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Third row of Pascal-(1,3,1) array A081578.
12

%I #51 Sep 08 2022 08:45:09

%S 1,9,33,73,129,201,289,393,513,649,801,969,1153,1353,1569,1801,2049,

%T 2313,2593,2889,3201,3529,3873,4233,4609,5001,5409,5833,6273,6729,

%U 7201,7689,8193,8713,9249,9801,10369,10953,11553,12169,12801,13449,14113

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

%C 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

%H Vincenzo Librandi, <a href="/A081585/b081585.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

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

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

%F a(n) = a(n-1) + 16*n - 8 with a(0)=1. - _Vincenzo Librandi_, Aug 08 2010

%F a(n) = sqrt(8*(A000217(2*n-1)^2 +A000217(2*n)^2) +1). - _J. M. Bergot_, Sep 04 2015

%F From _Amiram Eldar_, Jul 15 2020: (Start)

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

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

%F From _Amiram Eldar_, Feb 05 2021: (Start)

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

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

%F E.g.f.: (1 +8*x +8*x^2)*exp(x). - _G. C. Greubel_, May 26 2021

%p seq(1+8*n^2, n=0..100); # _Robert Israel_, Sep 04 2015

%t LinearRecurrence[{3,-3,1}, {1,9,33}, 40] (* _Vincenzo Librandi_, Feb 09 2012 *)

%o (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

%o (PARI) for(n=0, 50, print1(8*n^2+1", ")); \\ _Vincenzo Librandi_, Feb 09 2012

%o (Sage) [8*n^2 +1 for n in (0..40)] # _G. C. Greubel_, May 26 2021

%Y Cf. A016813, A081578, A081586, A157912.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Mar 23 2003