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A292397
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p-INVERT of the tribonacci numbers (A000073(k), k>=2), where p(S) = 1 - S - S^2 - S^3.
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1
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1, 3, 10, 33, 108, 352, 1144, 3714, 12050, 39084, 126752, 411041, 1332923, 4322363, 14016392, 45451793, 147389276, 477948252, 1549872500, 5025868667, 16297700769, 52849583211, 171378684824, 555740504324, 1802134907175, 5843896942499, 18950374573538
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OFFSET
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0,2
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COMMENTS
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Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
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LINKS
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FORMULA
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G.f.: -(((1 + x + x^2) (1 - 2 x + x^3 + x^4))/(-1 + 4 x - x^2 - 3 x^3 - 7 x^4 + 2 x^5 + 6 x^6 + 7 x^7 + 3 x^8 + x^9)).
a(n) = 4*a(n-1) - a(n-2) - 3*a(n-3) - 7*a(n-4) + 2*a(n-5) + 6*a(n-6) + 7*a(n-7) + 3*a(n-8) + a(n-9) for n >= 10.
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MATHEMATICA
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z = 60; s = x/(1 - x - x^2 - x^3); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000073 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292397 *)
LinearRecurrence[{4, -1, -3, -7, 2, 6, 7, 3, 1}, {1, 3, 10, 33, 108, 352, 1144, 3714, 12050}, 30] (* Vincenzo Librandi, Oct 13 2017 *)
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PROG
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(PARI) x='x+O('x^99); Vec(((1+x+x^2)*(1-2*x+x^3+x^4))/(1-4*x+x^2+3*x^3+7*x^4-2*x^5-6*x^6-7*x^7-3*x^8-x^9)) \\ Altug Alkan, Oct 04 2017
(Magma) I:=[1, 3, 10, 33, 108, 352, 1144, 3714, 12050]; [n le 9 select I[n] else 4*Self(n-1)-Self(n-2)-3*Self(n-3)-7*Self(n-4)+2*Self(n-5)+6*Self(n-6)+7*Self(n-7)+3*Self(n-8)+Self(n-9): n in [1..30]]; // Vincenzo Librandi, Oct 13 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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