A113253 -id:A113253 - cp's OEIS frontend

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.

-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

A113250 Expansion of g.f. -(1 - 48x^2 - 256x^3) / ((1 - 4x)(1 + 4x)(1 + 4x + 16x^2)).

Original entry on oeis.org

-1, 4, 32, 64, -256, 4096, -4096, 16384, 131072, 262144, -1048576, 16777216, -16777216, 67108864, 536870912, 1073741824, -4294967296, 68719476736, -68719476736, 274877906944, 2199023255552, 4398046511104, -17592186044416, 281474976710656, -281474976710656

Offset: 0

Creighton Dement, Nov 18 2005

Previous name was: Corresponds to m = 4 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(m-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.

Conjecture: a(m, 2*n+1) is a perfect square for all m (see A113249). Initial terms factored (without regards to sign): 1, 4, (2)^5, (2)^6,(2)^8, (2)^12, (2)^12, (2)^14, (2)^17, (2)^18, (2)^20, (2)^24, (2)^24, (2)^26, (2)^29, (2)^30, (2)^32, (2)^36.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,64,256).

Cf. A000302, A097948, A056450, A113249, A113251, A113252, A113253, A113254, A113255, A113256.

LinearRecurrence[{-4, 0, 64, 256}, {-1, 4, 32, 64}, 25] (* Robert P. P. McKone, Aug 25 2023 *)
CoefficientList[Series[-(1-48x^2-256x^3)/((1-4x)(1+4x)(1+4x+16x^2)),{x,0,30}],x] (* Harvey P. Dale, Aug 27 2025 *)

Vec(-(1 - 48*x^2 - 256*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + 4*x + 16*x^2)) + O(x^25)) \\ Colin Barker, May 19 2019

G.f.: -(1 - 48*x^2 - 256*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + 4*x + 16*x^2)). Corrected by Colin Barker, May 19 2019

New name using g.f. from Joerg Arndt, Aug 25 2023

A113251 Corresponds to m = 5 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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, 59, 289, -1381, 13924, 10079, 2209, 520439, 7628644, -23994301, 149401729, 490531859, 406344964, -1681645081, 149155846849, -249406479121, 1083427010884, 9530848465739, 30158362505569, -168169798384501, 2302905921914404, -239007146013841, 2988025311585889

Offset: 0

Creighton Dement, Nov 18 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,100,625).

Cf. A000302, A097948, A056450, A113249, A113250, A113252, A113253, A113254, A113255, A113256.

with(gfun): seriestolist(series((-1+75*x^2+625*x^3)/((5*x+1)*(1-5*x)*(25*x^2+4*x+1)), x=0,25));

LinearRecurrence[{-4,0,100,625},{-1,4,59,289},40] (* Harvey P. Dale, Jul 05 2021 *)

Vec(-(1 - 75*x^2 - 625*x^3) / ((1 - 5*x)*(1 + 5*x)*(1 + 4*x + 25*x^2)) + O(x^30)) \\ Colin Barker, May 20 2019

G.f.: (-1+75*x^2+625*x^3) / ((5*x+1)*(1-5*x)*(25*x^2+4*x+1)).
a(n) = -4*a(n-1) + 100*a(n-3) + 625*a(n-4) for n>3. - Colin Barker, May 20 2019
a(n) = 5^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/5)*(n+1)))/4. - Eric Simon Jacob, Jul 29 2023

A113252 Corresponds to m = 6 in a family of 4th order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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, 92, 784, -3856, 33856, 96704, 73984, -418048, 59474944, -101917696, 443355136, 6249181184, 37406654464, -217868812288, 2345945595904, 4101714673664, 699056521216, 52661959000064, 3420344569298944, -8264891921072128, 41548867031793664

Offset: 0

Creighton Dement, Nov 18 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,144,1296).

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113253, A113254, A113255, A113256.

LinearRecurrence[{-4, 0, 144, 1296}, {-1, 4, 92, 784}, 25] (* Paolo Xausa, Jun 10 2024 *)

Vec(-(1 - 108*x^2 - 1296*x^3) / ((1 - 6*x)*(1 + 6*x)*(1 + 4*x + 36*x^2)) + O(x^25)) \\ Colin Barker, May 20 2019

G.f.: (-1+108*x^2+1296*x^3)/((6*x+1)*(1-6*x)*(36*x^2+4*x+1)).

a(n) = -4*a(n-1) + 144*a(n-3) + 1296*a(n-4) for n>3. - Colin Barker, May 20 2019

A113254 Corresponds to m = 8 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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, 176, 3136, -15616, 123904, 1028096, 4734976, -51183616, 975437824, 1521483776, 205520896, 39241908224, 4227925540864, -10627091267584, 53396107165696, 1029499365883904, 10479050187341824, -71775363146973184, 769363745204862976

Offset: 0

Creighton Dement, Nov 18 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,256,4096).

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113253, A113255, A113256.

LinearRecurrence[{-4, 0, 256, 4096}, {-1, 4, 176, 3136}, 25] (* Paolo Xausa, Jun 10 2024 *)

Vec(-(1 - 192*x^2 - 4096*x^3) / ((1 - 8*x)*(1 + 8*x)*(1 + 4*x + 64*x^2)) + O(x^25)) \\ Colin Barker, May 20 2019

G.f.: (-1+192*x^2+4096*x^3) / ((8*x+1)*(1-8*x)*(64*x^2+4*x+1)).

a(n) = -4*a(n-1) + 256*a(n-3) + 4096*a(n-4) for n > 3. - Colin Barker, May 20 2019

A113255 Corresponds to m = 9 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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, 227, 5329, -26581, 206116, 2391479, 16785409, -174757993, 2826198244, 9824173259, 14210785681, -287742103741, 22876687229764, -22446053606113, 89792737665409, 5164999769137199, 122161424469552196, -606821408584323661, 4689875711360495569

Offset: 0

Creighton Dement, Nov 18 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,324,6561).

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113253, A113254, A113256.

LinearRecurrence[{-4, 0, 324, 6561}, {-1, 4, 227, 5329}, 25] (* Paolo Xausa, Jun 10 2024 *)

Vec(-(1 - 243*x^2 - 6561*x^3) / ((1 - 9*x)*(1 + 9*x)*(1 + 4*x + 81*x^2)) + O(x^20)) \\ Colin Barker, May 20 2019

G.f.: (-1+243*x^2+6561*x^3) / ((9*x+1)*(1-9*x)*(81*x^2+4*x+1)).

a(n) = -4*a(n-1) + 324*a(n-3) + 6561*a(n-4) for n > 3. - Colin Barker, May 20 2019

A113256 Corresponds to m = 10 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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, 284, 8464, -42256, 322624, 4935104, 47997184, -485499136, 7142278144, 39980801024, 125848981504, -2501476028416, 97421005963264, 60463578988544, 16045087719424, 13889461750267904, 942837644226985984, -3160296751934734336, 18357422585040338944

Offset: 0

Creighton Dement, Nov 18 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Colin Barker, Table of n, a(n) for n = 0..900
Index entries for linear recurrences with constant coefficients, signature (-4,0,400,10000).

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113253, A113254, A113255.

LinearRecurrence[{-4, 0, 400, 10000}, {-1, 4, 284, 8464}, 25] (* Paolo Xausa, Jun 10 2024 *)

Vec(-(1 - 300*x^2 - 10000*x^3) / ((1 - 10*x)*(1 + 10*x)*(1 + 4*x + 100*x^2)) + O(x^20)) \\ Colin Barker, May 20 2019

G.f.: (-1+300*x^2+10000*x^3) / ((10*x+1)*(1-10*x)*(100*x^2+4*x+1)).

a(n) = -4*a(n-1) + 400*a(n-3) + 10000*a(n-4) for n > 3. - Colin Barker, May 20 2019

cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A113250 Expansion of g.f. -(1 - 48*x^2 - 256*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + 4*x + 16*x^2)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A113251 Corresponds to m = 5 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(m-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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A113252 Corresponds to m = 6 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(m-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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A113254 Corresponds to m = 8 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(m-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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A113255 Corresponds to m = 9 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(m-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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A113256 Corresponds to m = 10 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(m-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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

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.

A113250 Expansion of g.f. -(1 - 48x^2 - 256x^3) / ((1 - 4x)(1 + 4x)(1 + 4x + 16x^2)).

A113251 Corresponds to m = 5 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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.

A113252 Corresponds to m = 6 in a family of 4th order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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.

A113254 Corresponds to m = 8 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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.

A113255 Corresponds to m = 9 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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.

A113256 Corresponds to m = 10 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-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.