2 !== 1 (mod q_j), q_i !== 1 (mod p_j), q_i !== 1 (mod q_j) for i != j. In this case there are 2^r groups of order m.
Note that the smallest abelian order with precisely 2^n groups must be the square of a squarefree number.
a(n) is the smallest square number with n distinct prime factors that is an abelian order.
a(n) is the smallest number of the form (p_1*p_2*...*p_n)^2 where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.
a(n) exists for all n.
Except for a(1) = 4, no term can be divisible by 2 or 3. Conjecture: lpf(a(n+1)) >= lpf(a(n)) for all n, where lpf = least prime factor.

			a(2) = 1225 = 35^2 since the smallest square number with 2 distinct prime factors that is an abelian order is 1225.
a(3) = 354025 = 595^2 since the smallest square number with 3 distinct prime factors that is an abelian order is 354025.

David A. Corneth, Table of n, a(n) for n = 0..18

Cf. A051532 (abelian orders), A264907, A350340.

isA051532(n) = my(f=factor(n), v=vector(#f[, 1])); for(i=1, #v, if(f[i, 2]>2, return(0), v[i]=f[i, 1]^f[i, 2])); for(i=1, #v, for(j=i+1, #v, if(v[i]%f[j, 1]==1 || v[j]%f[i, 1]==1, return(0)))); 1 \\ Charles R Greathouse IV's program for A051532
a(n) = for(k=1, oo, if(issquarefree(k) && omega(k)==n && isA051532(k^2), return(k^2)))

a(n) = A350340(n)^2.

a(7)-a(11) from David A. Corneth, Jan 02 2022

A100564 Normal sequence of primes with a(1) = 3.

Original entry on oeis.org

3, 5, 17, 23, 29, 53, 83, 89, 113, 149, 173, 197, 257, 263, 269, 293, 317, 353, 359, 383, 389, 419, 449, 467, 479, 503, 509, 557, 563, 569, 593, 617, 653, 659, 677, 683, 773, 797, 809, 827, 857, 863, 887, 947, 977, 983, 1049, 1097, 1109, 1217, 1223, 1229, 1283

Offset: 1

Franz Vrabec, Nov 28 2004

nonn

A sequence {a(1), a(2), a(3), ... } is called a "normal sequence of primes" if a(1) is prime and if for every n > 1 a(n) is the smallest prime greater than a(n-1) such that the primes a(1), a(2), ..., a(n-1) are not divisors of a(n)-1.
The existence of the primes a(n) is guaranteed by Dirichlet's theorem on primes in arithmetic progressions.
Erdős proved that the number of terms in this sequence which do not exceed x is ~ (1 + o(1)) x/(logx loglogx), and that the sum of their reciprocals diverges. - Amiram Eldar, May 15 2017
The sum of reciprocals diverges slowly: the sum exceeds 1 only after adding 159989 terms: 1/3 + 1/5 + ... + 1/11321273 = 1.0000000628... - Amiram Eldar, May 28 2017
The product a(1)*a(2)*...*a(n) gives a cyclic number (A003277) with n factors. For the smallest cyclic number with n prime factors, see A264907. - Jeppe Stig Nielsen, May 22 2021

			a(2) = 5 because a(1) = 3 is not a divisor of 4 = 5 - 1.
a(3) = 17 because a(1) = 3 is a divisor of 6 and 12 (so 7 and 13 are not possible for a(3)); a(2) = 5 is a divisor of 10 (so 11 is not possible for a(3)), but a(1) = 3 and a(2) = 5 both not divisors of 16 = 17 - 1.

S. W. Golomb, Problems in the Distribution of the Prime Numbers, Ph.D. dissertation, Dept. of Mathematics, Harvard University, May 1956. See page 8.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Paul Erdős, On a problem of G. [sic] Golomb, Journal of the Australian Mathematical Society, Vol 2, Issue 1 (1961), pp. 1-8.
Solomon W. Golomb, Sets of primes with intermediate density, Mathematica Scandinavica, Vol. 3 (1956), pp. 264-274.
H. G. Meijer, Sets of Primes with Intermediate Density, Mathematica Scandinavica, Vol. 34 (1974), pp. 37-43.
Erick Wong, Computations on Normal Families of Primes.

a:= proc(n) option remember; local p;
      if n=1 then 3
    else p:= a(n-1);
         do p:= nextprime(p);
            if {} = numtheory[factorset](p-1) intersect
               {seq(a(i), i=1..n-1)} then return p fi
         od
      fi
    end:
seq(a(n), n=1..70);  # Alois P. Heinz, Feb 05 2017

a[1] = 3; a[n_] := a[n] = Block[{k = PrimePi[a[n - 1]] + 1, t = Table[a[i], {i, n - 1}]}, While[ Union[ Mod[ Prime[k] - 1, t]][[1]] == 0, k++ ]; Prime[k]]; Table[ a[n], {n, 53}] (* Robert G. Wilson v, Dec 04 2004 *)

More terms from Robert G. Wilson v, Dec 04 2004

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A350340 a(n) is the smallest k such that k^2 is an abelian order with precisely 2^n groups.

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A350341 a(n) is the smallest abelian order with precisely 2^n groups.

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A100564 Normal sequence of primes with a(1) = 3.

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