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A056014
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a(n) = (Fibonacci(2n-1) - Fibonacci(n+1))/2.
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7
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0, 0, 0, 1, 4, 13, 38, 106, 288, 771, 2046, 5401, 14212, 37324, 97904, 256621, 672336, 1760997, 4611642, 12075526, 31617520, 82781215, 216732890, 567428401, 1485570024, 3889310328, 10182407328, 26657986681, 69791674108
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OFFSET
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0,5
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COMMENTS
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With a(0)=0, a(1)=1, a(2)=1, a(3)=2, this recurrence produces a(n)=A000045(n) (Fibonacci numbers).
Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 4. - Herbert Kociemba, Jun 16 2004
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REFERENCES
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H. Eriksson and M. Jonsson, Level sizes of the Bulgarian solitaire game tree, Fib. Q., 35:3 (2017), 243-251.
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LINKS
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FORMULA
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a(n) = 4*a(n-1) - 3*a(n-2) - 2*a(n-3) + a(n-4), a(0)=a(1)=a(2)=0, a(3)=1.
Convolution of Fibonacci numbers F(n) with F(2n). - Benoit Cloitre, Jun 07 2004
Binomial transform of x^3/(1-3x^2+x^4), or (essentially) F(2n) with interpolated zeros. a(n)=sum{k=0..n, binomial(n, k)((3/2-sqrt(5)/2)^(k/2)((sqrt(5)/20+1/4)(-1)^k-sqrt(5)/20-1/4)+ (sqrt(5)/2+3/2)^(k/2)((sqrt(5)/20-1/4)(-1)^k-sqrt(5)/20+1/4))}. - Paul Barry, Jul 26 2004
Convolution of the powers of 2 (A000079) with the number of positive rational knots with 2n+1 crossings (A051450), with three leading zeros. - Graeme McRae, Jun 28 2006
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MATHEMATICA
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Table[(Fibonacci[2n-1]-Fibonacci[n+1])/2, {n, 0, 40}] (* Harvey P. Dale, Mar 24 2011 *)
LinearRecurrence[{4, -3, -2, 1}, {0, 0, 0, 1}, 40] (* Vincenzo Librandi, Jun 23 2012 *)
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PROG
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(PARI) a(n)=(fibonacci(2*n-1)-fibonacci(n+1))/2
(Magma) I:=[0, 0, 0, 1]; [n le 4 select I[n] else 4*Self(n-1)-3*Self(n-2)-2*Self(n-3)+Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 23 2012
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Asher Auel (asher.auel(AT)reed.edu), Jun 06 2000
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STATUS
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approved
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