login
p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.
82

%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