%I #30 Dec 25 2024 21:00:24
%S 6,55,153,300,496,741,1035,1378,1770,2211,2701,3240,3828,4465,5151,
%T 5886,6670,7503,8385,9316,10296,11325,12403,13530,14706,15931,17205,
%U 18528,19900,21321,22791,24310,25878,27495,29161,30876,32640,34453,36315,38226,40186,42195
%N a(n) = 49*(n*(n+1)/2) + 6.
%C "If n is a triangular number, then so are 9*n+1, 25*n+3 and 49*n+6. (Euler, 1775)." Burton, p. 17.
%D D. M. Burton, Elementary Number Theory, Allyn and Bacon, Inc. Boston, MA, 1976, p. 17.
%H Harry J. Smith, <a href="/A061792/b061792.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=6, a(1)=55, a(2)=153. - _Harvey P. Dale_, Apr 11 2012
%F From _Elmo R. Oliveira_, Oct 23 2024: (Start)
%F G.f.: (6 + 37*x + 6*x^2)/(1 - x)^3.
%F E.g.f.: (6 + 49*x + 49*x^2/2)*exp(x).
%F a(n) = 6 + 49*A000217(n) = A177071(n)/2. (End)
%t 49*Accumulate[Range[0,40]]+6 (* or *) LinearRecurrence[{3,-3,1},{6,55,153},40] (* _Harvey P. Dale_, Apr 11 2012 *)
%o (PARI) a(n) = 49*(n*(n + 1)/2) + 6
%Y Cf. A000217, A177071.
%K easy,nonn
%O 0,1
%A _Jason Earls_, Jun 22 2001