A336883 a(n) = ((A002144(n) - 1)/2)! (mod A002144(n)) where A002144(n) is the n-th Pythagorean prime.

2, 5, 13, 12, 31, 9, 23, 11, 27, 34, 22, 91, 33, 15, 37, 44, 129, 80, 162, 81, 183, 122, 144, 64, 16, 187, 217, 53, 138, 288, 114, 189, 213, 42, 104, 274, 63, 381, 266, 29, 254, 382, 348, 48, 301, 286, 489, 439, 483, 24, 77, 125, 578, 423, 487, 149, 555, 615, 651, 135, 96, 380, 87, 39, 707

Offset

1,1

Comments

Let p(n) = A002144(n) be the n-th Pythagorean prime.

Pythagorean prime p can be divided into a pair of integers (a,b) such as p =a+b and a*b==1 mod p. And (p-2)!==1 mod p because of Wilson's Theorem (p-1)!==-1 mod p. It can be divided into two parts (a,b) such as {2*3*4*...*((p(n)-1)/2)==a(n) mod p(n)} and {((p(n)-1)/2+1)*...*(p(n)-4)*(p(n)-3)*(p(n)-2)==-a(n)==(p(n)-a(n)) mod p(n)}. The pair numbers make a(n)+(p(n)-a(n))=p(n) and a(n)*(p(n)-a(n))==1 mod p(n). The left integer of the pair numbers is a(n). The right integer (p(n)-a(n)) is A336884(n).

The set of selecting odd numbers from {a(n)} and A336884 is A206549. The set of selecting even numbers from {a(n)} and A336884 is A209874 except for the number 1. A256011 never appears in {a(n)} or A336884. It is related to nonexistence of numbers that the largest prime factor of n^2+1 is greater than n.

The odd number of the difference |a(n)-A336884(n)|=|a(n)-(p(n)-a(n))|=|2*a(n)-p(n)| is A186814(n). A282538 never appears in the set of the difference |a(n)-A336884(n)|.

If p(n) is unknown, p(n) can be derived from a(n) using following equation. From a*b==1 mod p, a*b=k*p+1. With p=a+b, it can transform to b(n)=(k*a(n)+1)/(a(n)-k), k is an odd integer parameter when the fraction makes an integer. If there are many k's, select the minimum k in those. Then a(n)+b(n)=p(n). b(n) is A336884(n).

Links

Hiroyuki Hara, Table of n, a(n) for n = 1..4783 [reformatted and restored by Georg Fischer, Oct 15 2020]

Example

p(1)=5: (5-2)!=2*3=a(1)*(5-a(1))==1 mod 5. 5=2+3.
p(2)=13: (13-2)!=(2*3*4*5*6)*(7*8*9*10*11)=(2*3*4*5*6)*((p-6)*(p-5)*(p-4)*(p-3)*(p-2))==5*(-5)==5*(13-5)=5*8==a(2)*(13-a(2))==1 mod 13. 13=5+8.
a(n)=13: b(n)=(k*13+1)/(13-k)=(3*13+1)/(13-3)=4, k=3. p(n)=13+4=17.
a(n)=12: b(n)=(k*12+1)/(12-k)=(7*12+1)/(12-7)=17, k=7. p(n)=12+17=29.

Mathematica

Map[Mod[((# - 1)/2)!, #] &, Select[4 Range[192] + 1, PrimeQ]] (* Michael De Vlieger, Oct 15 2020 *)

Prog

(PARI) my(v=select(p->p%4==1, primes(100))); apply(x->(((x-1)/2)! % x), v) \\ Michel Marcus, Aug 07 2020
(Python) n_start=5
n_end=n_start+10000
k = 1
for n in range(n_start, n_end, 4):
    c=(n-1)//2
    r=1
    for i in range(2, c+1):
        r=r*i % n
        if r==0:
            break
    if (n-r)*r % n ==1:
        print(k, r)
        k = k + 1
# modified by Georg Fischer, Oct 16 2020

Crossrefs

Cf. A336884, A002144 (Pythagorean primes), A206549, A209874, A256011, A186814, A282538.

Sequence in context: A173620 A319920 A166134 * A067365 A189993 A112838

Adjacent sequences: A336880 A336881 A336882 * A336884 A336885 A336886

Keyword

nonn

Author

Hiroyuki Hara, Aug 06 2020

Status

approved