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%I #37 Apr 06 2020 01:14:19
%S 1,4,14,47,156,517,1714,5684,18851,62520,207349,687676,2280686,
%T 7563923,25085844,83197513,275925586,915110636,3034975799,10065534960,
%U 33382471801,110713382644,367182309614,1217764693607,4038731742156,13394504020957,44423039068114
%N p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.
%C 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).
%C 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 Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
%C t(A000012) = t(1,1,1,1,1,1,1,...) = A001906
%C t(A000290) = t(1,4,9,16,25,36,...) = A289779
%C t(A000027) = t(1,2,3,4,5,6,7,8,...) = A289780
%C t(A000045) = t(1,2,3,5,8,13,21,...) = A289781
%C t(A000032) = t(2,1,3,4,7,11,14,...) = A289782
%C t(A000244) = t(1,3,9,27,81,243,...) = A289783
%C t(A000302) = t(1,4,16,64,256,...) = A289784
%C t(A000351) = t(1,5,25,125,625,...) = A289785
%C t(A005408) = t(1,3,5,7,9,11,13,...) = A289786
%C t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
%C t(A016777) = t(1,4,7,10,13,16,...) = A289789
%C t(A016789) = t(2,5,8,11,14,17,...) = A289790
%C t(A008585) = t(3,6,9,12,15,18,...) = A289795
%C t(A000217) = t(1,3,6,10,15,21,...) = A289797
%C t(A000225) = t(1,3,7,15,31,63,...) = A289798
%C t(A000578) = t(1,8,27,64,625,...) = A289799
%C t(A000984) = t(1,2,6,20,70,252,...) = A289800
%C t(A000292) = t(1,4,10,20,35,56,...) = A289801
%C t(A002620) = t(1,2,4,6,9,12,16,...) = A289802
%C t(A001906) = t(1,3,8,21,55,144,...) = A289803
%C t(A001519) = t(1,1,2,5,13,34,...) = A289804
%C t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
%C t(A008619) = t(1,1,2,2,3,3,4,4,...) = A289806
%C t(A080513) = t(1,2,2,3,3,4,4,5,...) = A289807
%C t(A133622) = t(1,2,1,3,1,4,1,5,...) = A289809
%C t(A000108) = t(1,1,2,5,14,42,...) = A081696
%C t(A081696) = t(1,1,3,9,29,97,...) = A289810
%C t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
%C t(A175676) = t(1,0,0,2,0,0,3,0,...) = A289844
%C t(A079977) = t(1,0,1,0,2,0,3,...) = A289845
%C t(A059841) = t(1,0,1,0,1,0,1,...) = A289846
%C t(A000040) = t(2,3,5,7,11,13,...) = A289847
%C t(A008578) = t(1,2,3,5,7,11,13,...) = A289828
%C t(A000142) = t(1!, 2!, 3!, 4!, ...) = A289924
%C t(A000201) = t(1,3,4,6,8,9,11,...) = A289925
%C t(A001950) = t(2,5,7,10,13,15,...) = A289926
%C t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
%C t(A000045*) = t(0,1,1,2,3,5,...) = A289975 (* indicates prepended 0's)
%C t(A000045*) = t(0,0,1,1,2,3,5,...) = A289976
%C t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
%C t(A290990*) = t(0,1,2,3,4,5,...) = A290990
%C t(A290990*) = t(0,0,1,2,3,4,5,...) = A290991
%C t(A290990*) = t(0,0,01,2,3,4,5,...) = A290992
%H Clark Kimberling, <a href="/A289780/b289780.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -7, 5, -1)
%F G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).
%F a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).
%e Example 1: s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
%e S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
%e p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
%e - p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
%e T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
%e t(s) = (1,3,8,21,...) = A001906.
%e ***
%e Example 2: s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
%e S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
%e p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
%e - p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
%e T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
%e t(s) = (1,4,14,47,...) = A289780.
%t z = 60; s = x/(1 - x)^2; p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289780 *)
%o (PARI) x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ _Altug Alkan_, Aug 13 2017
%o (GAP)
%o P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # _Muniru A Asiru_, Sep 03 2017
%Y Cf. A000027.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 10 2017
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