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A301383
Expansion of (1 + 3*x - 2*x^2)/(1 - 7*x + 7*x^2 - x^3).
4
1, 10, 61, 358, 2089, 12178, 70981, 413710, 2411281, 14053978, 81912589, 477421558, 2782616761, 16218279010, 94527057301, 550944064798, 3211137331489, 18715879924138, 109084142213341, 635788973355910, 3705649697922121, 21598109214176818, 125883005587138789, 733699924308655918
OFFSET
0,2
COMMENTS
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+2). The corresponding x values are listed in A075841.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+1) are in A002315, and A075870 gives the x values.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y) are in A046090, and A001653 gives the x values.
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, ...
First differences are in A106329.
FORMULA
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.
a(n) = A033539(2*n+2) = A241976(n+1) + 1 = 3*A001652(n) + 1 = 3*A046090(n) - 2.
a(n) = A053142(n+1) + 3*A053142(n) - 2*A053142(n-1), n>0.
2*a(n) = 3*A002315(n) - 1.
4*a(n) = 3*A077444(n+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
MAPLE
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):
map(f, [$0..50]); # Robert Israel, Mar 21 2018
MATHEMATICA
CoefficientList[Series[(1 + 3 x - 2 x^2)/(1 - 7 x + 7 x^2 - x^3), {x, 0, 30}], x]
PROG
(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
A301383List(23) |> println # Peter Luschny, Mar 21 2018
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Mar 20 2018
STATUS
approved