A190967 -id:A190967 - cp's OEIS frontend

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A113249 Corresponds to m = 3 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(n-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6(m-1) + 3(m-1)^2, a(m,3) = (-8+m^2)^2.

Original entry on oeis.org

-1, 4, 11, 1, 59, 484, -1009, 6241, -2761, 13924, 87251, 57121, 49139, 4072324, -7165609, 35058241, 10350959, 30492484, 559712411, 973502401, -1957852501, 30450948004, -41421000289, 174055005601, 241428053159, 9658565284, 2872244917091, 11300885699041, -25300162140061

Offset: 0

Creighton Dement, Oct 20 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m, n. Disregarding signs and/or initial term, we have: m = 0 (A000302), m = 1 (A097948), m = 2 (A056450), m = 3 (a(n)), m = 4 (A113250), m = 5 (A113251), m = 6 (A113252), m = 7 (A113253), m = 8 (A113254), m = 9 (A113255), m = 10 (A113256).

In this case, a(2n+1) = b(n)^2 where b(n) = Re((2+sqrt(-5))^(n+1)) satisfies the recurrence b(n) = 4*b(n-1) - 9*b(n-2) with b(0)=2, b(1)=-1. - Robert Israel, Oct 23 2017

			a(3, 13) = 93161710957356599364/((-2+i*sqrt(5))^14*(2+i*sqrt(5))^14) = 4072324 = 2^2*1009^2.

Colin Barker, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (-4,0,36,81).

Cf. A000302, A097948, A056450, A113250, A113251, A113252, A113253, A113254, A113255, A113256.

Cf. A176333, A190967.

f:= gfun:-rectoproc({a(n) = 81*a(n-4)+36*a(n-3)-4*a(n-1),a(0) = -1, a(1) = 4, a(2) = 11, a(3) = 1},a(n),remember):
map(f, [$0..30]); # Robert Israel, Oct 23 2017

LinearRecurrence[{-4, 0, 36, 81}, {-1, 4, 11, 1}, 29] (* Jean-François Alcover, Sep 25 2017 *)

Vec(-(1 - 27*x^2 - 81*x^3) / ((1 - 3*x)*(1 + 3*x)*(1 + 4*x + 9*x^2)) + O(x^30)) \\ Colin Barker, May 19 2019

G.f.: (-1+27*x^2+81*x^3)/((-3*x+1)*(3*x+1)*(9*x^2+4*x+1)).
a(2k+1) = (2*A176333(k)-3*A190967(k))^2. - Robert Israel, Oct 23 2017
a(n) = -4*a(n-1) + 36*a(n-3) + 81*a(n-4) for n>3. - Colin Barker, May 19 2019
a(n) = 3^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/3)*(n+1)))/4. - Eric Simon Jacob, Aug 06 2023

A176333 Expansion of (1-3x)/(1-4x+9*x^2).

Original entry on oeis.org

1, 1, -5, -29, -71, -23, 547, 2395, 4657, -2927, -53621, -188141, -269975, 613369, 4883251, 14012683, 12101473, -77708255, -419746277, -979610813, -140726759, 8253590281, 34280901955, 62841295291, -57162936431, -794223403343, -2662427185493, -3501698111885

Offset: 0

Paul Barry, Apr 15 2010

Hankel transform of A176332.

Michael De Vlieger, Table of n, a(n) for n = 0..2097
Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
Index entries for linear recurrences with constant coefficients, signature (4,-9).

Cf. A176332, A190958, A190967.

a:=[1,1];; for n in [3..30] do a[n]:=4*a[n-1]-9*a[n-2]; od; a; # G. C. Greubel, Dec 07 2019

I:=[1,1]; [n le 2 select I[n] else 4*Self(n-1) - 9*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 07 2019

seq(coeff(series((1-3*x)/(1-4*x+9*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 07 2019

CoefficientList[Series[(1-3x)/(1-4x+9x^2),{x,0,30}],x] (* or *) LinearRecurrence[{4,-9},{1,1},30] (* Harvey P. Dale, Sep 17 2012 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-4*x+9*x^2)) \\ G. C. Greubel, Dec 07 2019

def A176333_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-3*x)/(1-4*x+9*x^2) ).list()
A176333_list(30) # G. C. Greubel, Dec 07 2019

a(n) = 3^n*( cos(2*n*atan(1/sqrt(5))) - sin(2*n*atan(1/sqrt(5)))/sqrt(5) ).
a(0)=1, a(1)=1, a(n) = 4*a(n-1) - 9*a(n-2). - Harvey P. Dale, Sep 17 2012
a(n) = -3*A190967(n) + A190967(n+1). - R. J. Mathar, May 04 2013

cp's OEIS Frontend

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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A113249 Corresponds to m = 3 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4*a(n-4) + (2*m)^2*a(n-3) - 4*a(n-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6*(m-1) + 3*(m-1)^2, a(m,3) = (-8+m^2)^2.

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Maple

Mathematica

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Formula

A176333 Expansion of (1-3*x)/(1-4*x+9*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

A113249 Corresponds to m = 3 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(n-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6(m-1) + 3(m-1)^2, a(m,3) = (-8+m^2)^2.

A176333 Expansion of (1-3x)/(1-4x+9*x^2).