A178313 Absolute difference between prime factors of n-th semiprime mod n.

0, 1, 0, 3, 0, 2, 4, 1, 0, 1, 8, 3, 2, 3, 10, 5, 0, 14, 6, 16, 6, 7, 8, 20, 10, 4, 12, 12, 12, 26, 6, 28, 12, 14, 16, 34, 18, 19, 10, 0, 18, 38, 40, 12, 20, 44, 22, 2, 24, 21, 26, 25, 50, 16, 26, 0, 56, 29, 58, 32, 6, 33, 1, 35, 22, 36, 34, 8, 68, 35, 38, 24, 34, 70, 4, 35, 42, 76, 6, 0

Offset: 1

Author

Juri-Stepan Gerasimov, Dec 20 2010

Keywords

• nonn

Examples

			a(2)=1 because semiprime(2) = 6 = 3*2 and (3-2) mod 2 = 1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A001358, A178102, A109313.

Programs

Mathematica

semiPrimeQ[n_] := Plus @@ Last /@ FactorInteger@ n == 2; f[n_] := Subtract @@ Reverse@ Flatten[ Table[ #[[1]], {#[[2]]}] & /@ FactorInteger@ n]; t = Select[ Range@ 215, semiPrimeQ]; Table[ Mod[ f[ t[[n]]], n], {n, 80}]
f[{a_,b_}]:=Module[{c=FactorInteger[b][[;;,1]]},If[Length[c]==1,0,Mod[Differences[c][[1]],a]]]; Module[{nn=300,spr},spr=Select[Range[nn],PrimeOmega[#]==2&];f/@Thread[{Range[ Length[ spr]],spr}]] (* Harvey P. Dale, May 29 2024 *)

Python

from math import isqrt
from sympy import primepi, primerange, primefactors
from sympy.ntheory.primetest import is_square
def A178313(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+x+((t:=primepi(s:=isqrt(x)))*(t-1)>>1)-sum(primepi(x//p) for p in primerange(s+1)))
    return 0 if is_square(m:=bisection(f,n,n)) else ((p:=primefactors(m))[1]-p[0])%n # Chai Wah Wu, Apr 03 2025