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A290998 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^4.

%I A290998 #30 Jan 22 2024 05:47:53
%S A290998 0,0,1,4,10,21,43,92,205,462,1035,2301,5099,11303,25088,55728,123800,
%T A290998 274969,610628,1355970,3011157,6686979,14850196,32978725,73237462,
%U A290998 162641499,361184653,802098203,1781254927,3955712256,8784625824,19508406192,43323176177
%N A290998 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^4.
%C A290998 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).
%C A290998 See A291000 for a guide to related sequences.
%C A290998 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
%C A290998 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
%H A290998 Clark Kimberling, <a href="/A290998/b290998.txt">Table of n, a(n) for n = 0..1000</a>
%H A290998 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,5,-1).
%F A290998 a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - a(n-4) for n >= 5.
%F A290998 G.f.: x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4). - _Colin Barker_, Aug 22 2017
%F A290998 G.f.: 1/(x*(1-Sum_{k>=3} binomial(k,3)*x^k)) - 1/x. - _Enrique Navarrete_, Dec 26 2023
%e A290998 From _Enrique Navarrete_, Dec 25 2023: (Start)
%e A290998 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:
%e A290998  9 in 84 ways;
%e A290998  6+3 in 20 ways;
%e A290998  5+4 in 40 ways;
%e A290998  4+5 in 40 ways;
%e A290998  3+6 in 20 ways;
%e A290998  3+3+3 in 1 way, for a total of 205 ways. (End)
%t A290998 z = 60; s = x/(1 - x); p = 1 - s^3 - s^4;
%t A290998 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A290998 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* this sequence *)
%t A290998 LinearRecurrence[{4,-6,5,-1}, {0,0,1,4}, 41] (* _G. C. Greubel_, Apr 25 2023 *)
%o A290998 (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
%o A290998 (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
%o A290998 (SageMath)
%o A290998 def A290998_list(prec):
%o A290998     P.<x> = PowerSeriesRing(ZZ, prec)
%o A290998     return P( x^2/(1-4*x+6*x^2-5*x^3+x^4) ).list()
%o A290998 A290998_list(40) # _G. C. Greubel_, Apr 25 2023
%Y A290998 Cf. A000012, A033453, A289780, A290997, A291000.
%K A290998 nonn,easy
%O A290998 0,4
%A A290998 _Clark Kimberling_, Aug 22 2017