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A022096
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Fibonacci sequence beginning 1, 6.
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19
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1, 6, 7, 13, 20, 33, 53, 86, 139, 225, 364, 589, 953, 1542, 2495, 4037, 6532, 10569, 17101, 27670, 44771, 72441, 117212, 189653, 306865, 496518, 803383, 1299901, 2103284, 3403185, 5506469, 8909654, 14416123, 23325777, 37741900, 61067677, 98809577, 159877254
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OFFSET
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0,2
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COMMENTS
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a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(6;n-1-k,k), n>=1, with a(-1)=5. These are the sums of the SW-NE diagonals in P(6;n,k), the (6,1) Pascal triangle A093563. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of SW-NE diagonals in (1,5)-Pascal triangle A096940.
Subsequence of primes: 7, 13, 53, 139, 953, 44771, 189653, 1494692464747, ... - R. J. Mathar, Aug 09 2012
a(n) is the sum of seven consecutive Fibonacci numbers. a(n) = F(n-4) + F(n-3) + F(n-2) + F(n-1) + F(n) + F(n+1) + F(n+2), where F(n)=A000045(n), extended so that F(-1)=1, F(-2)=-1, F(-3)=2, and F(-4)=-3. - Graeme McRae, Apr 24 2014
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=6.
G.f.: (1+5*x)/(1-x-x^2).
a(n) = 5*Fibonacci(n+2) - 4*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010
a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-11 + sqrt(5)) + (1 + sqrt(5))^n*(11 + sqrt(5))))/sqrt(5). - Herbert Kociemba, Dec 18 2011
a(n) = Fibonacci(n+3) - Fibonacci(n-4). - Greg Dresden and Sam Neale, Mar 08 2022
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MATHEMATICA
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CoefficientList[Series[(1 + 5 x)/(1 - x - x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 25 2014 *)
LinearRecurrence[{1, 1}, {1, 6}, 40] (* Harvey P. Dale, Aug 07 2023 *)
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PROG
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(Magma) a0:=1; a1:=6; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
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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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EXTENSIONS
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STATUS
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approved
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