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A228873
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a(n) = F(n) * F(n+1) * F(n+2) * F(n+3), the product of four consecutive Fibonacci numbers, A000045.
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3
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6, 30, 240, 1560, 10920, 74256, 510510, 3495030, 23965920, 164237040, 1125770256, 7715953440, 52886430870, 362487682830, 2484530961360, 17029219589256, 116720030923320, 800010932051760, 5483356663145790, 37583485265670630, 257601041359736256
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OFFSET
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1,1
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COMMENTS
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Mohanty and Mohanty prove in Corollary 2.5 that these numbers are Pythagorean. The number a(n) is primitive Pythagorean if F(n) and F(n+1) have opposite parity. Every third number, starting at a(1) = 6, is not primitive Pythagorean.
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LINKS
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Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, The Fibonacci Quarterly, Vol. 28, No. 1 (1990), pp. 31-42.
H.-J. Seiffert, Problem B-1020, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 44, No. 4 (2006), p. 278; Two Fibonacci Identities, Solution to Problem B-1020 by Harris Kwong, ibid., Vol. 45, No. 2 (2007), pp. 182-184.
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FORMULA
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G.f.: -6*x/((x-1)*(x^2+3*x+1)*(x^2-7*x+1)). - Alois P. Heinz, Oct 02 2013
a(n+5) = 5*a(n+4)+15*a(n+3)-15*a(n+2)-5*a(n+1)+a(n). - Robert Israel, Apr 06 2015
a(n) = 3 * Sum_{k=1..n} 2^(n-k)*F(k)^2*F(k+1)*F(k+2) (Seiffert, 2006). - Amiram Eldar, Jan 11 2022
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MAPLE
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seq(mul(combinat:-fibonacci(i), i=n..n+3), n=1..30); # Robert Israel, Apr 06 2015
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MATHEMATICA
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Table[Fibonacci[n] Fibonacci[n+1] Fibonacci[n+2] Fibonacci[n+3], {n, 25}]
CoefficientList[Series[-6/((x - 1) (x^2 + 3 x + 1) (x^2 - 7 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 04 2013 *)
Times@@@Partition[Fibonacci[Range[30]], 4, 1] (* Harvey P. Dale, Dec 23 2013 *)
LinearRecurrence[{5, 15, -15, -5, 1}, {6, 30, 240, 1560, 10920}, 30] (* Harvey P. Dale, Jul 24 2021 *)
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PROG
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(Magma) [Fibonacci(n)*Fibonacci(n+1)*Fibonacci(n+2)*Fibonacci(n+3): n in [1..30]]; // Vincenzo Librandi, Oct 04 2013
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CROSSREFS
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Cf. A000045 (Fibonacci numbers), A228874 (similar sequence for Lucas numbers).
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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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