A100214 -id:A100214 - cp's OEIS frontend

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

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

cp's OEIS Frontend

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

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

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cp's OEIS Frontend

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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Magma

Maple

Mathematica

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