A323739 a(n) is the number of residues modulo (4*primorial(n)) of the squares of primes greater than or equal to prime(n+1).

2, 1, 1, 2, 6, 30, 180, 1440, 12960, 142560, 1995840, 29937600, 538876800, 10777536000, 226328256000, 5205549888000, 135344297088000, 3924984615552000, 117749538466560000, 3885734769396480000, 136000716928876800000, 4896025809439564800000, 190945006568143027200000

Offset: 0

Author

Jon E. Schoenfield, Feb 20 2019

Keywords

• nonn
• easy

Comments

Here, "primorial(n)" is A002110(n) = Product_{k=1..n} prime(k).

For n >= 1, a(n) is the number of coprime squares modulo 4*primorial(n). Note that 4*primorial(n) = A102476(n+1) is the smallest k such that rank((Z/kZ)*) = n+1 for n >= 1. (The rank of a finitely generated group rank(G) is defined to be the size of the minimal generating sets of G. In particular, rank((Z/kZ)*) = 0 if k <= 2 and A046072(k) otherwise.) - Jianing Song, Oct 18 2021

Examples

			a(3) = 2 because, for every prime p >= prime(3+1) = 7, p^2 mod (4*2*3*5 = 120) is one of the 2 values {1, 49}:
   7^2 mod 120 =  49 mod 120 = 49
  11^2 mod 120 = 121 mod 120 =  1
  13^2 mod 120 = 169 mod 120 = 49
  17^2 mod 120 = 289 mod 120 = 49
  19^2 mod 120 = 361 mod 120 =  1
  23^2 mod 120 = 529 mod 120 = 49
  29^2 mod 120 = 841 mod 120 =  1
  ...
.
   q=(n+1)st        b =          residues p^2 mod b
n    prime    4*primorial(n)         for p >= q         a(n)
=  =========  ===============  =======================  ====
0      2      4         =   4           {0,1}             2
1      3      4*2       =   8            {1}              1
2      5      4*2*3     =  24            {1}              1
3      7      4*2*3*5   = 120           {1,49}            2
4     11      4*2*3*5*7 = 840  {1,121,169,289,361,529}    6
		

Links

• Jianing Song, Table of n, a(n) for n = 0..200
• Jianing Song, Further terms: table of n, a(n) for n = 0..1000

Crossrefs

Cf. A002110, A005867, A240775, A046072, A102476.

Programs

PARI

a(n) = if(n==0, 2, my(t=1); forprime(p=3, , t*=(p-1)/2; if(n--<2, return(t)))) \\ Jianing Song, Oct 18 2021, following Charles R Greathouse IV's program for A078586

Formula

Conjecture: a(n) = 2^(1-n)*Product_{j=1..n} (prime(j)-1) for n >= 0, so a(n) = a(n-1)*(prime(n)-1)/2 for n >= 1.

From Charlie Neder, Feb 28 2019: (Start)

Conjecture is true. Since there exists a prime congruent to r modulo 4*primorial(n) for any r coprime to primorial(n), this set is precisely the set of coprime quadratic residues of 4*primorial(n). If n >= 1, each residue can be broken down into congruences modulo 8 and the first n-1 odd primes, each odd prime p has (p-1)/2 residue classes, and every combination eventually occurs, giving the formula. (End)

Extensions

More terms from Jianing Song, Oct 18 2021