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A301383
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Expansion of (1 + 3*x - 2*x^2)/(1 - 7*x + 7*x^2 - x^3).
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4
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1, 10, 61, 358, 2089, 12178, 70981, 413710, 2411281, 14053978, 81912589, 477421558, 2782616761, 16218279010, 94527057301, 550944064798, 3211137331489, 18715879924138, 109084142213341, 635788973355910, 3705649697922121, 21598109214176818, 125883005587138789, 733699924308655918
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OFFSET
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0,2
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COMMENTS
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Also, indices y for which 4*A000217(y) + 5 is a square. The next integers k such that k*A000217(y) + 5 is a square for infinitely many y values are 11, 20, 22, 29, 31, ...
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LINKS
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FORMULA
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O.g.f.: (1 + 3*x - 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) = 6*a(n-1) - a(n-2) + 2.
a(n) = (3/4)*((1 + sqrt(2))^(2*n + 1) + (1 - sqrt(2))^(2*n + 1)) - 1/2.
E.g.f.: (3*exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - cosh(x) - sinh(x))/2. - Stefano Spezia, Mar 06 2020
Let T(n) be the n-th triangular number, A000217(n). Then T(a(n)-3) + 2*T(a(n)-2) + 3*T(a(n)-1) + 4*T(a(n)) + 3*T(a(n)+1) + 2*T(a(n)+2) + T(a(n)+3) = (A001653(n) + A001653(n+2))^2. - Charlie Marion, Mar 16 2021
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MAPLE
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f:= gfun:-rectoproc({a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3), a(0)=1, a(1)=10, a(2)=61}, a(n), remember):
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MATHEMATICA
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CoefficientList[Series[(1 + 3 x - 2 x^2)/(1 - 7 x + 7 x^2 - x^3), {x, 0, 30}], x]
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PROG
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(PARI) Vec((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)+O(x^30))
(Maxima) makelist(coeff(taylor((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3), x, 0, n), x, n), n, 0, 30);
(Sage) m=30; L.<x> = PowerSeriesRing(ZZ, m); f=(1+3*x-2*x^2)/(1-7*x+7*x^2-x^3); print(f.coefficients())
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)));
(Julia)
using Nemo
function A301383List(len)
R, x = PowerSeriesRing(ZZ, len+2, "x")
f = divexact(1+3*x-2*x^2, 1-7*x+7*x^2-x^3)
[coeff(f, k) for k in 0:len]
end
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CROSSREFS
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Cf. A000217, A000290, A001652, A002315, A033539, A046090, A053142, A075841, A077444, A106329, A241976.
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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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