A103787 a(n) = number of k's that make primorial P(n)/A019565(k)+A019565(k) prime, A019565(k)^2<=P(n).

1, 2, 4, 8, 12, 21, 40, 70, 117, 263, 450, 703, 1385, 2423, 5501, 8617, 18249, 29352, 61970, 103568, 209309, 404977, 853279, 1609502, 3008915, 5342983, 10287184, 19087437, 38498011, 78520137, 145642314

Offset: 1

Author

Lei Zhou, Feb 15 2005

Keywords

• hard
• nonn

Comments

If we remove the restriction A019565(k)^2<=P(n), every term gets doubled.

Number of distinct primes of the form d + P(n)/d, where P(n) is the n-th primorial A002110(n) and d is a divisor of P(n).

Examples

			P(1)=2, A019565(0)=1, 2/1+1=3 is prime, a(1)=1;
P(2)=6, A019565(0)=1, 6/1+1=7; A019565(1)=2, 6/2+2=5; so a(2)=2.
		

Crossrefs

Cf. A002110, A019565, A103785, A103786.

Programs

Mathematica

npd = 1; Do[npd = npd*Prime[n]; tn = 0; tt = 1; cp = npd/tt + tt; ct = 0; While[IntegerQ[cp], If[(cp >= (tt*2)) && PrimeQ[cp], ct = ct + 1]; tn = tn + 1; tt = 1; k1 = tn; o = 1; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, tt = tt*Prime[o]]; k1 = (k1 - k2)/2; o = o + 1]; cp = npd/tt + tt]; Print[ct], {n, 1, 22}]
Table[ps=Prime[Range[n]]; cnt=0; Do[b=IntegerDigits[i,2,n]; p=Times@@(ps^b) + Times@@(ps^(1-b)); If[PrimeQ[p], cnt++], {i,0,2^(n-1)-1}]; cnt, {n,22}]

Formula

a(n) = A088627(A002110(n)/2).

Extensions

a(28)-a(31) from James G. Merickel, Aug 07 2015