A094705 Convolution of Jacobsthal(n) and 3^n.
0, 1, 4, 15, 50, 161, 504, 1555, 4750, 14421, 43604, 131495, 395850, 1190281, 3576304, 10739835, 32241350, 96767741, 290390604, 871346575, 2614389250, 7843866801, 23532998504, 70601791715, 211810967550, 635444087461, 1906354632004, 5719108635255, 17157415384250
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (4,-1,-6).
Programs
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Magma
[(3^(n+2) -2^(n+3) -(-1)^n)/12: n in [0..50]]; // G. C. Greubel, Jul 21 2022
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Mathematica
LinearRecurrence[{4,-1,-6},{0,1,4},30] (* Harvey P. Dale, Apr 02 2017 *) Jacob0[n_] := (2^n - (-1)^n)/3; a[n_] := First@ListConvolve[Table[Jacob0[i], {i, 0, n}], 3^Range[0, n]]; Table[a[x], {x, 0, 10}] (* Robert P. P. McKone, Nov 28 2020 *)
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PARI
concat(0, Vec(x/((1+x)*(1-2*x)*(1-3*x)) + O(x^50))) \\ Michel Marcus, Sep 13 2014
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SageMath
[(3^(n+1) - lucas_number1(n+3, 1, -2))/4 for n in (0..50)] # G. C. Greubel, Jul 21 2022
Formula
G.f.: x/((1+x)*(1-2*x)*(1-3*x)).
a(n) = (3^(n+2) - 2^(n+3) - (-1)^n)/12.
a(n) = 4*a(n-1) -a(n-2) -6*a(n-3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1)*2^(n-k-1)*(3/2)^k. - Paul Barry, Oct 25 2004
a(n) = (3^(n+1) - A001045(n+3))/4. - G. C. Greubel, Jul 21 2022
Comments