A306440 Number of different ways of expressing 2*n as (p - 1)*(q - 1), where p < q are primes.

0, 1, 1, 1, 1, 1, 2, 0, 1, 1, 1, 1, 2, 0, 1, 1, 1, 0, 2, 0, 2, 1, 1, 1, 1, 0, 1, 0, 1, 1, 3, 0, 1, 1, 0, 1, 4, 0, 0, 1, 1, 1, 1, 0, 2, 0, 1, 0, 2, 0, 1, 1, 1, 1, 2, 0, 2, 0, 1, 0, 3, 0, 0, 1, 0, 1, 2, 0, 1, 1, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 2, 1, 1, 1, 2, 0, 1, 0, 1, 1, 3, 0, 1, 0, 0, 1

Offset: 0

Author

Andres Cicuttin, Feb 15 2019

Keywords

• nonn

Comments

If 2*n-1 is an odd prime then a(n) > 0.

If a(n) > 0 then 2n+1 is in A323550.

Given any distinct odd primes p_1, ..., p_m, Dickson's conjecture implies that there are infinitely many m such that m/(p_i-1) + 1 is prime for all i. Thus the sequence should be unbounded. - Robert Israel, Mar 29 2019

Examples

			a(2) = 1 because 2*2 = 4 can only be expressed as (p - 1)*(q - 1) with primes p = 2 and q = 5.
a(6) = 2 because for 2*6 = 12, there are only two possible ordered pairs of distinct primes (p,q), (2,13) and (3,7), such that 12 = (p - 1)*(q - 1).
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A323550.

Programs

Maple

f:= proc(n) local t;
    nops(select(t -> t^2<2*n and isprime(t+1) and isprime(2*n/t+1), numtheory:-divisors(2*n)))
end proc:
map(f, [$0..200]); # Robert Israel, Mar 18 2019

Mathematica

a[n_]:=Module[{k=0},Do[Do[If[2n==(Prime[i]-1)*(Prime[j]-1),k++],{i,1,j-1}],{j,2,PrimePi[2n]+1}];Return[k]];
Table[a[j],{j,0,128}]

PARI

A306440(n,d,c)={forprime(p=2,sqrtint(-(n>0)+n*=2)+1,n%(p-1)==0 && isprime(n/(p-1)+1) && c++ && d && printf("%d-1=(%d-1)*(%d-1) [%d], ",n+1,p,n/(p-1)+1,c));c} \\ Give 1 as 2nd optional arg (d=debug) to get a list of all decompositions. - M. F. Hasler, Feb 25 2019

PARI

a(n) = if(n==0, return(0)); my(d=divisors(n<<1)); d+=vector(#d, i, 1); sum(i=1, #d\2, isprime(d[i]) && isprime(d[#d-i+1])) \\ for finding lots of terms or a(n) for large n. \\ David A. Corneth, Mar 18 2019

Formula

a(6k+1) = 0 for k > 0 because 12k+2 can't be written as (p-1)(q-1) except for k = 0 with p = 2, q = 3: If q > 3, then q-1 is congruent to 0 or 4 (mod 6), and no p = 2, p = 3 (=> q-1 = 6k+1) or p > 3 is possible. - M. F. Hasler, Feb 25 2019