|
%I #45 Sep 08 2022 08:44:43
%S 8,20,32,44,56,68,80,92,104,116,128,140,152,164,176,188,200,212,224,
%T 236,248,260,272,284,296,308,320,332,344,356,368,380,392,404,416,428,
%U 440,452,464,476,488,500,512,524,536,548,560,572,584,596,608,620,632
%N a(n) = 12*n + 8.
%C Also the number of cube units that frame a cube of edge length n+1. _Peter M. Chema_, Mar 27 2016
%H Vincenzo Librandi, <a href="/A017617/b017617.txt">Table of n, a(n) for n = 0..3000</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F a(n) = 2*a(n-1) - a(n-2). - _Vincenzo Librandi_, Jun 08 2011
%F A089911(a(n)) = 9. - _Reinhard Zumkeller_, Jul 05 2013
%F G.f.: 12*x/(1-x)^2 + 8/(1-x) = 4*(2+x)/(1-x)^2. (see the PARI program). - _Wolfdieter Lang_, Oct 11 2021
%F Sum_{n>=0} (-1)^n/a(n) = sqrt(3)*Pi/36 - log(2)/12. - _Amiram Eldar_, Dec 12 2021
%e For n=3; a(3)= 12*3+8 = 44.
%e Thus, there are 44 cube units that frame a cube of edge length 4. - _Peter M. Chema_, Mar 26 2016
%t 12*Range[0,200]+8 (* _Vladimir Joseph Stephan Orlovsky_, Feb 19 2011 *)
%o (Magma) [12*n+8: n in [0..60]]; // _Vincenzo Librandi_, Jun 08 2011
%o (Haskell)
%o a017617 = (+ 8) . (* 12) -- _Reinhard Zumkeller_, Jul 05 2013
%o (PARI) x='x+O('x^99); Vec(4*(2+x)/(1-x)^2) \\ _Altug Alkan_, Mar 27 2016
%Y Cf. A008594, A017533, A017545, A016957, A016789, A089911.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
|
