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A323232
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a(n) = 2^n*J(n, 1/2) where J(n, x) are the Jacobsthal polynomials as defined in A322942.
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3
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1, 3, 9, 51, 225, 1083, 5049, 23811, 111825, 525963, 2472489, 11625171, 54655425, 256967643, 1208146329, 5680180131, 26705711025, 125558574123, 590321410569, 2775432824691, 13048869758625, 61350071873403, 288441173689209, 1356124096054851, 6375901677678225
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OFFSET
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0,2
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COMMENTS
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Is it true that p prime and p not 2 or 5 implies that a(p) is squarefree?
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LINKS
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FORMULA
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a(n) = 3*a(n-1) + 8*a(n-2) for n >= 3.
a(n) is an odd integer and 3 | a(n) if n > 0.
a(n) = Sum_{k=0..n} 2^(n - k)*A322942(n, k).
a(n) = [x^n] (8*x^2 - 1)/(8*x^2 + 3*x - 1).
Let s = sqrt(41), u = -1/(s+3) and v = 1/(s-3); then
a(n) = (3/s)*16^n*(v^n - u^n) for n >= 1.
a(n) = n! [x^n](1 + (6*exp(3*x/2)*sinh(s*x/2))/s).
a(n) = n! [x^n](1 + (3/s)*(exp((3 + s)*x/2) - exp((3 - s)*x/2))).
a(n)/a(n+1) -> 2/(sqrt(41) + 3) = (sqrt(41) - 3)/16 for n -> oo.
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EXAMPLE
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The first few prime factorizations of a(n):
1| 3;
2| 3^2;
3| 3 * 17;
4| 3^2 * 5^2;
5| 3 * 19^2;
6| 3^3 * 11 * 17;
7| 3 * 7937;
8| 3^2 * 5^2 * 7 * 71;
9| 3 * 17 * 10313;
10| 3^2 * 19^2 * 761;
11| 3 * 3875057;
12| 3^3 * 5^2 * 11 * 17 * 433;
13| 3 * 85655881;
14| 3^2 * 13 * 1301 * 7937;
15| 3 * 17 * 19^2 * 308521;
16| 3^2 * 5^2 * 7 * 71 * 79 * 3023;
17| 3 * 67 * 624669523;
18| 3^4 * 11 * 17 * 3779 * 10313;
19| 3 * 419 * 2207981563;
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MAPLE
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a := proc(n) option remember:
if n < 3 then return [1, 3, 9][n+1] fi;
8*a(n-2) + 3*a(n-1) end:
seq(a(n), n=0..24);
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MATHEMATICA
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LinearRecurrence[{3, 8}, {1, 3, 9}, 25]
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PROG
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(Sage)
def a():
yield 1
yield 3
c = 3; b = 9
while True:
yield b
a = (b << 2) + (c << 3) - b
c = b
b = a
(Magma) [1] cat [n le 2 select 3^n else 3*Self(n-1) +8*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 27 2021
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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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