A081585 Third row of Pascal-(1,3,1) array A081578.

1, 9, 33, 73, 129, 201, 289, 393, 513, 649, 801, 969, 1153, 1353, 1569, 1801, 2049, 2313, 2593, 2889, 3201, 3529, 3873, 4233, 4609, 5001, 5409, 5833, 6273, 6729, 7201, 7689, 8193, 8713, 9249, 9801, 10369, 10953, 11553, 12169, 12801, 13449, 14113

Offset: 0

Author

Paul Barry, Mar 23 2003

Keywords

• easy
• nonn

Comments

The identity (8*n^2 +1)^2 - (64*n^2 +16)*n^2 = 1 can be written as a(n)^2 -A157912(n)*n^2 = 1 for n>0. - Vincenzo Librandi, Feb 09 2012

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A016813, A081578, A081586, A157912.

Programs

Magma

I:=[1,9,33]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 09 2012

Maple

seq(1+8*n^2, n=0..100); # Robert Israel, Sep 04 2015

Mathematica

LinearRecurrence[{3,-3,1}, {1,9,33}, 40] (* Vincenzo Librandi, Feb 09 2012 *)

PARI

for(n=0, 50, print1(8*n^2+1", ")); \\ Vincenzo Librandi, Feb 09 2012

Sage

[8*n^2 +1 for n in (0..40)] # G. C. Greubel, May 26 2021

Formula

a(n) = 8*n^2 + 1.

G.f.: (1+3*x)^2/(1-x)^3.

a(n) = a(n-1) + 16*n - 8 with a(0)=1. - Vincenzo Librandi, Aug 08 2010

a(n) = sqrt(8*(A000217(2*n-1)^2 +A000217(2*n)^2) +1). - J. M. Bergot, Sep 04 2015

From Amiram Eldar, Jul 15 2020: (Start)

Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(8))*coth(Pi/sqrt(8)))/2.

Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(8))*csch(Pi/sqrt(8)))/2. (End)

From Amiram Eldar, Feb 05 2021: (Start)

Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(8))*sinh(Pi/2).

Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(8))*csch(Pi/sqrt(8)). (End)

E.g.f.: (1 +8*x +8*x^2)*exp(x). - G. C. Greubel, May 26 2021