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.

Showing 1-3 of 3 results.

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

Views

Author

Jianing Song, Dec 25 2021

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

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

Programs

  • PARI
    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)))

Formula

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

Extensions

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

A350342 Numbers k such that k^2 is an abelian order.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 35, 37, 41, 43, 47, 53, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 97, 101, 103, 107, 109, 113, 115, 119, 127, 131, 133, 137, 139, 143, 149, 151, 157, 161, 163, 167, 173, 179, 181, 185, 187, 191, 193, 197, 199, 209
Offset: 1

Views

Author

Jianing Song, Dec 25 2021

Keywords

Comments

Such k must be squarefree. Actually, such k must be a cyclic number (A003277).
Number of the form p_1*p_2*...*p_r where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.
The smallest term with exactly n distinct prime factors is given by A350340.
From the term 5 on, no term can be divisible by 2 or 3.

Examples

			For primes p, p is a term since every group of order p^2 is abelian. Such group is isomorphic to either C_{p^2} or C_p X C_p.
For primes p, q, if p^2 !== 1 (mod q), q^2 !== 1 (mod p), then p*q is a term since every group of that order is abelian. Such group is isomorphic to C_{p^2*q^2}, C_p X C_{p*q^2}, C_q X C_{p^2*q} or C_{p*q} X C_{p*q}.
		

Crossrefs

Cf. A051532 (abelian orders), A003277 (cyclic numbers), A350343, A350340.
A350344 is the subsequence of composite numbers.

Programs

  • PARI
    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
    isA350342(n) = isA051532(n^2)

Formula

A350343(n) = a(n)^2.

A350344 Composite k such that k^2 is an abelian order.

Original entry on oeis.org

35, 65, 77, 85, 115, 119, 133, 143, 161, 185, 187, 209, 215, 217, 221, 235, 247, 259, 265, 299, 319, 323, 329, 335, 341, 365, 371, 377, 391, 403, 407, 413, 415, 427, 437, 451, 469, 481, 485, 493, 511, 515, 517, 527, 533, 535, 551, 553, 559, 565, 583, 589, 595
Offset: 1

Views

Author

Jianing Song, Dec 25 2021

Keywords

Comments

Numbers k such that k^2 is an abelian order with at least 4 groups.
Number of the form p_1*p_2*...*p_r where r > 1, the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.
The smallest number k such that k^2 is an abelian order with at least 8 groups is A350340(3) = 595.
No term can be divisible by 2 or 3.

Examples

			For primes p, q, if p^2 !== 1 (mod q), q^2 !== 1 (mod p), then p*q is a term since every group of that order is abelian. Such group is isomorphic to C_{p^2*q^2}, C_p X C_{p*q^2}, C_q X C_{p^2*q} or C_{p*q} X C_{p*q}.
95 is not a term since 95^2 = 5^2 * 19^2 is not an abelian order. Note that 95 itself is a cyclic number.
		

Crossrefs

Cf. A051532 (abelian orders), A050384, A350340.
Equals A350342 \ ({1} U A000040).

Programs

  • PARI
    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
    isA350344(n) = (n>1) && !isprime(n) && isA051532(n^2)

Formula

A350345(n) = a(n)^2.
Showing 1-3 of 3 results.