A098149 -id:A098149 - cp's OEIS frontend

A098111 Inverse binomial transform of A098149.

1, 0, -5, -25, -100, -375, -1375, -5000, -18125, -65625, -237500, -859375, -3109375, -11250000, -40703125, -147265625, -532812500, -1927734375, -6974609375, -25234375000, -91298828125, -330322265625, -1195117187500, -4323974609375, -15644287109375, -56601562500000

Offset: 0

Creighton Dement, Sep 23 2004

A030191(n) + 2*a(n) + A093129(n+2) = 4*A093129(n+1). - Creighton Dement, Oct 18 2004
From Wolfdieter Lang, Oct 02 2013: (Start)
These numbers a(n) and those of A030191(n) =: b(n), both interspersed with zeros, appear in the formula for nonnegative powers of the algebraic number rho(10) := 2*cos(pi/10) = phi*sqrt(3-phi), with the golden section phi, in terms of the power basis of the number field Q(rho(10)) of degree 4 (see A187360, n=10). In a (regular) decagon rho(10) is the length ratio of a smallest diagonal to the side. rho(10)^n = sum(A(n,k)*rho(10)^k, k=0..3), with A(2*k+1,0) = 0, A(2*k,0) = a(k), k >= 0; A(2*k,1) = 0, A(2*k+1,1) = a(k), k >= 0; A(2*k+1,2) = 0, k >= 0, A(0,2) = 0, A(2*k,2) = b(k-1), k >= 1; and A(2*k,3) = 0, k >= 0, A(1,3) = 0, A(2*k+1,3) = b(k-1), k >= 1. (End)

			Powers of rho(10) in the Q(rho(10)) power basis for n = 5: rho(10)^5 = 0*1 + a(2)*rho(10) + 0*rho(10)^2 + b(1)*rho(10)^3 = -5*rho(10) + 5*rho(10)^3.  - _Wolfdieter Lang_, Oct 02 2013

Index entries for linear recurrences with constant coefficients, signature (5,-5).
Index entries for sequences related to Chebyshev polynomials.

Cf. A030191, A049310, A098149, A187360.

LinearRecurrence[{5,-5},{1,0},40] (* Harvey P. Dale, Dec 08 2015 *)

G.f.: (1-5x)/(1-5x+5x^2).
From Wolfdieter Lang, Oct 02 2013: (Start)
a(n) = b(n) - 5*b(n-1), n >= 0,  with b(n) = A030191(n) = (sqrt(5))^n*S(n, sqrt(5)), with Chebyshev S-polynomials (see A049310).
a(n) = 5*(a(n-1) - a(n-2)), n >= 1, a(-1) = 1 = a(0). (End)

More terms from David Wasserman, Jan 16 2008

A121646 a(n) = Fibonacci(n-1)^2 - Fibonacci(n)^2.

Original entry on oeis.org

-1, 0, -3, -5, -16, -39, -105, -272, -715, -1869, -4896, -12815, -33553, -87840, -229971, -602069, -1576240, -4126647, -10803705, -28284464, -74049691, -193864605, -507544128, -1328767775, -3478759201, -9107509824, -23843770275, -62423800997, -163427632720

Offset: 1

Gary W. Adamson, Aug 13 2006

Negated first differences of A007598.
Real part of (F(n-1) + i*F(n))^2. Corresponding imaginary part = A079472(n); e.g., (3 + 5i)^2 = (-16 + 30i) where 30 = A079472(5). Consider a(n) and A079472(n) as legs of a Pythagorean triangle; then hypotenuse = corresponding n-th term in the sequence (1, 2, 5, 13, ...; i.e., odd-indexed Fibonacci terms). a(n)/a(n-1) tends to Phi^2.
3*A001654(n) - A001654(n+1) = A121646(n). - Vladimir Joseph Stephan Orlovsky, Nov 17 2009

			a(5) = -16 since Re(3 + 5i)^2 = (-16 + 30i).
a(5) = -16 = 3^2 - 5^2.

Daniele Corradetti, La Metafisica del Numero, 2008.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A079472.

List([1..40], n -> -Fibonacci(n-2)*Fibonacci(n+1)); # G. C. Greubel, Jan 07 2019

[-Fibonacci(n-2)*Fibonacci(n+1): n in [1..40]]; // G. C. Greubel, Jan 07 2019

A121646 := proc(n)
    combinat[fibonacci](n+1)*combinat[fibonacci](n-2) ;
    -% ;
end proc:
seq(A121646(n),n=1..10) ; # R. J. Mathar, Jun 22 2017

f[n_] := Re[(Fibonacci[n - 1] + I*Fibonacci[n])^2]; Array[f, 29] (* Robert G. Wilson v, Aug 16 2006 *)
lst={};Do[a1=Fibonacci[n]*Fibonacci[n+1];a2=Fibonacci[n+1]*Fibonacci[n+2];AppendTo[lst,3*a1-a2],{n,0,60}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)
Table[-Fibonacci[n-2]*Fibonacci[n+1],{n,1,40}] (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)
-Differences[Fibonacci[Range[0,30]]^2] (* Harvey P. Dale, Nov 01 2022 *)

a(n) = fibonacci(n-1)^2 - fibonacci(n)^2 \\ Charles R Greathouse IV, Jun 11 2015

[-fibonacci(n-2)*fibonacci(n+1) for n in (1..40)] # G. C. Greubel, Jan 07 2019

a(n) = Re(F(n-1) + F(n)*i)^2 = (F(n-1))^2 - (F(n))^2.
G.f.: (1-3*x)/((1+x)*(1 - 3*x + x^2)). - Paul Barry, Oct 13 2006
a(n) = -F(n+1)*F(n-2) where F=A000045. - Ron Knott, Jan 24 2009
a(n) = (4*(-1)^n - |A098149(n)|)/5. - R. J. Mathar, Jan 13 2011

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604

Offset: 0

Bruno Berselli, Jun 19 2017

a(n) mod 4 gives A101000.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A101000.
Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

[Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017

LucasL[4 Range[0, 21] + 3]
LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)

Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

from sympy import lucas
def a(n):  return lucas(4*n + 3)
print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021

def L():
    x, y = -1, 4
    while True:
        yield y
        x, y = y, 7*y - x
r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

A110034 Row sums of a characteristic triangle for the Fibonacci numbers.

Original entry on oeis.org

1, 0, -1, -4, -11, -30, -79, -208, -545, -1428, -3739, -9790, -25631, -67104, -175681, -459940, -1204139, -3152478, -8253295, -21607408, -56568929, -148099380, -387729211, -1015088254, -2657535551, -6957518400, -18215019649, -47687540548, -124847601995, -326855265438

Offset: 0

Paul Barry, Jul 08 2005

Rows sums of A110033.

Conjecture: |a(n)| = Sum_{k=1..n-1} A061646(k). - J. M. Bergot, Jun 10 2013

Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A061646, A110033.

CoefficientList[Series[(-2*z^3 + z^2 + 3*z - 1)/(z^4 - 3*z^3 + 3*z - 1), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011 *)

G.f.: (1-3x-x^2+2x^3)/((1-x^2)(1-3x+x^2)).
a(n) = a(n-1)-3a(n-3)+a(n-4).
a(n) = Sum_{k=0..n-1} F(k)*F(k+1) - F(2n) + 1.
a(n) = (5+(-1)^n-4*(-1)^n*A098149(n))/10. [R. J. Mathar, Jul 22 2010]

A098150 a(n) = 2(a(n-2) - a(n-1)) + a(n-3) where a(0)=-3, a(1)=11 & a(2)=-30.

Original entry on oeis.org

-3, 11, -30, 79, -207, 542, -1419, 3715, -9726, 25463, -66663, 174526, -456915, 1196219, -3131742, 8199007, -21465279, 56196830, -147125211, 385178803, -1008411198, 2640054791, -6911753175, 18095204734, -47373861027, 124026378347, -324705274014, 850089443695, -2225563057071, 5826599727518, -15254236125483

Offset: 0

Creighton Dement, Aug 29 2004

Sequence relates bisections of Lucas and Fibonacci numbers.

Pisano period lengths: 1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, ... - R. J. Mathar, Aug 10 2012

Indranil Ghosh, Table of n, a(n) for n = 0..2383
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (-3,-1).

Cf. A098149, A001519, A005248.

I:=[-3,11]; [n le 2 select I[n] else -3*Self(n-1)-Self(n-2): n in [1..35]]; // Vincenzo Librandi, Dec 26 2018

a[0] = -3; a[1] = 11; a[2] = -30; a[n_] := a[n] = 2(a[n - 2] - a[n - 1]) + a[n - 3]; Table[ a[n], {n, 0, 25}] (* Robert G. Wilson v, Sep 04 2004 *)
RecurrenceTable[{a[0]==-3,a[1]==11,a[2]==-30,a[n]==2(a[n-2]-a[n-1])+ a[n-3]},a,{n,30}] (* or *) LinearRecurrence[{-3,-1},{-3,11},30] (* Harvey P. Dale, Feb 05 2012 *)
Table[(-1)^(n+1)(3LucasL[2n+1]-Fibonacci[2n]), {n,0,20}] (* Rigoberto Florez, Dec 24 2018 *)

2*A098149(n) + a(n) = 8*(-1)^(n+1)*A001519(n) - (-1)^(n+1)*A005248(n+1).
a(n) = - 3a(n-1) - a(n-2). - Tanya Khovanova, Feb 02 2007
G.f.: (2x-3)/(1+3x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = (-1)^(n+1)*(3*L(2n+1)-F(2n)), where F(n) is the n-th Fibonacci number and L(n) is the n-th Lucas number. - Rigoberto Florez, Dec 24 2018

More terms from Robert G. Wilson v, Sep 04 2004

A109794 a(2n) = A001906(n+1), a(2n+1) = A002878(n).

Original entry on oeis.org

1, 1, 3, 4, 8, 11, 21, 29, 55, 76, 144, 199, 377, 521, 987, 1364, 2584, 3571, 6765, 9349, 17711, 24476, 46368, 64079, 121393, 167761, 317811, 439204, 832040, 1149851, 2178309, 3010349, 5702887, 7881196, 14930352, 20633239, 39088169

Offset: 0

Creighton Dement, Aug 14 2005

Sequence relates bisections of Lucas and Fibonacci numbers (see also A098149).

Floretion Algebra Multiplication Program, FAMP code: 4jesleftforsumseq[ + .25'i + .25i' + .25'ii' + .25'jj' + .25'kk' + .25'jk' + .25'kj' + .25e], vesleftforsumseq = A000045, sumtype: (Y[15], *, inty*sum) (internal program code)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1)

Cf. A001906, A002878, A098149, A000045, A189761.

a:=[1,1,3,4];; for n in [5..40] do a[n]:=3*a[n-2]-a[n-4]; od; a; # Muniru A Asiru, Aug 09 2018

a:= n-> (<<0|1>, <-1|3>>^iquo(n, 2, 'r'). <<1, 3+r>>)[1, 1]:
seq(a(n), n=0..50);  # Alois P. Heinz, May 02 2011

LinearRecurrence[{0, 3, 0, -1}, {1, 1, 3, 4}, 40] (* Robert G. Wilson v, Aug 06 2018 *)
CoefficientList[Series[(1+x+x^3)/((1+x-x^2)(1-x-x^2)),{x,0,40}],x] (* Harvey P. Dale, Aug 10 2021 *)

G.f.: (1+x+x^3)/((1+x-x^2)*(1-x-x^2)).
a(n) = ((3/20)*sqrt(5) + 3/4)*(1/2 + (1/2)*sqrt(5))^n + (-(3/20)*sqrt(5) + 3/4)*(1/2 - (1/2)*sqrt(5))^n + (-(3/20)*sqrt(5) - 1/4)*(-1/2 + (1/2)*sqrt(5))^n + ((3/20)*sqrt(5) - 1/4) *(-1/2 - (1/2)*sqrt(5))^n.
a(n) = 3*a(n-2) - a(n-4), n >= 4; a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 4. - Daniel Forgues, May 07 2011

A105671 a(2n) = Lucas(2n+3)^2, a(2n+1) = Lucas(2n+1)^2.

Original entry on oeis.org

16, 1, 121, 16, 841, 121, 5776, 841, 39601, 5776, 271441, 39601, 1860496, 271441, 12752041, 1860496, 87403801, 12752041, 599074576, 87403801, 4106118241, 599074576, 28143753121, 4106118241, 192900153616, 28143753121

Offset: 0

Creighton Dement, Apr 17 2005

This sequence is related to several other sequences on squares. In reference to the link "Sequences in Context", (a(n)) = vessigcycseq. Note that the identity "vessigcyc = jessigcyc + lessigcyc + tessigcyc" holds.

Floretion Algebra Multiplication Program, FAMP Code: 1vessigcycseq[ + 4.75'i - .75'j - .25'k + 4.75i' - .75j' - .25k' - 1.75'ii' - 3.75'jj' + 4.25'kk' - .25'ij' + 1.25'ik' - .25'ji' + 2.75'jk' + 1.25'ki' + 2.75'kj' + 2.25e]

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-1,1).

Cf. A081071.

a[n_?EvenQ] := LucasL[n+3]^2; a[n_?OddQ] := LucasL[n]^2; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Sep 28 2011 *)

Vec((16 - 15*x + 8*x^2 + x^4) / ((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)) + O(x^40)) \\ Colin Barker, May 01 2019

G.f.: (-x^4-8x^2+15x-16)/((x-1)(x^4-7x^2+1)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n>4. - Colin Barker, May 01 2019
a(n) = 2*A098149(n+2) +5*A001519(n+2)-2. - R. J. Mathar, Sep 11 2019

A247526 a(n) = L(n+1) * L(n) * L(n-1) * L(n-2) / 6, where L(n) = Lucas numbers (A000032).

Original entry on oeis.org

-1, -1, 4, 14, 154, 924, 6699, 44979, 310764, 2123554, 14571974, 99833524, 684385079, 4690541639, 32150245204, 220358978774, 1510368355474, 10352204457804, 70955102255139, 486333408161979, 3333379024971324, 22847319059525674, 156597856242950654

Offset: 0

Michael Somos, Sep 19 2014

			G.f. = -1 - x + 4*x^2 + 14*x^3 + 154*x^4 + 924*x^5 + 6699*x^6 + 44979*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000032, A085696.

m:=60; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-(1-4*x-24*x^2+6*x^3-4*x^4)/((1-x)*(1+3*x+x^2)*(1 - 7*x+x^2)))); // G. C. Greubel, Aug 05 2018

CoefficientList[Series[-(1-4*x-24*x^2+6*x^3-4*x^4)/((1-x)*(1+3*x+x^2)*(1 - 7*x+x^2)), {x, 0, 60}], x] (* G. C. Greubel, Aug 05 2018 *)
Times@@#/6&/@Partition[LucasL[Range[-2,30]],4,1] (* or *) LinearRecurrence[{5,15,-15,-5,1},{-1,-1,4,14,154},30] (* Harvey P. Dale, Apr 20 2022 *)

{a(n) = my(u = fibonacci(n), v = fibonacci(n-1)); (3*u + v) * (u + 2*v) * (2*u - v) * (-u + 3*v) / 6};

{a(n) = if( n<1, n=1-n); polcoeff( - (1 - 4*x - 24*x^2 + 6*x^3 - 4*x^4) / ((1 - x) * (1 + 3*x + x^2) * (1 - 7*x +x^2)) + x * O(x^n), n)};

G.f.: -(1 - 4*x - 24*x^2 + 6*x^3 - 4*x^4)/((1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2)).
a(n) = a(1-n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) for all n in Z.
0 = a(n)*(+a(n+1) - 2*a(n+2)) + a(n+1)*(-5*a(n+1) + a(n+2)) for all n in Z.
From Klaus Purath, Oct 02 2020: (Start)
a(n) = (L(n-2)*L(n-1)^3 - L(n-1)*L(n-2)^3)/6 where L = Lucas.
a(n) = f(n-3)*f(n-2)*f(n-1)*f(n)/96 where f = A022112.
a(n) = (f(n-2)*f(n-1)^3 - f(n-1)*f(n-2)^3)/96 where f = A022112.
(End)
2*a(n) = A098149(n) +A004187(n+1)-6*A004187(n) -2 . - R. J. Mathar, Sep 24 2021

cp's OEIS Frontend

A098111 Inverse binomial transform of A098149.

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A121646 a(n) = Fibonacci(n-1)^2 - Fibonacci(n)^2.

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A288913 a(n) = Lucas(4*n + 3).

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A110034 Row sums of a characteristic triangle for the Fibonacci numbers.

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A098150 a(n) = 2(a(n-2) - a(n-1)) + a(n-3) where a(0)=-3, a(1)=11 & a(2)=-30.

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A109794 a(2n) = A001906(n+1), a(2n+1) = A002878(n).

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A105671 a(2n) = Lucas(2n+3)^2, a(2n+1) = Lucas(2n+1)^2.

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