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a(n) = (n+1)*(2-n)/2.
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%I #60 Sep 26 2022 05:43:00

%S 1,1,0,-2,-5,-9,-14,-20,-27,-35,-44,-54,-65,-77,-90,-104,-119,-135,

%T -152,-170,-189,-209,-230,-252,-275,-299,-324,-350,-377,-405,-434,

%U -464,-495,-527,-560,-594,-629,-665,-702,-740,-779,-819,-860,-902,-945,-989,-1034,-1080,-1127,-1175,-1224,-1274,-1325,-1377

%N a(n) = (n+1)*(2-n)/2.

%C Coefficient of x in the polynomial C(n,0)+C(n+1,1)x+C(n+2,2)x(x-1)/2.

%C Equals A154990 * [1,2,3,...]. - _Gary W. Adamson_ & _Mats Granvik_, Jan 19 2009

%C a(n) is essentially the case 1 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_1(n) = n*(3-n)/2 and a(n) = P_1(n+1). See A005563 for the case k=0. - _Peter Luschny_, Jul 08 2011

%C This is the case k=-1 of the formula (k*m*(m+1)-(-1)^k+1)/2. See similar sequences listed in A262221. - _Bruno Berselli_, Sep 17 2015

%H Vincenzo Librandi, <a href="/A080956/b080956.txt">Table of n, a(n) for n = 0..10000</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 2*(C(n+1, 1)-C(n+2, 2)) = (n+1)*(2-n)/2.

%F G.f.: (1-2*x)/(1-x)^3. - _R. J. Mathar_, Jun 11 2009

%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-1,2), for n>=3. - _Milan Janjic_, Dec 20 2008

%F E.g.f.: exp(x)*(1-x^2/2). - _Zerinvary Lajos_, Apr 05 2009, _R. J. Mathar_, Jun 11 2009

%F a(n) = - A214292(n,1) for n > 0. - _Reinhard Zumkeller_, Jul 12 2012

%F Recurrence: a(0)=1, a(n+1) = a(n) - n. Also a(n)=(n+1)-Sum[k=1..n](k). Also a(n) = A000027(n+1) - A000217(n). Also, for n>1, a(n) = - A000096(n-2). - _Stanislav Sykora_, Feb 19 2014

%F Sum_{n>=3} 1/a(n) = -11/9. - _Amiram Eldar_, Sep 26 2022

%e a(5) = 6-(1+2+3+4+5). - _Stanislav Sykora_, Feb 19 2014

%p G(x):=exp(x)*(x-x^2/2): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=1..54 ); # _Zerinvary Lajos_, Apr 05 2009

%t f[n_] := n; lst = {}; Do[a = f[n]; Do[a -= f[m], {m, n - 1, 1, -1}]; AppendTo[lst, a], {n, 46}]; lst (* _Vladimir Joseph Stephan Orlovsky_, Feb 10 2010 *)

%t FoldList[#1 - #2 &, 1, Range[0, 44]] (* _Arkadiusz Wesolowski_, May 26 2013 *)

%t LinearRecurrence[{3,-3,1},{1,1,0},60] (* _Harvey P. Dale_, Nov 29 2019 *)

%o (Magma) [(n+1)*(2-n)/2: n in [0..80]]; // _Vincenzo Librandi_, Jul 08 2011

%o (PARI) a(n)=(n+1)*(2-n)/2;

%Y Cf. A000027, A000096, A000217, A154990, A214292, A262221.

%K sign,easy

%O 0,4

%A _Paul Barry_, Mar 01 2003

%E Lajos e.g.f. adapted to offset zero by _R. J. Mathar_, Jun 11 2009