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A051942 Truncated triangular numbers: a(n) = n*(n+1)/2-3*t*(t+1)/2, t = 5. 10

%I

%S 0,10,21,33,46,60,75,91,108,126,145,165,186,208,231,255,280,306,333,

%T 361,390,420,451,483,516,550,585,621,658,696,735,775,816,858,901,945,

%U 990,1036,1083,1131,1180,1230,1281,1333,1386,1440,1495,1551,1608,1666

%N Truncated triangular numbers: a(n) = n*(n+1)/2-3*t*(t+1)/2, t = 5.

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

%F a(n) = (n^2+n-90)/2 = (n-9)*(n+10)/2 = n*(n+1)/2 - 45.

%F a(n) = +3*a(n-1) -3*a(n-2) +1*a(n-3), n>=13.

%F a(n) = n+a(n-1) (with a(9)=0). - _Vincenzo Librandi_, Aug 06 2010

%F G.f.: x^10*(10-9*x)/(1-x)^3.

%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+9) = -f(n,n-1,10), for n>=1. - _Milan Janjic_, Dec 20 2008

%F a(n) = 10n - floor(n/2) + floor(n^2/2). - _Wesley Ivan Hurt_, Jun 15 2013

%e a(10) = 10 + 0 = 10; a(11) = 11 + 10 = 21; a(12) = 12 + 21 = 33.

%p A051942:=n->(n^2+n-90)/2: seq(A051942(n), n=9..80); # _Wesley Ivan Hurt_, Jan 28 2017

%t Table[n (n + 1)/2 - 45, {n, 9, 100}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 15 2011 *)

%t #-45&/@Drop[Accumulate[Range[60]],8] (* _Harvey P. Dale_, Jul 24 2011 *)

%t LinearRecurrence[{3,-3,1},{0,10,21},60] (* _Harvey P. Dale_, Mar 25 2015 *)

%o (PARI) a(n)=(n-9)*(n+10)/2;

%Y a(n) = A000217(n)-45, n>8.

%Y Cf. A000096, A056121, A079664, A001477.

%K easy,nice,nonn

%O 9,2

%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 21 1999

%E More terms from _Zerinvary Lajos_, Oct 01 2006

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Last modified July 20 14:21 EDT 2019. Contains 325185 sequences. (Running on oeis4.)