A180507 Numbers k such that k^2 + 1 = p*q, p and q prime with p == q (mod k).

3, 8, 12, 144, 1020, 8040, 13860, 34840, 729180, 1728240, 3232060, 17576520, 39279240, 85184880, 117649980, 778689840, 884737920, 1225045140, 1771563420, 3723878100, 3869896140, 4574299320, 7762395960, 12487172640, 14348911860, 14886940920, 21484957560, 24137574780

Offset

1,1

Comments

q - p = k with k = 3, 8, 144.

The next terms with q - p = k are F(432) = 85738...5984 and F(570) where F(n) is the n-th Fibonacci number. All such terms are in A001906; the next such term, if one exists, has more than 25000 decimal digits. - Charles R Greathouse IV, Jan 21 2011

Links

Jinyuan Wang, Table of n, a(n) for n = 1..10000

Example

a(3) = 12 because 12^2 + 1 = 5*29 and 29 - 5 = 2*12;
a(8) = 34840 because 34840^2 + 1 = 4289 * 283009 and 283009 - 4289 = 278720 = 8*34840.

Maple

with(numtheory):for k from 1 to 40000 do: x:=k^2+1:y:=factorset(x):yy:=bigomega(x):if
  yy=2 and irem(y[2], k) =y[1] then printf(`%d, `, k):else fi:od:

Prog

(PARI) w(m, r) = Vec(x*(1-x)/(1-(m^2+2)*x+x^2) + O(x^r));
isok(s, t) = isprime(s) && isprime(s+t);
lista(nn) = {my(g, k, m=1, r, u=w(1, nn), v=List([])); for(i=2, r=#u, g=k=(u[i]+sqrtint(5*u[i]^2-4))/2; if(isok(u[i], k), listput(v, k))); while(r>2, u=w(m++, r); for(i=2, #u, k=(m*u[i]+sqrtint((m^2+4)*u[i]^2-4))/2; if(k<g, if(isok(u[i], m*k), listput(v, k)), r=i; break))); Set(v); } \\ Jinyuan Wang, Mar 29 2020

Crossrefs

Subset of A085722.

Cf. A001906, A002496, A005574, A027862, A134406, A134407.

Sequence in context: A032142 A032046 A247396 * A218889 A131213 A320260

Adjacent sequences: A180504 A180505 A180506 * A180508 A180509 A180510

Keyword

nonn

Author

Michel Lagneau, Jan 20 2011

Extensions

More terms from Charles R Greathouse IV, Jan 24 2011

Missing terms inserted and more terms from Jinyuan Wang, Mar 30 2020

Status

approved