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a(n) = 3 + binomial(n-1,2).
3

%I #45 Dec 13 2022 03:01:52

%S 3,3,4,6,9,13,18,24,31,39,48,58,69,81,94,108,123,139,156,174,193,213,

%T 234,256,279,303,328,354,381,409,438,468,499,531,564,598,633,669,706,

%U 744,783,823,864,906,949,993,1038,1084,1131,1179,1228,1278,1329,1381

%N a(n) = 3 + binomial(n-1,2).

%C a(1)=3; then add 0 to the first number, then 1,2,3,4,... and so on.

%H Muniru A Asiru, <a href="/A152949/b152949.txt">Table of n, a(n) for n = 1..5000</a>

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

%F a(n) = a(n-1) + n - 2 (with a(1)=3). - _Vincenzo Librandi_, Nov 27 2010

%F G.f.: x*(3-6*x+4*x^2)/(1-x)^3. - _Nikita Gogin_, Jul 24 2013

%F a(n) = A016028(n+1) for n >= 2. - _Georg Fischer_, Oct 28 2018

%F Sum_{n>=1} 1/a(n) = 1/3 + 2*Pi*tanh(sqrt(23)*Pi/2)/sqrt(23). - _Amiram Eldar_, Dec 13 2022

%p seq(coeff(series(x*(4*x^2-6*x+3)/(1-x)^3,x,n+1), x, n), n = 1 .. 55); # _Muniru A Asiru_, Oct 28 2018

%t s=3;lst={3};Do[s+=n;AppendTo[lst,s],{n,0,5!}];lst

%t Table[Binomial[n-1,2],{n,60}]+3 (* _Harvey P. Dale_, Feb 27 2013 *)

%o (Sage) [3+binomial(n,2) for n in range(0, 54)] # _Zerinvary Lajos_, Mar 12 2009

%o (PARI) Vec( x*(3-6*x+4*x^2)/(1-x)^3 + O(x^66) ) \\ _Joerg Arndt_, Jul 24 2013

%o (GAP) List([1..55],n->3+Binomial(n-1,2)); # _Muniru A Asiru_, Oct 28 2018

%Y Cf. A000217, A016028, A152947, A000124, A152948, A152950.

%K nonn,easy

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, Dec 15 2008