A174685 Indices i such that 2*A000326(i)+1 is also a pentagonal number.

0, 75, 244, 86359, 281384, 99658019, 324716700, 115005267375, 374722790224, 132715978892539, 432429775201604, 153154124636722439, 499023585859860600, 176739727114798801875, 575872785652503930604, 203957491936353180641119

Offset: 1

Author

Jonathan Vos Post, Mar 27 2010

Keywords

• nonn
• easy

Examples

			Using P(n) = A000326(n) we have:
a(1) = 0 because 2*P(0)+1 = 2*0+1 = 1 = P(1).
a(2) = 75 because 2*P(75)+1 = 2*8400+1 = 16801 = P(106).
a(3) = 244 because 2*P(244)+1 = 2*89182+1 = 178365 = P(345).
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..653 (terms to 1000 digits)
• Eric Weisstein's World of Mathematics, Pentagonal Number.
• Eric Weisstein's World of Mathematics, Sophie Germain prime.
• Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Crossrefs

Cf. A000326, A005384, A124174, A260937.

Programs

Mathematica

LinearRecurrence[{1,1154,-1154,-1,1},{0,75,244,86359,281384},30] (* Harvey P. Dale, Jul 17 2014 *)

PARI

/* Jack Brennen, who extended this sequence, found that, other than the trivial pair (0,1), you need to solve: u^2 - 2*v^2 = 23. And take only those values where u and v are both congruent to 5 mod 6. Then you have P((u+1)/6) = 2*P((v+1)/6)+1. The following PARI/GP will give the nontrivial answers: */
{forstep(i=7,47,8,
  print(Vec(lift(Mod((x+1),x^2-2)^i*(4*x-3)+(x+1))/6));
  print(Vec(lift(Mod((x+1),x^2-2)^i*(4*x+3)+(x+1))/6)))}

Formula

G.f.: x^2*(-75-169*x+435*x^2+x^3) / ( (x-1)*(x^2-34*x+1)*(x^2+34*x+1) ) with a(n) = 1/6 -7*A029547(n) +239*A029547(n-1) +35*A029547(n)*(-1)^n/6 +1183*A029547(n-1)*(-1)^(n-1)/6. - R. J. Mathar, Oct 25 2011

a(n)= a(n-1)+ 1154*a(n-2)-1154*a(n-3)-a(n-4)+a(n-5). - Harvey P. Dale, Jul 17 2014