%I #25 Apr 25 2024 09:30:02
%S 1,18,525,17340,586177,19896030,675781821,22956120408,779829016225,
%T 26491211221770,899921240562957,30570830315362260,1038508305678375841,
%U 35278711540581704598,1198437683944896688125,40711602541832856049200,1382996048733983114022337
%N If b(n) is A011900(n) and c(n) is A001109(n), then a(n) = b(n)*c(n) = b(n) + (b(n)+1) + (b(n)+2) + ... + c(n).
%D Mario Velucchi "From the desk of ... Mario Velucchi" in 'Mathematics and Informatics quarterly' volume 7 - 2/1997, p. 81.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (41,-246,246,-41,1).
%F From _R. J. Mathar_, Apr 15 2010: (Start)
%F G.f.: x*(-1+23*x-33*x^2+3*x^3)/((x-1)*(x^2-34*x+1)*(1-6*x+x^2)).
%F a(n) = 41*a(n-1) -246*a(n-2) +246*a(n-3) -41*a(n-4) +a(n-5). (End)
%F Lim_{n -> infinity} a(n)/a(n-1) = A156164. - _César Aguilera_, Jul 17 2020
%e a(3) = 525 = 15*35 = 15 + 16 + ... + 35.
%p A011900 := proc(n) coeftayl( (1-4*x+x^2)/((1-x)*(1-6*x+x^2)),x=0,n) ; end proc: A001109 := proc(n) coeftayl( x/(1-6*x+x^2),x=0,n) ; end proc: A011906 := proc(n) A001109(n)*A011900(n-1) ; end proc: seq(A011906(n),n=1..30) ; # _R. J. Mathar_, Apr 15 2010
%t LinearRecurrence[{41, -246, 246, -41, 1}, {1, 18, 525, 17340, 586177}, 20] (* _Paul Cleary_, Dec 05 2015 *)
%t CoefficientList[Series[(-1 + 23*x - 33*x^2 + 3*x^3)/((x - 1)*(x^2 - 34*x + 1)*(1 - 6*x + x^2)), {x, 0, 20}], x] (* _Wesley Ivan Hurt_, Sep 16 2017 *)
%Y Cf. A001109, A011900.
%K nonn,easy
%O 1,2
%A Mario Velucchi (mathchess(AT)velucchi.it)
%E More terms from _R. J. Mathar_, Apr 15 2010