A123880
Inverse of number triangle A123878.
Original entry on oeis.org
1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 5, 3, 5, 0, 1, 11, 18, 5, 7, 0, 1, 41, 39, 35, 7, 9, 0, 1, 120, 157, 75, 56, 9, 11, 0, 1, 421, 459, 325, 119, 81, 11, 13, 0, 1, 1381, 1668, 950, 553, 171, 110, 13, 15, 0, 1, 4840
Offset: 0
Number triangle begins
1,
0, 1,
1, 0, 1,
1, 3, 0, 1,
5, 3, 5, 0, 1,
11, 18, 5, 7, 0, 1,
41, 39, 35, 7, 9, 0, 1,
120, 157, 75, 56, 9, 11, 0, 1
A123879
Expansion of (1-2*x+2*x^2-x^3)/(1-3*x+5*x^2-3*x^3+x^4).
Original entry on oeis.org
1, 1, 0, -3, -7, -7, 5, 32, 57, 33, -95, -311, -416, -11, 1209, 2745, 2573, -2368, -12943, -22015, -11007, 40593, 123712, 157165, -14279, -498119, -1075179, -934944, 1090985, 5220257, 8476193, 3535193, -17205600
Offset: 0
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a:=[1,1,0,-3];; for n in [5..40] do a[n]:=3*a[n-1]-5*a[n-2]+3*a[n-3]-a[n-4]; od; a; # G. C. Greubel, Aug 08 2019
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R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-2*x+2*x^2-x^3)/(1-3*x+5*x^2-3*x^3+x^4) )); // G. C. Greubel, Aug 08 2019
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seq(coeff(series((1-2*x+2*x^2-x^3)/(1-3*x+5*x^2-3*x^3+x^4), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Aug 08 2019
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LinearRecurrence[{3,-5,3,-1}, {1,1,0,-3}, 40] (* G. C. Greubel, Aug 08 2019 *)
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my(x='x+O('x^40)); Vec((1-2*x+2*x^2-x^3)/(1-3*x+5*x^2-3*x^3+x^4)) \\ G. C. Greubel, Aug 08 2019
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def A123879_list(prec):
P. = PowerSeriesRing(ZZ, prec)
return P((1-2*x+2*x^2-x^3)/(1-3*x+5*x^2-3*x^3+x^4)).list()
A123879_list(40) # G. C. Greubel, Aug 08 2019
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