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 A320534 a(n) = ((1 + sqrt(4*n^2 + 1))^n + (1 - sqrt(4*n^2 + 1))^n)/2^n. 3
 2, 1, 9, 28, 577, 3251, 105193, 857501, 37831169, 403541596, 22550351001, 297238464799, 20106709638337, 315569447182601, 25059144736026633, 456277507970965876, 41600491470425952257, 862007599260004863571, 88733427132980061934777, 2061632980592377284802309 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS a(0) = 2 assuming 0^0 = 1, or using the limit for n -> 0 (assuming n is a real variable); the same value for a(0) arises from other formulae for this sequence. LINKS Eric Weisstein's World of Mathematics, Lucas Polynomial Wikipedia, Fibonacci polynomials FORMULA a(n) = 2^(1 - n) * Sum_{k=0..floor(n/2)} binomial(n, 2*k)*(4*n^2 + 1)^k. a(n) = 2^(1 - n) * hypergeom([(1 - n)/2, -n/2], [1/2], 4*n^2 + 1). For n > 0, a(n) = n^n * L_n(1/n), where L_n(x) is the Lucas polynomial. For n > 0, a(n) = 2*(-i*n)^n*cos(n*arcsin(sqrt(4*n^2+1)/(2*n))). - Peter Luschny, Oct 14 2018 MATHEMATICA Table[2^(1 - n) Hypergeometric2F1[(1 - n)/2, -n/2, 1/2, 4 n^2 + 1], {n, 0, 19}] (* or *) a[0] = Limit[n^n LucasL[n, 1/n], n -> 0]; (* a[0] = 2 *) a[n_] := a[n] = n^n LucasL[n, 1/n]; Table[a[n], {n, 0, 19}] CROSSREFS Cf. A084844, A320519. Sequence in context: A144244 A079582 A259872 * A012892 A013071 A155756 Adjacent sequences:  A320531 A320532 A320533 * A320535 A320536 A320537 KEYWORD nonn AUTHOR Vladimir Reshetnikov, Oct 14 2018 STATUS approved

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Last modified January 26 17:27 EST 2020. Contains 331280 sequences. (Running on oeis4.)