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%I #28 Jan 11 2021 02:49:38
%S 0,15,31,48,66,85,105,126,148,171,195,220,246,273,301,330,360,391,423,
%T 456,490,525,561,598,636,675,715,756,798,841,885,930,976,1023,1071,
%U 1120,1170,1221,1273,1326,1380,1435,1491,1548,1606,1665
%N a(n) = n*(n+29)/2.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F If we define f(n,i,a) = sum_{k=0..n-i} (binomial(n,k)*stirling1(n-k,i)*product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,15), for n>=1. - _Milan Janjic_, Dec 20 2008
%F a(n) = n + a(n-1) + 14 with n>0, a(0)=0. - _Vincenzo Librandi_, Aug 03 2010
%F From _Colin Barker_, Mar 18 2012: (Start)
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f.: x*(15-14*x)/(1-x)^3. (End)
%F a(n) = 15n - floor(n/2) + floor(n^2/2). - _Wesley Ivan Hurt_, Jun 15 2013
%F From _Amiram Eldar_, Jan 11 2021: (Start)
%F Sum_{n>=1} 1/a(n) = 2*A001008(29)/(29*A002805(29)) = 9227046511387/33771798660600.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/29 - 236266661971/4824542665800. (End)
%t i=-14;s=0;lst={};Do[s+=n+i;If[s>=0, AppendTo[lst, s]], {n, 0, 6!, 1}];lst (* _Vladimir Joseph Stephan Orlovsky_, Oct 29 2008 *)
%o (PARI) a(n)=n*(n+29)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A000217, A001008, A002805, A056126.
%K easy,nonn
%O 0,2
%A _Omar E. Pol_, Aug 28 2007
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