|
%I #37 Feb 21 2023 02:11:58
%S 0,14,42,84,140,210,294,392,504,630,770,924,1092,1274,1470,1680,1904,
%T 2142,2394,2660,2940,3234,3542,3864,4200,4550,4914,5292,5684,6090,
%U 6510,6944,7392,7854,8330,8820,9324,9842,10374,10920,11480,12054,12642,13244,13860
%N 14 times triangular numbers.
%C Sequence found by reading the line from 0, in the direction 0, 14, ... and the same line from 0, in the direction 0, 42, ..., in the square spiral whose vertices are the generalized 16-gonal numbers. - _Omar E. Pol_, Oct 03 2011
%H Vincenzo Librandi, <a href="/A163756/b163756.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 7*n*(n+1) = 14*A000217(n).
%F G.f.: 14*x/(1-x)^3.
%F a(n) = 7*A002378(n) = 2*A024966(n) = A069127(n+1) - 1. - _Omar E. Pol_, Oct 03 2011
%F E.g.f.: 7*x*(x + 2)*exp(x). - _G. C. Greubel_, Aug 02 2017
%F From _Amiram Eldar_, Feb 21 2023: (Start)
%F Sum_{n>=1} 1/a(n) = 1/7.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/7.
%F Product_{n>=1} (1 - 1/a(n)) = -(7/Pi)*cos(sqrt(11/7)*Pi/2).
%F Product_{n>=1} (1 + 1/a(n)) = (7/Pi)*cos(sqrt(3/7)*Pi/2). (End)
%t Table[7*n*(n-1),{n,100}] (* _Vladimir Joseph Stephan Orlovsky_, Jul 06 2011 *)
%t 14*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,14,42},50] (* _Harvey P. Dale_, May 11 2021 *)
%o (PARI) a(n)=7*n*(n+1) \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A000217, A002378, A069127.
%Y Cf. A274978 (generalized 16-gonal numbers).
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Aug 03 2009
%E Extended by _R. J. Mathar_, Aug 06 2009
|
