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A220360
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a(n) = Fibonacci(n-1) * Fibonacci(n+1) * Fibonacci(n+2).
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2
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0, 6, 15, 80, 312, 1365, 5712, 24310, 102795, 435744, 1845360, 7817849, 33115680, 140282310, 594242103, 2517255280, 10663255848, 45170290605, 191344398960, 810547917686, 3433536019155, 14544692076096, 61612304191200, 260993909055025, 1105587940064832
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OFFSET
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1,2
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COMMENTS
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An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons.
Type A have sides A066259(n+1), a(n+1), A066259(n+1), a(n+1), A066259(n+1), and opposite diagonals A056570(n+2), A056570(n+2), A220361(n+2), A056570(n+2), A056570(n+2), for n=1,2,...
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, D20.
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LINKS
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FORMULA
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G.f.: (6*x - 3*x^2 - x^3)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4); a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4). [Ron Knott, Jun 27 2013]
Sum {n >= 2} 1/a(n) = 1/4. - Peter Bala, Nov 30 2013
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MATHEMATICA
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Table[Fibonacci[n - 1]*Fibonacci[n + 1]*Fibonacci[n + 2], {n, 30}] (* T. D. Noe, Dec 13 2012 *)
#[[1]]#[[3]]#[[4]]&/@Partition[Fibonacci[Range[0, 30]], 4, 1] (* Harvey P. Dale, Apr 08 2022 *)
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PROG
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(PARI) a(n) = fibonacci(n-1) * fibonacci(n+1) * fibonacci(n+2); \\ Michel Marcus, Mar 26 2016
(PARI) x='x+O('x^99); concat(0, Vec((6*x-3*x^2-x^3)/(1-3*x-6*x^2+3*x^3+x^4))) \\ Altug Alkan, Mar 26 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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