A258780 a(n) is the least k such that k^2 + 1 is a semiprime p*q, p < q, and (q - p)/2^n is prime.

8, 12, 140, 64, 2236, 196, 1300, 1600, 6256, 5084, 248756, 246196, 484400, 36680, 887884, 821836, 1559116, 104120, 126072244, 9586736, 4156840, 542759984, 1017981724, 2744780140, 405793096, 148647496, 1671024916

Offset: 2

Author

Michel Lagneau, Jun 10 2015

Keywords

• nonn
• more

Comments

The corresponding primes are 2, 3, 71, 7, 1069, 7, 5, 5, 59, 2, 368471, 180463, 12421, 2, 29, 125683, 226169, 5, 369704891, 197, 5, 263, 7444559, 239621423, 594271, 2, 474359, ...

All terms are even, in order for k^2+1 to be odd. Otherwise, with k^2+1 being even, p-q would be odd and hence not a multiple of 2^n. - Michel Marcus, Apr 13 2019

Examples

			a(2)=8 because 8^2+1 = 5*13 and (13-5)/2^2 = 2 is prime. The number 8 is the first term of the sequence 8, 22, 34, 46, 50, 58, ...
a(3)=12 because 12^2+1 = 5*29 and (29-5)/2^3 = 3 is prime. The number 12 is the first term of the sequence 12, 28, 44, 52, 76, 80, ...
a(4)=140 because 140^2+1 = 17*1153 and (1153-17)/2^4 = 71 is prime. The number 140 is the first term of the sequence 140, 296, 404, 604, ...
		

Crossrefs

Cf. A085722, A144255.

Programs

Mathematica

lst={};Do[k=2;While[!(Plus@@Last/@FactorInteger[k^2+1]==2&&PrimeQ[(FactorInteger[k^2+1][[-1,1]]-FactorInteger[k^2+1][[1,1]])/2^n]),k=k+2];Print[n," ",k],{n,2,19}];lst

PARI

isok(k, n) = my(kk=k^2+1, f=factor(kk)[,1]~); (bigomega(kk) == 2) && (#f == 2) && (p=f[1]) && (q=f[2]) && (qq=(q-p)/2^n) && !frac(qq) && isprime(qq);
a(n) = my(k=2); while (!isok(k,n), k+=2); k; \\ Michel Marcus, Apr 13 2019

Extensions

Name edited by Jon E. Schoenfield, Sep 12 2017

a(20)-a(22) from Daniel Suteu, Apr 13 2019

a(23)-a(28) from Daniel Suteu, Nov 09 2019