A323741 a(n) = m-p where m = (2n+1)^2 and p is the largest prime < m.

2, 2, 2, 2, 8, 2, 2, 6, 2, 2, 6, 6, 2, 2, 8, 2, 2, 2, 10, 12, 2, 8, 2, 2, 8, 6, 2, 20, 12, 2, 2, 6, 6, 2, 2, 6, 2, 2, 12, 8, 6, 6, 8, 2, 8, 2, 12, 6, 10, 8, 2, 22, 2, 14, 20, 6, 6, 2, 2, 2, 8, 6, 2, 8, 2, 6, 2, 12, 2, 14, 6, 2, 8, 8, 14, 10, 2, 18, 20, 2, 8, 14, 6, 2, 10, 2, 32, 2, 12, 12, 2, 8, 6, 44, 2, 6, 14, 6, 20, 14

Offset

1,1

Comments

a(n) cannot be a square: suppose a(n) = k^2; then p=m-a(n) could be factored as (2n+k-1)*(2n-k-1); hence it would not be a prime.

Legendre's conjecture implies a(n) <= 4*n. Oppermann's conjecture implies a(n) <= 2*n. - Robert Israel, Sep 04 2019

All terms are even. - Alois P. Heinz, Sep 04 2019

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Formula

a(n) = A049711(A016754(n)).

Example

When n=4, m=81, p=79, so a(4) = 81-79 = 2.

Maple

seq((2*n+1)^2-prevprime((2*n+1)^2), n=1..100); # Robert Israel, Sep 04 2019

Mathematica

mp[n_]:=Module[{m=(2n+1)^2}, m-NextPrime[m, -1]]; Array[mp, 100] (* Harvey P. Dale, Feb 03 2022 *)

Prog

(VBA/Excel)
Sub A323741()
For n = 1 To 100
   Cells(n, 1) = (2 * n + 1) ^ 2
   k = Cells(n, 1) - 2
   k1 = (2 * n - 1) ^ 2 + 2
   For p = k To k1 Step -2
      IsPrime = True
      For i = 2 To Int(Sqr(p))
         If p mod i = 0 Then
            IsPrime = False
            Exit For
         End If
      Next i
      If IsPrime Then
         Cells(n, 2) = p
         Cells(n, 3) = Cells(n, 1) - Cells(n, 2)
         Exit For
      End If
   Next p
Next n
End Sub
(PARI) a(n) = (2*n+1)^2 - precprime((2*n+1)^2 - 1); \\ Michel Marcus, Sep 05 2019

Crossrefs

Cf. A016754, A049711, A088572, A089166, A151799.

Sequence in context: A327641 A066874 A087577 * A168281 A203610 A294873

Adjacent sequences: A323738 A323739 A323740 * A323742 A323743 A323744

Keyword

nonn

Author

Ali Sada, Sep 03 2019

Status

approved