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A229339
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GCD of all sums of n consecutive Lucas numbers.
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3
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1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29, 2, 105, 1, 76, 1, 275, 2, 199, 1, 720, 1, 521, 2, 1885, 1, 1364, 1, 4935, 2, 3571, 1, 12920, 1, 9349, 2, 33825, 1, 24476, 1, 88555, 2, 64079, 1, 231840, 1, 167761, 2, 606965, 1, 439204, 1, 1589055, 2, 1149851, 1, 4160200, 1, 3010349, 2
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OFFSET
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1,3
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COMMENTS
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The sum of two consecutive Lucas number is the sum of four consecutive Fibonacci numbers, which is verified easily enough with the identity L(n) = F(n - 1) + F(n + 1). Therefore a(1) = a(2) = A210209(4).
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).
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FORMULA
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a(n) = 3*a(n-4) + a(n-6) - a(n-8) - 3*a(n-10) + a(n-14) for n > 14. - Giovanni Resta, Oct 04 2013
G.f.: x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1). - Colin Barker, Nov 09 2014
a(n) = gcd(L(n+1)-1, L(n+2)-3).
a(n) = Lcm_{A106291(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)
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EXAMPLE
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a(3) = 2 because any sum of three consecutive Lucas numbers is an even number.
a(4) = 5 because all sums of four consecutive Lucas numbers are divisible by 5.
a(5) = 1 because some sums of five consecutive Lucas numbers are coprime.
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MATHEMATICA
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a[n_] := a[n] = If[n <= 14, {1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29}[[n]], 3*a[n - 4] + a[n - 6] - a[n - 8] - 3*a[n - 10] + a[n - 14]]; Array[a, 64] (* Giovanni Resta, Oct 04 2013 *)
CoefficientList[Series[(x^12 - x^11 + 2 x^10 - 5 x^9 - 2 x^8 - x^7 - 6 x^6 + x^5 - 2 x^4 + 5 x^3 + 2 x^2 + x + 1) / (-x^14 + 3 x^10 + x^8 - x^6 - 3 x^4 + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 09 2014 *)
LinearRecurrence[{0, 0, 0, 3, 0, 1, 0, -1, 0, -3, 0, 0, 0, 1}, {1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29}, 70] (* Harvey P. Dale, Jul 21 2021 *)
Table[GCD[LucasL[n + 1] - 2, LucasL[n] + 1], {n, 0, 50}] (* Horst H. Manninger, Dec 25 2021 *)
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PROG
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(PARI) Vec(x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1) + O(x^100)) \\ Colin Barker, Nov 09 2014
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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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STATUS
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approved
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