A152740 11 times triangular numbers.
0, 11, 33, 66, 110, 165, 231, 308, 396, 495, 605, 726, 858, 1001, 1155, 1320, 1496, 1683, 1881, 2090, 2310, 2541, 2783, 3036, 3300, 3575, 3861, 4158, 4466, 4785, 5115, 5456, 5808, 6171, 6545, 6930, 7326, 7733, 8151, 8580, 9020, 9471, 9933, 10406, 10890, 11385, 11891
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..5000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Magma
[11*n*(n+1)/2 : n in [0..60]]; // Wesley Ivan Hurt, Dec 22 2015
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Maple
A152740:=n->11*n*(n+1)/2: seq(A152740(n), n=0..60); # Wesley Ivan Hurt, Dec 22 2015
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Mathematica
Table[11*n*(n - 1)/2, {n, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 06 2011 *) LinearRecurrence[{3, -3, 1}, {0, 11, 33}, 100] (* G. C. Greubel, Dec 22 2015 *)
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PARI
my(x='x+O('x^100)); concat(0, Vec(11*x/(1-x)^3)) \\ Altug Alkan, Dec 23 2015
Formula
a(n) = 11*n*(n+1)/2 = 11*A000217(n).
a(n) = a(n-1) + 11*n with n > 0, a(0)=0. - Vincenzo Librandi, Nov 26 2010
a(n) = A069125(n+1) - 1. - Omar E. Pol, Oct 03 2011
From Philippe Deléham, Mar 27 2013: (Start)
G.f.: 11*x/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2, a(0)=0, a(1)=11, a(2)=33.
a(n) = A218530(11*n+10).
a(n) = Sum_{i=5*n..6*n} i. - Wesley Ivan Hurt, Dec 22 2015
From Amiram Eldar, Feb 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2/11.
Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/11.
Product_{n>=1} (1 - 1/a(n)) = -(11/(2*Pi))*cos(sqrt(19/11)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (11/(2*Pi))*cos(sqrt(3/11)*Pi/2). (End)
E.g.f.: 11*exp(x)*x*(2 + x)/2. - Elmo R. Oliveira, Dec 25 2024
Comments