A038503 -id:A038503 - cp's OEIS frontend

A090408 a(n) = Sum_{k=0..n} binomial(4n+3,4k).

1, 36, 496, 8256, 130816, 2098176, 33550336, 536887296, 8589869056, 137439215616, 2199022206976, 35184376283136, 562949936644096, 9007199321849856, 144115187807420416, 2305843010287435776, 36893488143124135936, 590295810375885520896, 9444732965670570950656

Offset: 0

Paul Barry, Nov 29 2003

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (12,64).

Cf. A070775, A001025, A090407, A038503.

Table[Sum[Binomial[4n+3,4k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Feb 12 2020 *)

a(n) = 2^(1+4*n) - (-1)^n*4^n. - R. J. Mathar, Nov 27 2014

A091917 Coefficient array of polynomials (z-1)^n-1.

Original entry on oeis.org

1, -2, 1, 0, -2, 1, -2, 3, -3, 1, 0, -4, 6, -4, 1, -2, 5, -10, 10, -5, 1, 0, -6, 15, -20, 15, -6, 1, -2, 7, -21, 35, -35, 21, -7, 1, 0, -8, 28, -56, 70, -56, 28, -8, 1, -2, 9, -36, 84, -126, 126, -84, 36, -9, 1, 0, -10, 45, -120, 210, -252, 210, -120, 45, -10, 1, -2, 11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1

Offset: 0

Paul Barry, Feb 13 2004

The first element has been changed to 1 to produce an invertible matrix. Alternatively, this is the coefficient array for the polynomials P(z,n) = Product_{j=0..n-1} (z-(1+w(n)^j)) where w(n) = e^(2*Pi*i/n), i=sqrt(-1).
The row entries determine interesting recurrences. For instance, a(n) = 4a(n-1) + 6a(n-2) + 4a(n-3), a(0)=a(1)=a(2)=1, gives A038503. Sequences of the form a(n) = Sum_{k=0..n} (binomial(n,k) if k mod m = r, otherwise 0), for r=0..m-1, result. Equivalently, a(n) = Sum_{j=0..n-1} 2^n*(cos(Pi*j/m))^n*cos((n-2r)Pi*j/m)/m, r=0..m-1. These include A024493, A024494, A024495, A038503, A038504, A038505. The inverse matrix is A091918.
Triangle T(n,k), 0 <= k <= n, read by rows given by [ -2, 2, 1/2, -1/2, 0, 0, 0, 0, 0, ...] DELTA [1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 11 2007

			Rows begin:
  { 1},
  {-2,  1},
  { 0, -2,  1},
  {-2,  3, -3,  1},
  { 0, -4,  6, -4,  1},
  ...

Alois P. Heinz, Rows n = 0..140, flattened

T:= n-> `if`(n=0, 1, (p-> seq(coeff(p,z,i), i=0..n))((z-1)^n-1)):
seq(T(n), n=0..12);  # Alois P. Heinz, May 23 2015

Table[If[n == 0, 1, CoefficientList[(z-1)^n-1, z]], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 08 2016 *)

row(n) = if (n==0, 1, Vecrev((z-1)^n-1)); \\ Michel Marcus, May 23 2015

T(n,k) = T(n-1,k-1) + T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,1) = T(2,2) = 1, T(1,0) = T(2,1) = -2, T(2,0) = 0, T(n,k) = 0 for k > n or for k < 0. - Philippe Deléham, May 23 2015

G.f.: (1-2*x-x^2+x^2*y)/((x-1)*(-x+x*y-1)). - R. J. Mathar, Aug 11 2015

A094266 LQTL Lean Quaternary Temporal Logic: a terse form of temporal logic created by assigning four descriptors such that false, becoming true, true and becoming false are represented and become a linear sequence. In a branching tree two alternative are open, change or no change. The integer sequence above is the count of the row possibilities of the four states over successive iterations.

Original entry on oeis.org

1, 1, 0, 0, 1, 2, 1, 0, 1, 3, 3, 1, 2, 4, 6, 4, 6, 6, 10, 10, 16, 12, 16, 20, 36, 28, 28, 36, 72, 64, 56, 64, 136, 136, 120, 120, 256, 272, 256, 240, 496, 528, 528, 496, 992, 1024, 1056, 1024, 2016, 2016, 2080, 2080, 4096, 4032, 4096, 4160, 8256, 8128, 8128, 8256, 16512

Offset: 0

Robert H Barbour and L. D. Painter, Jun 01 2004

This is a table read by rows of length 4. Every row is formed from the previous one by the circular Pascal triangle-like rule: a, b, c, d -> d+a, a+b, b+c, c+d. Consider a labeled binary tree such that the root has label 0 and every node labeled k has children labeled k and (k+1) mod 4; the n-th row of this sequence counts nodes on the level n+1 with labels 0, 1, 2, 3, while the n-th row of A099423 counts nodes up to level n. - Andrey Zabolotskiy, Jan 06 2023

Robert H. Barbour, 2D Four-Color Cellular Automaton, NKS 2006 Wolfram Science Conference.
Robert H. Barbour, Two-dimensional Four Color Cellular Automaton: Surface Explorations, Complex Systems, 17 (2007), 103-112.

Cf. A000749, A005418, A038503, A038504, A038505, A020522, A063376.

Algorithm available from Robert H Barbour

Appears to satisfy a 12-degree linear recurrence. - Ralf Stephan, Dec 04 2004

A100213 Expansion of g.f.: x(4-7x+2x^2-8x^4+16x^5-16x^6)/((1-2x) (1-2x^2) (1-2x+2x^2) * (1+2*x^2)).

Original entry on oeis.org

4, 9, 14, 18, 32, 64, 128, 256, 544, 1104, 2144, 4128, 8192, 16384, 32768, 65536, 131584, 263424, 525824, 1049088, 2097152, 4194304, 8388608, 16777216, 33562624, 67129344, 134242304, 268443648, 536870912, 1073741824, 2147483648, 4294967296, 8590065664

Offset: 1

Creighton Dement, Nov 11 2004

The sequence can be created applying the pos operator (which sums over the positive coefficients) to the n-th power of the Floretion element (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e).

			a(5) = 32 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^5 = - 2 'j - 2 'k - 2 j' - 2 k' + 6 'ii' + 10 'jj' + 10 'kk' + 6 e,
and the sum of all positive coefficients is 6+10+10+6 = 32.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Creighton Dement, Floretion Online Multiplier.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,4,-16,24,-16).

Cf. A000079, A009116, A009545, A038503, A077957, A077966, A100212, A100215, A100216.

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( x*(4-7*x+2*x^2-8*x^4+16*x^5-16*x^6)/((1-4*x+6*x^2-4*x^3)*(1-4*x^4)) )); // G. C. Greubel, Mar 29 2024

Rest[CoefficientList[Series[x(4-7x+2x^2-8x^4+16x^5-16x^6)/((1-2x)(1-2x^2)(1-2x+2x^2)(1+2x^2)),{x,0,40}],x]] (* or *) LinearRecurrence[{4,-6,4,4,-16,24,-16},{4,9,14,18,32,64,128},40] (* Harvey P. Dale, Aug 23 2015 *)

def A100213_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(4-7*x+2*x^2-8*x^4+16*x^5-16*x^6)/((1-4*x+6*x^2-4*x^3)*(1-4*x^4)) ).list()
a=A100213_list(51); a[1:] # G. C. Greubel, Mar 29 2024

a(n) = A100215(n) - A100212(n).
a(n) = (-1)^n*A009116(n+3) + A100216 + A038503(n+1).
Equation above in Floretian Algebra operator speak: (pos) + (neg) = (ves) = (jes) + (les) + (tes)
a(n-1) = A000079(n+1) + (5*A077957(n) + 6*A077957(n-1))/4 + A009545(n)/2 + A009545(n+1) + A077966(n-1) - A077966(n)/4. - R. J. Mathar, May 07 2008
From G. C. Greubel, Mar 29 2024: (Start)
a(n) = (1/16)*( 2^(n+4) - 2*((1+5*i)*(1+i)^n + (1-5*i)*(1-i)^n) + (1 - (-1)^n)*2^((n+1)/2)*(5+i^(n+1)) + (1+(-1)^n)*2^(1+n/2)*(3-2*i^n) ).
a(2*n-1) = 2^(n-3)*( 2^(n+2) + 5 + (-1)^n - 6*cos(n*Pi/2) + 4*sin(n*Pi/2) ), for n >= 1.
a(2*n) = 2^(n-2)*( 2^(n+2) + 3 - 2*(-1)^n - cos(n*Pi/2) + 5*sin(n*Pi/2) ), n >= 1.
E.g.f.: -1 + exp(2*x) + (1/8)*(6*cosh(sqrt(2)*x) + 5*sqrt(2)* sinh(sqrt(2)*x) - (4*cos(sqrt(2)*x) + sqrt(2)*sin(sqrt(2)*x)) - 2*exp(x)*(cos(x) - 5*sin(x)) ). (End)

Replaced definition with generating function, changed offset to 1. - R. J. Mathar, Mar 12 2010

A100215 Expansion of (4 - 7x + 2x^2)/((1-2x)(1 - 2x + 2x^2)).

Original entry on oeis.org

4, 9, 14, 18, 24, 44, 104, 248, 544, 1104, 2144, 4128, 8064, 16064, 32384, 65408, 131584, 263424, 525824, 1049088, 2095104, 4189184, 8382464, 16775168, 33562624, 67129344, 134242304, 268443648, 536838144

Offset: 0

Creighton Dement, Nov 11 2004

a(n) = (-1)^n*A009116(n+3) + A100216(n) + A038503(n+1), where A009116, A100216 and A038503 can be generated by the operators jes, les and tes of the Floretion algebra, which is a product factor space Q x Q /{(1,1), (-1,-1)}.

Binomial transform of the sequence 4,5,0,-1 (repeated with period length 4). - R. J. Mathar, Apr 18 2009

			a(2) = 14 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 = 1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e and the sum of these coefficients is 1 + 1 + 1 + 1 + 3 + 2 + 2 + 1 + 1 + 1 = 14 (see comment).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Creighton Dement, Floretion Online Multiplier.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A009116, A038503, A099087, A100213, A100214, A100216.

I:=[4, 9, 14]; [n le 3 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3): n in [1..35]]; // Vincenzo Librandi, Jun 25 2012

LinearRecurrence[{4,-6,4},{4,9,14},40] (* Vincenzo Librandi, Jun 25 2012 *)

A099087=BinaryRecurrenceSequence(2,-2,1,2)
def A100215(n): return 2^(n+1) + 2*A099087(n) + A099087(n-1)
[A100215(n) for n in range(41)] # G. C. Greubel, Mar 29 2024

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3).
a(n) = (-1)^n*A009116(n+3) + A100216(n) + A038503(n+1).
a(n) = vesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e), where ves sums over all floretion basis vector coefficients.
a(n) = 2^(n+1) + 2*A099087(n) + A099087(n-1). - R. J. Mathar, Apr 18 2009

Definition replaced with the more precise g.f. by R. J. Mathar, Nov 17 2010

A100216 Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).

Original entry on oeis.org

1, 4, 9, 16, 26, 44, 84, 176, 376, 784, 1584, 3136, 6176, 12224, 24384, 48896, 98176, 196864, 393984, 787456, 1573376, 3144704, 6288384, 12578816, 25163776, 50335744, 100675584, 201342976, 402661376, 805289984, 1610563584, 3221159936

Offset: 0

Creighton Dement, Nov 11 2004

A100215(n) (ves) = ((-1)^n)*A009116(n+3) (jes) + a(n) (les) + A038503(n+1) (tes) (Sn, below, corresponds to the generating function from above). Coefficients of Sn(z)*(1-z)/(1+z) gives match to A038504 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1"). Coefficients of Sn(z)/(1+z) gives match to A038505 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 2"). Coefficients of Sn(z)/(1-z^2) gives match to A000749 (Number of strings over Z_2 of length n with trace 1 and subtrace 1). The elements 'i, 'j, 'k, i', j', k', 'ii', 'jj', 'kk', 'ij', 'ik', 'ji', 'jk', 'ki', 'kj', e ("floretions") are members of the quaternion product factor space Q x Q /{(1,1), (-1,-1)}. "les" sums over coefficients belonging to basis vectors which squared give the unit e (excluding e itself).

This sequence is identical to its 4th differences. - Jean-François Alcover, Nov 07 2013

			a(2) = 9 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 =
1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e
and the sum of the coefficients belonging to basis vectors which squared give the unit e (excluding e itself) is 3+2+2+1+1 = 9 (see comment).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A000749, A009116, A009545, A038503, A038504, A038505, A100213, A100214, A100215.

[n le 3 select n^2 else 4*Self(n-1) -6*Self(n-2) +4*Self(n-3): n in [1..40]]; // G. C. Greubel, Mar 28 2024

a:= n-> (<<0|1|0>, <0|0|1>, <4|-6|4>>^n. <<1, 4, 9>>)[1, 1]:
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 07 2013

d = 4; nmax = 31; a[n_ /; n < d] := (n + 1)^2; seq = Table[a[n], {n, 0, nmax}]; seq /. Solve[ Thread[ Take[seq, nmax - d + 1] == Differences[seq, d]]] // First (* Jean-François Alcover, Nov 07 2013 *)
LinearRecurrence[{4,-6,4}, {1,4,9}, 41] (* G. C. Greubel, Mar 28 2024 *)

@CachedFunction
def a(n): # a = A100216
    if n<3: return (n+1)^2
    else: return 4*a(n-1) -6*a(n-2) +4*a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Mar 28 2024

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), with a(0) = 1, a(1) = 4, a(2) = 9.
G.f.: (1-x^2)/((1-2*x)*(1-2*x+2*x^2)).
(a(n)) = lesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e).
2*a(n) = 3*2^n - A009545(n+1) + 4*A009545(n). - R. J. Mathar, May 21 2019
E.g.f.: (1/2)*exp(x)*(3*sin(x) - cos(x) + 3*exp(x)). - G. C. Greubel, Mar 28 2024

A323346 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = 1 (including e = 1) in Clifford algebra Cl(p,q)(R).

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 2, 6, 10, 10, 6, 6, 12, 16, 20, 16, 12, 16, 28, 28, 36, 36, 28, 28, 36, 64, 56, 64, 72, 64, 56, 64, 72, 136, 120, 120, 136, 136, 120, 120, 136, 136, 272, 256, 240, 256, 272, 256, 240, 256, 272, 256, 528, 528, 496, 496, 528, 528, 496, 496, 528, 528, 496

Offset: 0

Jianing Song, Jan 12 2019

See A323100 for an introduction of Clifford algebras.

			Table begins
p\q|  0   1   2    3    4    5  ...
---+-------------------------------
0  |  1,  1,  1,   2,   6,  16, ...
1  |  2,  3,  4,   6,  12,  28, ...
2  |  3,  6, 10,  16,  28,  56, ...
3  |  4, 10, 20,  36,  64, 120, ...
4  |  6, 16, 36,  72, 136, 256, ...
5  | 12, 28, 64, 136, 272, 528, ...
...
See A323100 for an example that shows T(1,3) = 6.

Jianing Song, Antidiagonals n = 0..99, flattened
Wikipedia, Clifford algebras

Cf. A038503(n+1) (first row), A038504(n+1) (first column), A007582 (main diagonal).

A323100 is the complement sequence.

s := sqrt(2): h := n -> [ 0, -s, -2, -s, 0, s, 2,  s][1 + modp(n+1, 8)]:
T := proc(n, k) option remember;
if n = 0 then return 2^k*(1 - 1/2) - 2^((k - 3)/2)*h(k + 2) fi;
if k = 0 then return 2^n*(1 - 1/2) - 2^((n - 3)/2)*h(n) fi;
T(n, k-1) + T(n-1, k) end:
for n from 0 to 9 do seq(T(n, k), k=0..9) od; # Peter Luschny, Jan 12 2019

T[n_, k_] := 2^(n + k) - Sum[Binomial[n, i] Binomial[k, j] Mod[Binomial[i - j, 2], 2], {i, 0, n}, {j, 0, k}];
Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jun 19 2019 *)

T(p,q) = sum(i=0, p, sum(j=0, q, binomial(p, i)*binomial(q, j)*!(binomial(i-j, 2)%2)))

T(p,q) = Sum_{i=0..p} Sum_{j=0..q} binomial(p, i)*binomial(q, j)*(1 - (binomial(i - j, 2) mod 2)).

T(p,q) = 2^(p+q) - A323100(p,q).

A373906 a(n) = Sum_{k=0..floor(n/4)} binomial(n+2k,n-4k).

Original entry on oeis.org

1, 1, 1, 1, 2, 8, 29, 85, 212, 476, 1016, 2172, 4825, 11213, 26763, 64095, 151851, 354737, 820328, 1889968, 4361521, 10106859, 23509678, 54793282, 127709888, 297336790, 691382201, 1606284377, 3731020629, 8668253125, 20146856893, 46840732201, 108918637566, 253262275888

Offset: 0

Seiichi Manyama, Jun 22 2024

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-14,6,-1).

Cf. A003269, A003522, A005252, A038503, A099131.

Cf. A107025, A373904, A373905.

a(n) = sum(k=0, n\4, binomial(n+2*k, n-4*k));

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6).

G.f.: 1/(1 - x - x^4/(1 - x)^5).

A081276 Floor(n^3/8).

Original entry on oeis.org

0, 0, 1, 3, 8, 15, 27, 42, 64, 91, 125, 166, 216, 274, 343, 421, 512, 614, 729, 857, 1000, 1157, 1331, 1520, 1728, 1953, 2197, 2460, 2744, 3048, 3375, 3723, 4096, 4492, 4913, 5359, 5832, 6331, 6859, 7414, 8000, 8615, 9261, 9938, 10648, 11390, 12167, 12977

Offset: 0

Paul Barry, Mar 15 2003

a(2n) = n^3.

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

Cf. A002620, A011863, A038503.

[Floor(n^3/8): n in [0..50]]; // Vincenzo Librandi, Aug 07 2013

Floor[Range[0,50]^3/8] (* or *) LinearRecurrence[ {3,-3,1,0,0,0,0,1,-3,3,-1},{0,0,1,3,8,15,27,42,64,91,125},50] (* Harvey P. Dale, Jan 27 2012 *)

a(n) = floor(n^3/8).
G.f.: x^2*(-x^3-2*x^5+3*x^4+1+4*x^6+2*x^2-2*x^7+x^8)/((-1+x)^4*(1+x)*(1+x^2)*(x^4+1)). - R. J. Mathar, Jun 26 2009
a(0)=0, a(1)=0, a(2)=1, a(3)=3, a(4)=8, a(5)=15, a(6)=27, a(7)=42, a(8)=64, a(9)=91, a(10)=125, a(n)=3*a(n-1)-3*a(n-2)+a(n-3)+a(n-8)- 3*a(n-9)+ 3*a(n-10)-a (n-11). - Harvey P. Dale, Jan 27 2012

A098173 Triangle T(n,k) with diagonals T(n,n-k) = binomial(n, 4k).

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 5, 1, 0, 0, 0, 0, 0, 15, 1, 0, 0, 0, 0, 0, 0, 35, 1, 0, 0, 0, 0, 0, 0, 1, 70, 1, 0, 0, 0, 0, 0, 0, 0, 9, 126, 1, 0, 0, 0, 0, 0, 0, 0, 0, 45, 210, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 165, 330, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 495, 495, 1

Offset: 0

Paul Barry, Aug 30 2004

Row sums are A038503.

			Rows begin
  {1},
  {0,1},
  {0,0,1},
  {0,0,0,1},
  {0,0,0,1,1},
  {0,0,0,0,5,1},
  ...

Seiichi Manyama, Rows n = 0..139, flattened

Cf. A098158, A098172.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n, 4*(n-k)) ))); # G. C. Greubel, Mar 15 2019

[[Binomial(n, 4*(n-k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Mar 15 2019

Table[Binomial[n, 4(n-k)], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 15 2019 *)

{T(n, k) = binomial(n, 4*(n-k))}; \\ G. C. Greubel, Mar 15 2019

[[binomial(n, 4*(n-k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 15 2019

Triangle T(n, k) = binomial(n, 4(n-k)).

cp's OEIS Frontend

A090408 a(n) = Sum_{k=0..n} binomial(4n+3,4k).

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A091917 Coefficient array of polynomials (z-1)^n-1.

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A100213 Expansion of g.f.: x*(4-7*x+2*x^2-8*x^4+16*x^5-16*x^6)/((1-2*x) * (1-2*x^2) * (1-2*x+2*x^2) * (1+2*x^2)).

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A100215 Expansion of (4 - 7*x + 2*x^2)/((1-2*x)*(1 - 2*x + 2*x^2)).

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A100216 Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).

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A323346 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = 1 (including e = 1) in Clifford algebra Cl(p,q)(R).

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A100213 Expansion of g.f.: x(4-7x+2x^2-8x^4+16x^5-16x^6)/((1-2x) (1-2x^2) (1-2x+2x^2) * (1+2*x^2)).

A100215 Expansion of (4 - 7x + 2x^2)/((1-2x)(1 - 2x + 2x^2)).

A373906 a(n) = Sum_{k=0..floor(n/4)} binomial(n+2k,n-4k).