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 A274520 a(n) = ((1 + sqrt(7))^n - (1 - sqrt(7))^n)/sqrt(7). 3
 0, 2, 4, 20, 64, 248, 880, 3248, 11776, 43040, 156736, 571712, 2083840, 7597952, 27698944, 100985600, 368164864, 1342243328, 4893475840, 17840411648, 65041678336, 237125826560, 864501723136, 3151758405632, 11490527150080, 41891604733952, 152726372368384 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Number of zeros in substitution system {0 -> 111, 1 -> 1001} at step n from initial string "1" (see example). LINKS Ilya Gutkovskiy, Illustration (substitution system {0 -> 111, 1 -> 1001}) Eric Weisstein's World of Mathematics, Substitution System Index entries for linear recurrences with constant coefficients, signature (2,6) FORMULA O.g.f.: 2*x/(1 - 2*x - 6*x^2). E.g.f.: 2*exp(x)*sinh(sqrt(7)*x)/sqrt(7). Dirichlet g.f.: (PolyLog(s,1+sqrt(7)) - PolyLog(s,1-sqrt(7)))/sqrt(7), where PolyLog(s,x) is the polylogarithm function. a(n) = 2*a(n-1) + 6*a(n-2). a(n) = 2*A083099(n). Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(7) = 1 + A010465. EXAMPLE Evolution from initial string "1": 1 -> 1001 -> 10011111111001 -> 1001111111100110011001100110011001100110011111111001 -> ... Therefore, number of zeros at step n: a(0) = 0; a(1) = 2; a(2) = 4; a(3) = 20, etc. MATHEMATICA LinearRecurrence[{2, 6}, {0, 2}, 27] PROG (PARI) a(n)=([0, 1; 6, 2]^n*[0; 2])[1, 1] \\ Charles R Greathouse IV, Jul 26 2016 CROSSREFS Cf. A010465, A083099. Sequence in context: A137697 A192380 A009336 * A238229 A192377 A286344 Adjacent sequences:  A274517 A274518 A274519 * A274521 A274522 A274523 KEYWORD nonn,easy AUTHOR Ilya Gutkovskiy, Jun 26 2016 STATUS approved

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Last modified October 16 21:57 EDT 2018. Contains 316275 sequences. (Running on oeis4.)