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%I #43 Feb 21 2023 02:12:24
%S 0,15,45,90,150,225,315,420,540,675,825,990,1170,1365,1575,1800,2040,
%T 2295,2565,2850,3150,3465,3795,4140,4500,4875,5265,5670,6090,6525,
%U 6975,7440,7920,8415,8925,9450,9990,10545,11115,11700,12300,12915,13545,14190,14850,15525
%N 15 times triangular numbers.
%C Sequence found by reading the line from 0, in the direction 0, 15, ... and the same line from 0, in the direction 0, 45, ..., in the square spiral whose vertices are the generalized 17-gonal numbers.
%C Sum of the numbers from 7n to 8n. - _Wesley Ivan Hurt_, Dec 23 2015
%C Also the number of 4-cycles in the (n+6)-triangular honeycomb obtuse knight graph. - _Eric W. Weisstein_, Jul 28 2017
%H Vincenzo Librandi, <a href="/A194715/b194715.txt">Table of n, a(n) for n = 0..10000</a>
%H M. Janjic and B. Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv 1301.4550 [math.CO], 2013.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GraphCycle.html">Graph Cycle</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 15*n*(n+1)/2 = 15*A000217(n) = 5*A045943(n) = 3*A028895(n) = A069128(n+1) - 1.
%F From _Wesley Ivan Hurt_, Dec 23 2015: (Start)
%F G.f.: 15*x/(1-x)^3.
%F a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>2.
%F a(n) = Sum_{i=7n..8n} i. (End)
%F From _Amiram Eldar_, Feb 21 2023: (Start)
%F Sum_{n>=1} 1/a(n) = 2/15.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/15.
%F Product_{n>=1} (1 - 1/a(n)) = -(15/(2*Pi))*cos(sqrt(23/15)*Pi/2).
%F Product_{n>=1} (1 + 1/a(n)) = (15/(2*Pi))*cos(sqrt(7/15)*Pi/2). (End)
%p A194715:=n->15*n*(n+1)/2: seq(A194715(n), n=0..60); # _Wesley Ivan Hurt_, Dec 23 2015
%t 15*Accumulate[Range[0, 60]] (* _Harvey P. Dale_, Feb 12 2012 *)
%t Table[15 n (n + 1)/2, {n, 0, 60}] (* _Wesley Ivan Hurt_, Dec 23 2015 *)
%t 15 Binomial[Range[20], 2] (* _Eric W. Weisstein_, Jul 28 2017 *)
%t 15 PolygonalNumber[Range[0, 20]] (* _Eric W. Weisstein_, Jul 28 2017 *)
%o (Magma) [15*n*(n+1)/2: n in [0..50]]; // _Vincenzo Librandi_, Oct 04 2011
%o (PARI) a(n)=15*n*(n+1)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A000217, A028895, A035008, A045943, A069128, A163756.
%Y Cf. A001105 (3-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - _Eric W. Weisstein_, Jul 29 2017
%K nonn,easy
%O 0,2
%A _Omar E. Pol_, Oct 03 2011
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