A056450 -id:A056450 - cp's OEIS frontend

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.

-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

A321391 Array read by antidiagonals: T(n,k) is the number of achiral rows of n colors using up to k colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 3, 4, 1, 0, 1, 5, 4, 9, 4, 1, 0, 1, 6, 5, 16, 9, 8, 1, 0, 1, 7, 6, 25, 16, 27, 8, 1, 0, 1, 8, 7, 36, 25, 64, 27, 16, 1, 0, 1, 9, 8, 49, 36, 125, 64, 81, 16, 1, 0, 1, 10, 9, 64, 49, 216, 125, 256, 81, 32, 1, 0

Offset: 0

Robert A. Russell, Nov 08 2018

The antidiagonals go from top-right to bottom-left.

			The array begins with T(0,0):
1 1  1   1    1     1     1      1      1      1       1       1 ...
0 1  2   3    4     5     6      7      8      9      10      11 ...
0 1  2   3    4     5     6      7      8      9      10      11 ...
0 1  4   9   16    25    36     49     64     81     100     121 ...
0 1  4   9   16    25    36     49     64     81     100     121 ...
0 1  8  27   64   125   216    343    512    729    1000    1331 ...
0 1  8  27   64   125   216    343    512    729    1000    1331 ...
0 1 16  81  256   625  1296   2401   4096   6561   10000   14641 ...
0 1 16  81  256   625  1296   2401   4096   6561   10000   14641 ...
0 1 32 243 1024  3125  7776  16807  32768  59049  100000  161051 ...
0 1 32 243 1024  3125  7776  16807  32768  59049  100000  161051 ...
0 1 64 729 4096 15625 46656 117649 262144 531441 1000000 1771561 ...
For T(3,3)=9, the rows are AAA, ABA, ACA, BAB, BBB, BCB, CAC, CBC, and CCC.

Columns 0-6 are A000007, A000012, A060546, A056449, A056450, A056451, A056452.

Cf. A003992 (oriented), A277504 (unoriented), A293500 (chiral).

Table[If[n>0, (n-k)^Ceiling[k/2], 1], {n, 0, 12}, {k, 0, n}] // Flatten

T(n,k) = [n==0] + [n>0] * k^ceiling(n/2).

The generating function for column k is (1+k*x) / (1-k*x^2).

A032087 Number of reversible strings with n beads of 4 colors. If more than 1 bead, not palindromic.

Original entry on oeis.org

4, 6, 24, 120, 480, 2016, 8064, 32640, 130560, 523776, 2095104, 8386560, 33546240, 134209536, 536838144, 2147450880, 8589803520, 34359607296, 137438429184, 549755289600, 2199021158400, 8796090925056, 35184363700224, 140737479966720, 562949919866880

Offset: 1

Christian G. Bower

From Petros Hadjicostas, Jun 30 2018: (Start)
Using the formulae in C. G. Bower's web link below about transforms, it can be proved that, for k >= 2, the BHK[k] transform of sequence (c(n): n >= 1), which has g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has generating function B_k(x) = (1/2)*(C(x)^k - C(x^2)^{k/2}) if k is even, and B_k(x) = C(x)*B_{k-1}(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. For k=1, Bower assumes that the BHK[k=1] transform of (c(n): n >= 1) is itself, which means that the g.f. of the output sequence is C(x). (This assumption is not accepted by all mathematicians because a sequence of length 1 is not only reversible but palindromic as well.)
Since a(m) = BHK(c(n): n >= 1)(m) = Sum_{k=1..m} BHK[k](c(n): n >= 1)(m) for m = 1,2,3,..., it can be easily proved (using sums of infinite geometric series) that the g.f. of BHK(c(n): n >= 1) is A(x) = (C(x)^2 - C(x^2))/(2*(1-C(x))*(1-C(x^2))) + C(x). (The extra C(x) is due of course to the special assumption made for the BHK[k=1] transform.)
Here, BHK(c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK and the input sequence is (c(n): n >= 1). Similarly, BHK[k](c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK[k] (i.e., with k boxes) and the input sequence is (c(n): n >= 1).
For the current sequence, c(1) = 4, and c(n) = 0 for all n >= 2, and thus, C(x) = 4*x. Substituting into the above formula for A(x), and doing the algebra, we get A(x) = 2*x*(2-5*x-8*x^2+32*x^3) / ((1+2*x)*(1-2*x)*(1-4*x)), which is R. J. Mathar's formula below.
(End)
The formula for a(n) for this sequence was Ralf Stephan's conjecture 72. It was solved by Elizabeth Wilmer (see Proposition 1 in one of the links below). She does not accept Bower's assertion that a string of length 1 is not palindromic. - Petros Hadjicostas, Jul 05 2018

Colin Barker, Table of n, a(n) for n = 1..1000
C. G. Bower, Transforms (2)
Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.
Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74
Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74 [cached copy]
Index entries for linear recurrences with constant coefficients, signature (4,4,-16).

Column 4 of A293500 for n>1.

Cf. A000302, A026337 (bisection), A032121, A056450, A088037.

A032087:= func< n | n eq 1 select 4 else 2^(2*n-1) -(3-(-1)^n)*2^(n-2) >;
[A032087(n): n in [1..30]]; // G. C. Greubel, Oct 02 2024

Join[{4}, LinearRecurrence[{4, 4, -16}, {6, 24, 120}, 24]] (* Jean-François Alcover, Oct 11 2017 *)

Vec(2*x*(2 - 5*x - 8*x^2 + 32*x^3) / ((1 - 2*x)*(1 + 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Mar 08 2017

def A032087(n): return 2^(2*n-1) -3*2^(n-2) +(-2)^(n-2) +4*int(n==1)
[A032087(n) for n in range(1,31)] # G. C. Greubel, Oct 02 2024

"BHK" (reversible, identity, unlabeled) transform of 4, 0, 0, 0, ...
a(2*n+1) = 2^(4*n+1) - 2^(2*n+1), a(2*n) = 2^(4*n-1) - 2^(2*n) + 2^(2*n-1), a(1)=4.
a(n) = (A000302(n) - A056450(n))/2 for n > 1.
From R. J. Mathar, Mar 20 2009: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) for n > 4.
G.f.: 2*x*(2-5*x-8*x^2+32*x^3)/((1-2*x)*(1+2*x)*(1-4*x)). (End)
From Colin Barker, Mar 08 2017: (Start)
a(n) = 2^(n-1) * (2^n-1) for n > 1 and even.
a(n) = 2^(2*n-1) - 2^n for n > 1 and odd. (End)
E.g.f.: (1/4)*( exp(-2*x) - 3*exp(2*x) + 2*exp(4*x) ) + 4*x. - G. C. Greubel, Oct 02 2024

A056460 Number of primitive (aperiodic) palindromes using a maximum of four different symbols.

Original entry on oeis.org

4, 0, 12, 12, 60, 48, 252, 240, 1008, 960, 4092, 4020, 16380, 16128, 65460, 65280, 262140, 261072, 1048572, 1047540, 4194036, 4190208, 16777212, 16772880, 67108800, 67092480, 268434432, 268419060, 1073741820, 1073675280, 4294967292, 4294901760, 17179865076, 17179607040, 68719476420

Offset: 1

Marks R. Nester

nonn

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Column 4 of A284823.

Cf. A056450.

a(n) = sumdiv(n, d, moebius(d)*(3*2^(n/d)-(-2)^(n/d))/2);
for(n=1, 40, print1(a(n), ", ")); \\ Petros Hadjicostas, Apr 24 2020

a(n) = Sum_{d|n} mu(d)*A056450(n/d).

More terms from Petros Hadjicostas, Apr 24 2020

A164907 a(n) = (3*3^n-(-1)^n)/2.

Original entry on oeis.org

1, 5, 13, 41, 121, 365, 1093, 3281, 9841, 29525, 88573, 265721, 797161, 2391485, 7174453, 21523361, 64570081, 193710245, 581130733, 1743392201, 5230176601, 15690529805, 47071589413, 141214768241, 423644304721, 1270932914165

Offset: 0

Klaus Brockhaus, Aug 31 2009

Interleaving of A096053 and A083884 without initial term 1.
Partial sums are (essentially) in A080926.
First differences are (essentially) in A105723.
a(n)+a(n+1) = A008776(n+1) = A099856(n+1) = A110593(n+2).
Binomial transform of A056450. Inverse binomial transform of A164908.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (2,3).

Equals A046717 without initial term 1 and A080925 without initial term 0. Equals A084182 / 2 from second term onward.

Cf. A096053, A083884, A080926, A105723, A008776, A099856, A110593, A056450, A164908.

Magma
```
[ (3*3^n-(-1)^n)/2: n in [0..25] ];
```

A164907:=n->(3*3^n - (-1)^n)/2; seq(A164907(n), n=0..30); # Wesley Ivan Hurt, Mar 21 2014

Table[(3*3^n - (-1)^n)/2, {n, 0, 30}] (* Wesley Ivan Hurt, Mar 21 2014 *)
LinearRecurrence[{2,3},{1,5},50] (* Harvey P. Dale, Oct 31 2018 *)

a(n) = 2*a(n-1)+3*a(n-2) for n > 1; a(0) = 1, a(1) = 5.
G.f.: (1+3*x)/((1+x)*(1-3*x)).
a(n) = 3*a(n-1)+2*(-1)^n. - Carmine Suriano, Mar 21 2014

A242026 Number of non-palindromic n-tuples of 4 distinct elements.

Original entry on oeis.org

0, 12, 48, 240, 960, 4032, 16128, 65280, 261120, 1047552, 4190208, 16773120, 67092480, 268419072, 1073676288, 4294901760, 17179607040, 68719214592, 274876858368, 1099510579200, 4398042316800, 17592181850112, 70368727400448, 281474959933440, 1125899839733760

Offset: 1

Mikk Heidemaa, Aug 12 2014

Non-palindromic vs palindromic (DNA) sequences (e.g., {a,c,a,c} is a non-palindromic sequence but {a,c,c,a} is palindromic). Useful in bioinformatics.

			For n=2 the a(2)=12 solutions (non-palindromic 2-tuples over 4 distinct elements {a,c,g,t}) are: {a,c}, {a,g}, {a,t}, {c,a}, {c,g}, {c,f}, {g,a},{g,c}, {g,t}, {t,a}, {t,c}, {t,g}.

Index entries for linear recurrences with constant coefficients, signature (4,4,-16).

Cf. A000302, A056450, A233411, A242278, A240437.

Table[2^(n-1) * (2^(n+1) + (-1)^n - 3), {n, 66}]
LinearRecurrence[{4,4,-16},{0,12,48},30] (* Harvey P. Dale, May 24 2023 *)

a(n) = ((-1)^n - 3)*2^(n-1) + 4^n; \\ Michel Marcus, Aug 12 2014

concat(0, Vec(12*x^2 / ((2*x-1)*(2*x+1)*(4*x-1)) + O(x^100))) \\ Colin Barker, Aug 12 2014

a(n) = 2^(n-1) * (2^(n+1) + (-1)^n - 3).
a(n) = 4^n - 4^ceiling(n/2) = A000302(n) - A056450(n).
From Colin Barker, Aug 12 2014: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3).
G.f.: 12*x^2 / ((2*x-1)*(2*x+1)*(4*x-1)). (End)

Typos in formula fixed by Colin Barker, Aug 12 2014

cp's OEIS Frontend

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.

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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.

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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.

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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.

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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.

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A321391 Array read by antidiagonals: T(n,k) is the number of achiral rows of n colors using up to k colors.

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A032087 Number of reversible strings with n beads of 4 colors. If more than 1 bead, not palindromic.

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A056460 Number of primitive (aperiodic) palindromes using a maximum of four different symbols.

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.