A245014 Least prime p such that 2n*4^n divides p + 4n^2 + 1.

3, 47, 347, 6079, 10139, 147311, 687931, 18874111, 37748411, 104857199, 276823579, 805305791, 29662117211, 30064770287, 64424508539, 2473901161471, 11098195491707, 7421703486191, 83562883709531, 527765581330879, 369435906930971, 27866022694353007, 19421773393033147

Offset: 1

Author

R. J. Cano Sep 17 2014

Keywords

• nonn

Comments

All those terms such that 2n*4^n is equal to p + 4n^2 + 1 belong to A247024.

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A247024.

Programs

Mathematica

a[n_] := With[{k = n*2^(2*n+1)}, p = -4*n^2-1; While[!PrimeQ[p += k]]; p]; Table[a[n], {n, 1, 23}] (* Jean-François Alcover, Oct 09 2014, translated from Charles R Greathouse IV's PARI code *)

PARI

search(u)={ /* Slow, u must be a small integer. */
  my(log2=log(2),q,t,t0,L1=List());
  forprime(y=3,prime(10^u),
    t=log(y+1)\log2;
    while(t>t0,
      q=4*t^2+y+1;
      if(q%(t*(2^(2*t+1)))==0,
        listput(L1,[t,y]);
        t0=t;
        break
      ,
        t--
      )));
  L1
}

PARI

a(n)=my(k=n<<(2*n+1),p=-4*n^2-1); while(!isprime(p+=k),); p \\ Charles R Greathouse IV, Sep 18 2014

Formula

a(n) << n^5*1024^n by Xylouris' version of Linnik's theorem. - Charles R Greathouse IV, Sep 18 2014

Extensions

a(10)-a(23) from Charles R Greathouse IV, Sep 18 2014