A113067 -id:A113067 - cp's OEIS frontend

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960

Offset: 0

Clark Kimberling, Aug 15 2017

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).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S)                 t(1,2,3,4,5,...)
- S                    A001906
- S^2                  A290890; see A113067 for signed version
- S^3                  A290891
- S^4                  A290892
- S^5                  A290893
- S^6                  A290894
- S^7                  A290895
- S^8                  A290896
- S - S^2              A289780
- S - S^3              A290897
- S - S^4              A290898
- S^2 - S^4            A290899
- S^2 - S^3            A290900
- S^3 - S^4            A290901
- 2S                   A052530;  (1/2)*A052530 = A001353
- 3S                   A290902;  (1/3)*A290902 = A004254
- 4S                   A003319;  (1/4)*A003319 = A001109
- 5S                   A290903;  (1/5)*A290903 = A004187
- 2*S^2                A290904;  (1/2)*A290904 = A290905
- 3*S^2                A290906;  (1/3)*A290906 = A290907
- 4*S^2                A290908;  (1/4)*A290908 = A099486
- 5*S^2                A290909;  (1/5)*A290909 = A290910
- 6*S^2                A290911;  (1/6)*A290911 = A290912
- 7*S^2                A290913;  (1/7)*A290913 = A290914
- 8*S^2                A290915;  (1/8)*A290915 = A290916
(1 - S)^2                A290917
(1 - S)^3                A290918
(1 - S)^4                A290919
(1 - S)^5                A290920
(1 - S)^6                A290921
- S - 2*S^2            A290922
- 2*S - 2*S^2          A290923;  (1/2)*A290923 = A290924
- 3*S - 2*S^2          A290925
(1 - S^2)^2              A290926
(1 - S^2)^3              A290927
(1 - S^3)^2              A290928
(1 - S)(1 - S^2)         A290929
(1 - S^2)(1 - S^4)       A290930
- 3 S + S^2            A291025
- 4 S + S^2            A291026
- 5 S + S^2            A291027
- 6 S + S^2            A291028
- S - S^2 - S^3        A291029
- S - S^2 - S^3 - S^4  A201030
- 3 S + 2 S^3          A291031
- S - S^2 - S^3 + S^4  A291032
- 6 S                  A291033
- 7 S                  A291034
- 8 S                  A291181
- 3 S + 2 S^3          A291031
- 3 S + 2 S^2          A291182
- 4 S + 2 S^3          A291183
- 4 S + 3 S^3          A291184

			(See the examples at A289780.)

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -5, 4, -1)

Cf. A000027, A113067, A289780, A113067 (signed version of same sequence).

z = 60; s = x/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)

G.f.: x/(1 - 4 x + 5 x^2 - 4 x^3 + x^4).

a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).

A113066 Expansion of (1 + x)^2/((1 + x + x^2)(1 + 3x + x^2)).

Original entry on oeis.org

1, -2, 4, -10, 27, -72, 189, -494, 1292, -3382, 8855, -23184, 60697, -158906, 416020, -1089154, 2851443, -7465176, 19544085, -51167078, 133957148, -350704366, 918155951, -2403763488, 6293134513, -16475640050, 43133785636, -112925716858, 295643364939, -774004377960

Offset: 0

Creighton Dement, Oct 13 2005

Binomial transform gives signed version of A093040.
The positive sequence has g.f. (1 - x)^2/((1 - x + x^2)(1 - 3*x + x^2)) and a(n) = Sum_{k=0..n} binomial(n+k+1, n-k)*(1+(-1)^k)/2. - Paul Barry, Jul 06 2009
Floretion Algebra Multiplication Program, FAMP Code: 2basei[C*F]; C = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'; F = + .5'i + .5'ii' + .5'ij' + .5'ik'

C. Dement, Floretion Integer Sequences (work in progress).

Muniru A Asiru, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (-4,-5,-4,-1).

Cf. A113067, A113068, A093040, A109961, A001906.

a:=[1,-2,4,-10];; for n in [5..35] do a[n]:=-4*a[n-1]-5*a[n-2]-4*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 11 2018

I:=[1,-2,4,-10]; [n le 4 select I[n] else -4*Self(n-1)-5*Self(n-2)- 4*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Sep 12 2018

seq(coeff(series((1+x)^2/((1+x+x^2)*(1+3*x+x^2)),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Sep 11 2018

LinearRecurrence[{-4, -5, -4, -1}, {1, -2, 4, -10}, 40] (* Vincenzo Librandi, Sep 12 2018 *)
CoefficientList[Series[(1 + x)^2/((1 + x + x^2)*(1 + 3 x + x^2)), {x, 0, 50}], x] (* Stefano Spezia, Sep 12 2018 *)

x='x+O('x^99); Vec((1+x)^2/((1+x+x^2)*(1+3*x+x^2))) \\ Altug Alkan, Sep 11 2018

a(n) + a(n+1) = (-1)^(n+1)*A109961(n+1).
a(n) + a(n+1) + a(n+2) = (-1)^n*A001906(n+2) = (-1)^n*F(2*n+4).
a(n) = A049347(n)/2 + (-1)^n*A001906(n+1)/2. - R. J. Mathar, Nov 10 2009
Lim_{n -> inf} a(n)/a(n-1) = -(1 + A001622). - A.H.M. Smeets, Sep 11 2018
a(n) = -4*a(n-1) - 5*a(n-2) - 4*a(n-3) - a(n-4). - Muniru A Asiru, Sep 11 2018

A285935 Square array a(n, m) read by antidiagonals whose g.f. is 1 / ((1-x)^2 * (1-y)^2 - x*y).

Original entry on oeis.org

1, 2, 2, 3, 5, 3, 4, 10, 10, 4, 5, 18, 26, 18, 5, 6, 30, 58, 58, 30, 6, 7, 47, 116, 153, 116, 47, 7, 8, 70, 214, 354, 354, 214, 70, 8, 9, 100, 371, 746, 931, 746, 371, 100, 9, 10, 138, 612, 1464, 2204, 2204, 1464, 612, 138, 10, 11, 185, 969, 2714, 4816, 5794

Offset: 0

Michael Somos, Jun 14 2017

			a(n,m) 0   1   2   3
----+--- --- --- ---
0   |  1   2   3   4
1   |  2   5  10  18
2   |  3  10  26  58
3   |  4  18  58 153

Cf. A113067, A177787.

a[n_, m_] := SeriesCoefficient[1/((1-x)^2*(1-y)^2-x*y), {x, 0, n}, {y, 0, m}];
Table[a[n-m, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jun 15 2017 *)

{a(n, m) = if( n<0 || m<0, 0, polcoeff( polcoeff( -1/(x*y-sqr(1-x-y+x*y))*(1+x*O(x^n))*(1+y*O(y^k)), n), m))};

G.f. Sum_{n>=0, m>=0} a(n, m) * x^n * y^m = 1 / ((1-x)^2 * (1-y)^2 - x*y).
T(n, k) := a(n-k, k) where 0 <= k <= n.
a(n, m) = a(m, n) = T(n+m, n), T(n, 0) = a(n, 0) = n+1, if n>=0, m>=0.
Row sums are (-1)^(n+1) * A113067(n+1).
T(n, 1) = A177787(n+1).

cp's OEIS Frontend

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

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A113066 Expansion of (1 + x)^2/((1 + x + x^2)(1 + 3x + x^2)).

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A285935 Square array a(n, m) read by antidiagonals whose g.f. is 1 / ((1-x)^2 * (1-y)^2 - x*y).

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cp's OEIS Frontend

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

Views

Author

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Examples

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Programs

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Formula

A113066 Expansion of (1 + x)^2/((1 + x + x^2)*(1 + 3*x + x^2)).

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Author

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Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Formula

A285935 Square array a(n, m) read by antidiagonals whose g.f. is 1 / ((1-x)^2 * (1-y)^2 - x*y).

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Author

Keywords

Examples

Crossrefs

Programs

Mathematica

PARI

Formula

A113066 Expansion of (1 + x)^2/((1 + x + x^2)(1 + 3x + x^2)).