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 A011906 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). 2
 1, 18, 525, 17340, 586177, 19896030, 675781821, 22956120408, 779829016225, 26491211221770, 899921240562957, 30570830315362260, 1038508305678375841, 35278711540581704598, 1198437683944896688125, 40711602541832856049200, 1382996048733983114022337 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 REFERENCES Mario Velucchi "From the desk of ... Mario Velucchi" in 'Mathematics and Informatics quarterly' volume 7 - 2/1997, p. 81. LINKS Table of n, a(n) for n=1..17. Index entries for linear recurrences with constant coefficients, signature (41,-246,246,-41,1). FORMULA From R. J. Mathar, Apr 15 2010: (Start) G.f.: x*(-1+23*x-33*x^2+3*x^3)/((x-1)*(x^2-34*x+1)*(1-6*x+x^2)). a(n) = 41*a(n-1) -246*a(n-2) +246*a(n-3) -41*a(n-4) +a(n-5). (End) Lim_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jul 17 2020 EXAMPLE a(3) = 525 = 15*35 = 15 + 16 + ... + 35. MAPLE 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 MATHEMATICA LinearRecurrence[{41, -246, 246, -41, 1}, {1, 18, 525, 17340, 586177}, 20] (* Paul Cleary, Dec 05 2015 *) 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 *) CROSSREFS Cf. A001109, A011900. Sequence in context: A180791 A126276 A035277 * A255859 A334179 A183498 Adjacent sequences: A011903 A011904 A011905 * A011907 A011908 A011909 KEYWORD nonn,easy AUTHOR Mario Velucchi (mathchess(AT)velucchi.it) EXTENSIONS More terms from R. J. Mathar, Apr 15 2010 STATUS approved

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