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 A295683 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 2, a(1) = 1, a(2) = 0, a(3) = 1. 1
 2, 1, 0, 1, 4, 5, 6, 11, 20, 31, 48, 79, 130, 209, 336, 545, 884, 1429, 2310, 3739, 6052, 9791, 15840, 25631, 41474, 67105, 108576, 175681, 284260, 459941, 744198, 1204139, 1948340, 3152479, 5100816, 8253295, 13354114, 21607409, 34961520, 56568929, 91530452 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth-rate of the Fibonacci numbers (A000045). LINKS Clark Kimberling, Table of n, a(n) for n = 0..2000 Index entries for linear recurrences with constant coefficients, signature (1,0,1,1). FORMULA G.f.: (2 - x - x^2 - x^3)/(1 - x - x^3 - x^4). a(n) = (2/5) * A000045(n) + (4/5) * A000045(n-1) + (6/5) * A056594(n) + (3/5) * A056594(n-1) for n >= 1. - Robert Israel, Apr 03 2019 MAPLE f:= gfun:-rectoproc(a(n) = a(n-1) + a(n-3) + a(n-4), a(0) = 2, a(1) = 1, a(2) = 0, a(3) = 1}, a(n), remember): map(f, [\$0..100]); # Robert Israel, Apr 03 2019 MATHEMATICA LinearRecurrence[{1, 0, 1, 1}, {2, 1, 0, 1}, 45] PROG (PARI) my(x='x+O('x^45)); Vec((-2 + x + x^2 + x^3)/(-1 + x + x^3 + x^4)) \\ Georg Fischer, Apr 03 2019 (MAGMA) I:=[2, 1, 0, 1]; [n le 4 select I[n] else Self(n-1) +Self(n-3) +Self(n-4): n in [1..45]]; // G. C. Greubel, Apr 03 2019 (Sage) ((2-x-x^2-x^3)/(1-x-x^3-x^4)).series(x, 45).coefficients(x, sparse=False) # G. C. Greubel, Apr 03 2019 CROSSREFS Cf. A001622, A000045, A056594. Sequence in context: A262495 A336703 A323174 * A165519 A266972 A266493 Adjacent sequences:  A295680 A295681 A295682 * A295684 A295685 A295686 KEYWORD nonn,easy AUTHOR Clark Kimberling, Nov 29 2017 EXTENSIONS a(0) = 2 amended by Georg Fischer, Apr 03 2019 STATUS approved

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Last modified August 4 09:03 EDT 2020. Contains 336201 sequences. (Running on oeis4.)