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A290998
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p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^4.
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6
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0, 0, 1, 4, 10, 21, 43, 92, 205, 462, 1035, 2301, 5099, 11303, 25088, 55728, 123800, 274969, 610628, 1355970, 3011157, 6686979, 14850196, 32978725, 73237462, 162641499, 361184653, 802098203, 1781254927, 3955712256, 8784625824, 19508406192, 43323176177
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OFFSET
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0,4
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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).
See A291000 for a guide to related sequences.
For n >= 1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 3 blocks (i.e., subintervals) from each interval. For example, for n=9, a(8)=205 since the number of ways to split [9] into intervals and then select 3 blocks from each interval is C(9,3) + C(6,3)*C(3,3) + C(5,3)*C(4,3) + C(4,3)*C(5,3) + C(3,3)*C(6,3) + C(3,3)*C(3,3)*C(3,3) for a total of 205 ways. - Enrique Navarrete, Dec 23 2023
a(n-1) is also the number of compositions of n using parts of size at least 3 where there are binomial(i,3) types of i, n>=1, i>=3 (see example). - Enrique Navarrete, Dec 25 2023
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LINKS
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FORMULA
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a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - a(n-4) for n >= 5.
G.f.: x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4). - Colin Barker, Aug 22 2017
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EXAMPLE
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Since there are binomial(3,3) = 1 type of 3, binomial(4,3) = 4 types of 4, binomial(5,3) = 10 types of 5, binomial(6,3) = 20 types of 6, and binomial(9,3) = 84 types of 9, we can write 9 in the following ways:
9 in 84 ways;
6+3 in 20 ways;
5+4 in 40 ways;
4+5 in 40 ways;
3+6 in 20 ways;
3+3+3 in 1 way, for a total of 205 ways. (End)
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MATHEMATICA
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z = 60; s = x/(1 - x); p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* this sequence *)
LinearRecurrence[{4, -6, 5, -1}, {0, 0, 1, 4}, 41] (* G. C. Greubel, Apr 25 2023 *)
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PROG
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(PARI) concat(vector(2), Vec(x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4) + O(x^50))) \\ Colin Barker, Aug 22 2017
(Magma) I:=[0, 0, 1, 4]; [n le 4 select I[n] else 4*Self(n-1) -6*Self(n-2) +5*Self(n-3) -Self(n-4): n in [1..41]]; // G. C. Greubel, Apr 25 2023
(SageMath)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( x^2/(1-4*x+6*x^2-5*x^3+x^4) ).list()
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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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