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A163756 14 times triangular numbers.

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