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a(n) = 3 + n*(n-1)/2.
16

%I #43 Dec 13 2022 02:15:16

%S 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 744,783,823,864,906,949,993,1038,1084,1131,1179,1228,1278,1329,1381,1434

%N a(n) = 3 + n*(n-1)/2.

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

%C Numbers m such that 8m-23 is a square. - _Bruce J. Nicholson_, Jul 25 2017

%H Michael De Vlieger, <a href="/A152950/b152950.txt">Table of n, a(n) for n = 1..10000</a>

%H Ângela Mestre and José Agapito, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Mestre/mestre2.html">Square Matrices Generated by Sequences of Riordan Arrays</a>, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.

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

%F a(n) = A152949(n+1) = 3+A000217(n-1). - _R. J. Mathar_, Jan 03 2009

%F a(n) = 3+C(n,2), n>=1. - _Zerinvary Lajos_, Mar 12 2009

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

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

%p A152950:=n->3 + n*(n-1)/2; seq(A152950(n), n=1..100); # _Wesley Ivan Hurt_, Jan 28 2014

%t s=3;lst={3};Do[s+=n;AppendTo[lst,s],{n,1,5!}];lst

%t Table[3 + n*(n-1)/2, {n, 100}] (* _Wesley Ivan Hurt_, Jan 28 2014 *)

%o (Sage) [3+binomial(n,2) for n in range(1, 55)] # _Zerinvary Lajos_, Mar 12 2009

%o (PARI) a(n)=3+n*(n-1)/2 \\ _Charles R Greathouse IV_, Oct 07 2015

%o (Magma) [3+n*(n-1)/2 : n in [1..50]]; // _Wesley Ivan Hurt_, Mar 25 2020

%Y Cf. A000217, A152947, A000124, A152948, A152949.

%K nonn,easy

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, Dec 15 2008