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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). 4
2, 1, 1, 2, 6, 30, 180, 1440, 12960, 142560, 1995840, 29937600, 538876800, 10777536000, 226328256000, 5205549888000, 135344297088000, 3924984615552000, 117749538466560000, 3885734769396480000, 136000716928876800000, 4896025809439564800000, 190945006568143027200000 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
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
LINKS
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)
EXAMPLE
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
PROG
(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
CROSSREFS
Sequence in context: A319031 A022874 A022873 * A122160 A058316 A082386
KEYWORD
nonn,easy
AUTHOR
Jon E. Schoenfield, Feb 20 2019
EXTENSIONS
More terms from Jianing Song, Oct 18 2021
STATUS
approved

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Last modified April 18 15:48 EDT 2024. Contains 371780 sequences. (Running on oeis4.)