A323739 - OEIS

%I #32 Nov 13 2021 13:54:51

%S 2,1,1,2,6,30,180,1440,12960,142560,1995840,29937600,538876800,

%T 10777536000,226328256000,5205549888000,135344297088000,

%U 3924984615552000,117749538466560000,3885734769396480000,136000716928876800000,4896025809439564800000,190945006568143027200000

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

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

%C 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

%H Jianing Song, <a href="/A323739/b323739.txt">Table of n, a(n) for n = 0..200</a>

%H Jianing Song, <a href="/A323739/a323739.txt">Further terms: table of n, a(n) for n = 0..1000</a>

%F 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.

%F From _Charlie Neder_, Feb 28 2019: (Start)

%F 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)

%e 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}:

%e    7^2 mod 120 =  49 mod 120 = 49

%e   11^2 mod 120 = 121 mod 120 =  1

%e   13^2 mod 120 = 169 mod 120 = 49

%e   17^2 mod 120 = 289 mod 120 = 49

%e   19^2 mod 120 = 361 mod 120 =  1

%e   23^2 mod 120 = 529 mod 120 = 49

%e   29^2 mod 120 = 841 mod 120 =  1

%e   ...

%e .

%e    q=(n+1)st        b =          residues p^2 mod b

%e n    prime    4*primorial(n)         for p >= q         a(n)

%e =  =========  ===============  =======================  ====

%e 0      2      4         =   4           {0,1}             2

%e 1      3      4*2       =   8            {1}              1

%e 2      5      4*2*3     =  24            {1}              1

%e 3      7      4*2*3*5   = 120           {1,49}            2

%e 4     11      4*2*3*5*7 = 840  {1,121,169,289,361,529}    6

%o (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

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

%K nonn,easy

%O 0,1

%A _Jon E. Schoenfield_, Feb 20 2019

%E More terms from _Jianing Song_, Oct 18 2021