cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A261803 a(n) is the smallest number satisfying a(n)^2+1 = p(n)*q(n), p(n) < q(n) both prime, such that q(n+1)/p(n+1) < q(n)/p(n) with the initial condition q(1)/p(1) < 3/2.

Original entry on oeis.org

50, 334, 516, 670, 844, 1164, 1250, 1800, 2450, 9800, 14450, 20000, 24200, 101250, 105800, 135200, 162450, 168200, 204800, 304200, 336200, 451250, 480200, 490050, 530450, 696200, 924800, 966050, 1008200, 1125000, 1155200, 1428050, 1805000, 2332800, 2420000
Offset: 1

Views

Author

Michel Lagneau, Sep 01 2015

Keywords

Comments

The sequence is probably infinite. [This follows from Schinzel's hypothesis H, for example. - Charles R Greathouse IV, Aug 31 2021]
A majority of numbers in the sequence are of the form 2*q^2 with q = 5, 25, 30, 35, 70, 85, 100, 110, 225, 230, 260, 285, 290, 320, 390, 410, ... So, it seems that {a(n)} = {334, 516, 670, 844, 1164} union {2*A109306(n)^2} where A109306 are the numbers k such that k^2 + (k-1)^2 and k^2 + (k+1)^2 are both primes.

Examples

			a(1) = 50 because 50^2+1 = 41*61 => 61/41 = 1.4878... < 1.5
a(2) = 334 because 334^2+1 = 281*397 => 397/281 = 1.4128... < 1.4878...
a(3) = 516 because 516^2+1 = 449*593 => 593/449 = 1.3207... < 1.4128...
a(4) = 670 because 670^2+1 = 593*757 => 757/593 = 1.2765... < 1.3207...
		

Crossrefs

Subsequence of A085722.

Programs

  • Maple
    with(numtheory):nn:=100:d:=1.5:
    for n from 1 to nn do:
      x:=factorset(n^2+1):n0:=bigomega(n^2+1):
       if n0=2
       then
       q:=evalf(x[2]/x[1]):
       if q
    				
  • Mathematica
    (* Assumption: n>7 ==> a(n)=0 mod 50 *) a[n_] := a[n] = For[k = Which[n==1, 0, n <= 7, a[n-1]+1, True, a[n-1] + 50], True, k = Which[n <= 7, k+1, k == a[n-1]+1, k+49, True, k+50], f = FactorInteger[k^2+1]; If[Length[f] == 2, If[f[[All, 2]] == {1, 1}, {p1, q1} = f[[All, 1]]; If[q1/p1 < If[n == 1, 3/2, q[n-1]/p[n-1]], p[n] = p1; q[n] = q1; Return[k]]]]]; Table[Print["a(", n, ") = ", a[n], "  p = ", p[n], "  q = ", q[n],  "  q/p = ", N[q[n]/p[n], 10], "  q-p = ", q[n]-p[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Sep 28 2015 *)