A097948 -id:A097948 - cp's OEIS frontend

A264341 T(n,k)=Number of (n+1)X(k+1) arrays of permutations of 0..(n+1)*(k+1)-1 with each element having index change +-(.,.) 0,0 0,1 or 1,2.

4, 13, 8, 49, 55, 16, 181, 490, 233, 32, 676, 3567, 4900, 987, 64, 2521, 28925, 70669, 49000, 4181, 128, 9409, 223356, 1243225, 1399783, 490000, 17711, 256, 35113, 1759250, 20386617, 53429620, 27726581, 4900000, 75025, 512, 131044, 13750304

Offset: 1

R. H. Hardin, Nov 11 2015

Table starts
....4......13.........49...........181..............676.................2521
....8......55........490..........3567............28925...............223356
...16.....233.......4900.........70669..........1243225.............20386617
...32.....987......49000.......1399783.........53429620...........1855980772
...64....4181.....490000......27726581.......2296230561.........168990466353
..128...17711....4900000.....549201567......98684484373.......15386771913704
..256...75025...49000000...10878455069....4241136597604.....1400983500645217
..512..317811..490000000..215477871383..182270189212469...127561175981852920
.1024.1346269.4900000000.4268134837381.7833376999538689.11614593343457551705

			Some solutions for n=3 k=4
..7..8..9..3..4....1..0..3..2..4....7..8..2..3..4....1..2..9..4..3
.12..5..0..1..2...12..6..7..9..8....5..6..0..1..9...12..6..0..7..8
.17.10.13..6.14...11.10..5.14.13...11.10.12.13.14...10.18..5.14.13
.15.16.18.11.19...16.15.17.18.19...15.17.16.18.19...15.17.16.11.19

R. H. Hardin, Table of n, a(n) for n = 1..143

Column 1 is A000079(n+1).
Column 2 is A033887(n+1).
Row 1 is A097948.

Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 4*a(n-1) +a(n-2)
k=3: a(n) = 10*a(n-1)
k=4: a(n) = 19*a(n-1) +16*a(n-2)
k=5: a(n) = 43*a(n-1) -43*a(n-3) +a(n-4)
k=6: a(n) = 87*a(n-1) +374*a(n-2) -470*a(n-3) +207*a(n-4) +3*a(n-5)
k=7: a(n) = 191*a(n-1) +1102*a(n-2) -7594*a(n-3) -38349*a(n-4) +38507*a(n-5)
Empirical for row n:
n=1: a(n) = 4*a(n-1) -4*a(n-3) +a(n-4)
n=2: [order 14]
n=3: [order 34]

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

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

A113253 Corresponds to m = 7 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, 131, 1681, -8341, 68644, 369431, 923521, -10266601, 278289124, -45142549, 385690321, 28351798019, 545917055044, -2216460177409, 15348835582081, 113677067503919, 421612384372804, -3999798649362349, 75132454060794001

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 (-4,0,196,2401).

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113254, A113255, A113256.

LinearRecurrence[{-4, 0, 196, 2401}, {-1, 4, 131, 1681}, 25] (* Paolo Xausa, Jun 10 2024 *)

Vec(-(1 - 147*x^2 - 2401*x^3) / ((1 - 7*x)*(1 + 7*x)*(1 + 4*x + 49*x^2)) + O(x^25)) \\ Colin Barker, May 20 2019

G.f.: (-1+147*x^2+2401*x^3) / ((7*x+1)*(1-7*x)*(49*x^2+4*x+1)).
a(n) = -4*a(n-1) + 196*a(n-3) + 2401*a(n-4) for n > 3. - Colin Barker, May 20 2019
a(n) = 7^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/7)*(n+1)))/4. - Eric Simon Jacob, Jul 30 2023

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

A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1).

Original entry on oeis.org

1, -3, 13, -48, 181, -675, 2521, -9408, 35113, -131043, 489061, -1825200, 6811741, -25421763, 94875313, -354079488, 1321442641, -4931691075, 18405321661, -68689595568, 256353060613, -956722646883, 3570537526921, -13325427460800, 49731172316281

Offset: 0

Creighton Dement, Jul 21 2005

In reference to program code, 2baseiseq[X](n) = ((-1)^n)*A001353(n) (a(n)^2 + 1 is a perfect square.) 1tesseq[X](n) = (-1^(n+1))*A097948(n).

Floretion Algebra Multiplication Program, FAMP Code: 1ibaseiseq[X] with X = .5'i + .5i' + 'ii' - .5'jj' + 1.5'kk' - 1 (* Corrected by Creighton Dement, Dec 11 2009 *)

Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (-4,0,4,1).

Cf. A007654, A001570, A076139. See also A117808, A122571 (same except for signs).

Cf. A001353, A097948.

/* By definition: */
m:=15; R:=PowerSeriesRing(Integers(), m);
A001570:=Coefficients(R!((1-x)/(1-14*x+x^2)));
A007654:=Coefficients(R!(-3*x^2*(1+x)/(-1+x)/(1-14*x+x^2)));
&cat[[A001570[i],-A007654[i]]: i in [1..m-2]]; // Bruno Berselli, Feb 05 2013

seriestolist(series((x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)), x=0,25));

LinearRecurrence[{-4,0,4,1},{1,-3,13,-48},30] (* Harvey P. Dale, Jun 15 2018 *)

G.f.: (x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)).
Floor(((2 + sqrt(3))^n + (2 - sqrt(3))^n)/4) produces this sequence with a different offset and without signs. - James R. Buddenhagen, May 20 2010
Define c(n) = a(n) - 4*a(n+1) - a(n+2) and d(n) = -a(n) - 4*a(n+1) - a(n+2); Conjectures: I: c(2n) = 24*A076139(n); (Triangular numbers that are one-third of another triangular number) II: c(2n+1) = -A011943(n+1); (Numbers n such that any group of n consecutive integers has integral standard deviation) III: d(2n) = -2; IV: d(2n+1) = -1

cp's OEIS Frontend

A264341 T(n,k)=Number of (n+1)X(k+1) arrays of permutations of 0..(n+1)*(k+1)-1 with each element having index change +-(.,.) 0,0 0,1 or 1,2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

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

A113253 Corresponds to m = 7 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

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.

A113253 Corresponds to m = 7 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.