A009116 -id:A009116 - 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

A120743 a(n) = (1/2)(1 + 3i)^n + (1/2)(1 - 3i)^n where i = sqrt(-1).

Original entry on oeis.org

1, -8, -26, 28, 316, 352, -2456, -8432, 7696, 99712, 122464, -752192, -2729024, 2063872, 31417984, 42197248, -229785344, -881543168, 534767104, 9884965888, 14422260736, -70005137408, -284232882176, 131585609728, 3105500041216

Offset: 1

Creighton Dement, Jun 11 2007

From R. J. Mathar, Jun 15 2007: (Start)
These are the row sums of the triangle A013610 after every 2nd column is deleted, then every 2nd column reversed in sign, creating an intermediate irregular triangle with entries C(n,2*k)*(-9)^k, k = 0..floor(n/2):
  1;
  1,   -9;
  1,  -27;
  1,  -54,    81;
  1,  -90,   405;
  1, -135,  1215,    -729;
  1, -189,  2835,   -5103;
  1, -252,  5670,  -20412,   6561;
  1, -324, 10206,  -61236,  59049;
  1, -405, 17010, -153090, 295245, -59049; (End)
Floretion Algebra Multiplication Program, FAMP Code: 2tesseq[A*B] with A = + 1.5i' + .5j' + .5k' + .5e and B = 'ji' + e

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Index entries for linear recurrences with constant coefficients, signature (2, -10).

Cf. A006495

[ n eq 1 select 1 else n eq 2 select -8 else 2*Self(n-1) -10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 24 2011

LinearRecurrence[{2,-10}, {1,-8}, 30] (* G. C. Greubel, Nov 09 2018 *)

x='x+O('x^30); Vec((1-10*x)/(1-2*x+10*x^2)) \\ G. C. Greubel, Nov 09 2018

a(n) = 2*a(n-1) - 10*a(n-2).
G.f.: x*(1-10*x)/(10*x^2 - 2*x + 1).
a(n) mod 9 = 1. - Paul Curtz, Apr 20 2011
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(9*k+1)/(x*(9*k+10) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013
E.g.f.: exp(x)*cos(3*x). - Sergei N. Gladkovskii, May 29 2013
a(n) = A190958(n)-10*A190958(n-1). - R. J. Mathar, Dec 13 2022

a(0)=1: a(n) is main diagonal of A009116(n). - Paul Curtz, Jul 22 2011

Edited by Jon E. Schoenfield, Nov 09 2018

A100212 Expansion of 4x^4(2 + x)/(1 - 2x + 2x^2 - 4x^4 + 8x^5 - 8*x^6).

Original entry on oeis.org

0, 0, 0, 0, 8, 20, 24, 8, 0, 0, 0, 0, 128, 320, 384, 128, 0, 0, 0, 0, 2048, 5120, 6144, 2048, 0, 0, 0, 0, 32768, 81920, 98304, 32768, 0, 0, 0, 0, 524288, 1310720, 1572864, 524288, 0, 0, 0, 0, 8388608, 20971520, 25165824, 8388608, 0, 0, 0, 0, 134217728, 335544320

Offset: 0

Creighton Dement, Nov 08 2004

a(n) = 0 iff n == {0, 1, 2 or 3} (mod 8). - Robert G. Wilson v, Nov 12 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,0,4,-8,8).

Cf. A038503, A009116, A100213.

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0] cat Coefficients(R!( 4*x^4*(2+x)/(1-2*x+2*x^2-4*x^4+8*x^5-8*x^6) )); // G. C. Greubel, Apr 01 2024

CoefficientList[ Series[4*x^4*(2+x)/(1-2*x+2*x^2-4*x^4+8*x^5-8*x^6), {x, 0, 55}], x] (* Robert G. Wilson v, Nov 12 2004 *)
LinearRecurrence[{2,-2,0,4,-8,8},{0,0,0,0,8,20},60] (* Harvey P. Dale, Oct 10 2012 *)

Vec(4*x^4*(2+x)/(1-2*x+2*x^2-4*x^4+8*x^5-8*x^6)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

def A100212_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 4*x^4*(2+x)/(1-2*x+2*x^2-4*x^4+8*x^5-8*x^6) ).list()
A100212_list(60) # G. C. Greubel, Apr 01 2024

a(8n+4) = a(8n+7) = 2^(4n+3), a(8n+5) = (5/2)*2^(4n+3), a(8n+6) = 3*2^(4n+3), a(8n+8) = 0, a(8n+9) = 0, a(8n+10) = 0, a(8n+11) = 0.
(a(n)) = negseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)
a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=8, a(5)=20, a(n) = 2*a(n-1) - 2*a(n-2) + 4*a(n-4) - 8*a(n-5) + 8*a(n-6). - Harvey P. Dale, Oct 10 2012

More terms from Robert G. Wilson v, Nov 12 2004

A108086 Triangle, read by rows, where T(0,0) = 1, T(n,k) = (-1)^(n+k)*T(n-1,k) + T(n-1,k-1); a signed version of Pascal's triangle.

Original entry on oeis.org

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

Offset: 0

Gerald McGarvey, Jun 05 2005

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000984, A001700, A002054, A007318, A009116.

Cf. A090132, A108520, A260192, A333378.

A108086:= func< n,k | (-1)^Floor((n-k+1)/2)*Binomial(n,k) >;
[A108086(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 02 2022

A108086[n_, k_]:= (-1)^(Floor[(n-k+1)/2])*Binomial[n, k];
Table[A108086[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 02 2022 *)

def A108086(n,k): return (-1)^int((n-k+1)/2)*binomial(n,k)
flatten([[A108086(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 02 2022

T(n,k) = (-1)^(n+k)*T(n-1,k) + T(n-1,k-1), with T(0, 0) = 1.
T(n, k) = (-1)^floor((n-k+1)/2) * A007318(n, k).
From G. C. Greubel, Dec 02 2022: (Start)
T(2*n, n) = (-1)^binomial(n+1,2) * A000984(n).
T(2*n, n+1) = (-1)^binomial(n,2) * A001791(n), n >= 1.
T(2*n, n-1) = (-1)^binomial(n+2,2) * A001791(n).
T(2*n+1, n-1) = (-1)^binomial(n-1,2) * A002054(n).
T(2*n+1, n+1) = (-1)^binomial(n+1,2) * A001700(n+1).
Sum_{k=0..n} T(n, k) = (-1)^n * A090132(n).
Sum_{k=0..n} (-1)^k * T(n, k) = A108520(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = (-1)^n * A260192(n-1).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = A333378(n+1). (End)

A133212 a(n) = 4a(n-1) - 6a(n-2) + 4*a(n-3), n > 3; a(0) = 1, a(1) = 4, a(2) = 12, a(3) = 32.

Original entry on oeis.org

1, 4, 12, 32, 72, 144, 272, 512, 992, 1984, 4032, 8192, 16512, 33024, 65792, 131072, 261632, 523264, 1047552, 2097152, 4196352, 8392704, 16781312, 33554432, 67100672, 134201344, 268419072, 536870912, 1073774592, 2147549184

Offset: 0

Paul Curtz, Oct 11 2007

Conjecture: a(n) = 2*A038503(n+3) if n > 0. - R. J. Mathar, Oct 23 2007

Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A009116, A038503, A099087.

A133212 := proc(n) option remember ; if n <= 3 then op(n+1,[1,4,12,32]) ; else 4*A133212(n-1)-6*A133212(n-2)+4*A133212(n-3) ; fi ; end: seq(A133212(n),n=0..50) ; # R. J. Mathar, Oct 23 2007

Join[{1},LinearRecurrence[{4, -6, 4},{4, 12, 32},29]] (* Ray Chandler, Sep 23 2015 *)

Sequence is identical to its fourth differences.
From R. J. Mathar, Nov 18 2007: (Start)
G.f.: -(1 + 2*x^2 + 4*x^3)/((2*x - 1)*(2*x^2 - 2*x + 1)). - [Corrected by Georg Fischer, May 12 2019]
a(n) = -2*(-1)^n*A009116(n)+3*2^n. (End)
E.g.f.: exp(x)*(3*cosh(x) - 2*(cos(x) + sin(x)) + 5*sinh(x)). - Stefano Spezia, Jan 03 2023

More terms from R. J. Mathar, Oct 23 2007

A134813 a(n) = b(n+1)-2b(n) where b() is A134812.

Original entry on oeis.org

1, -3, 3, 0, -6, 12, -12, 0, 24, -48, 48, 0, -96, 192, -192, 0, 384, -768, 768, 0, -1536, 3072, -3072, 0, 6144, -12288, 12288, 0, -24576, 49152, -49152, 0, 98304, -196608, 196608, 0, -393216, 786432, -786432, 0, 1572864, -3145728, 3145728, 0, -6291456

Offset: 0

Paul Curtz, Jan 28 2008

sign

Index entries for linear recurrences with constant coefficients, signature (-2,-2)

Cf. A003945.

LinearRecurrence[{-2,-2},{1,-3,3},50] (* Harvey P. Dale, Mar 22 2018 *)

G.f. 1 -3*x*(1+x) / ( 1+2*x+2*x^2 ). - R. J. Mathar, Aug 11 2012

a(n) = -3*A009116(n-1), n>0. - R. J. Mathar, Aug 11 2012

More terms from Harvey P. Dale, Mar 22 2018

A138231 A009545 alternated with its first differences.

Original entry on oeis.org

0, 1, 1, 1, 2, 0, 2, -2, 0, -4, -4, -4, -8, 0, -8, 8, 0, 16, 16, 16, 32, 0, 32, -32, 0, -64, -64, -64, -128, 0, -128, 128, 0, 256, 256, 256, 512, 0, 512, -512, 0, -1024, -1024, -1024, -2048, 0, -2048, 2048, 0, 4096, 4096, 4096, 8192, 0, 8192, -8192, 0, -16384, -16384

Offset: 0

Paul Curtz, May 05 2008

sign

The unsigned sequence contains 3 copies of the strictly positive powers of 2.

Index entries for linear recurrences with constant coefficients, signature (0, 2, 0, -2).

Cf. A000079.

With[{c=LinearRecurrence[{2,-2},{0,1},40]},Riffle[c,Differences[c]]] (* or *) LinearRecurrence[{0,2,0,-2},{0,1,1,1},80] (* Harvey P. Dale, Oct 15 2017 *)

a(2n)= A009545(n). a(2n+1)= A009545(n+1)-A009545(n) = (-1)^(n+1)*A009116(n).
a(n) = 2a(n-2)-2a(n-4). a(n) = -4a(n-8), n > 7.
O.g.f.: x(1+x-x^2)/(1-2x^2+2x^4). - R. J. Mathar, Jul 08 2008

Edited by R. J. Mathar, Jul 08 2008

A009769 Expansion of tanh(log(1+1/x)).

Original entry on oeis.org

1, 0, -4, 24, -96, 0, 5760, -80640, 645120, 0, -116121600, 2554675200, -30656102400, 0, 11158821273600, -334764638208000, 5356234211328000, 0, -3278015337332736000, 124564582818643968000, -2491291656372879360000, 0

Offset: 0

R. H. Hardin

Cf. A009014, A009116.

nn = 26; Range[0, nn]! CoefficientList[Series[Tanh[Log[1 + 1/x]], {x, 0, nn}], x] (* T. D. Noe, Oct 05 2011 *)

E.g.f.: (2*x+1)/(2*x^2+2*x+1) = 1-4*x^2/2!+24*x^3/3!-96*x^4/4!+....
Recurrence: a(n) = -2*n*a(n-1)-2*n*(n-1)*a(n-2), a(0) = 1, a(1) = 0.
a(n) = -n!/2*((-1+i)^(n+1) + (-1-i)^(n+1)) = -n!*sqrt(2)^(n+1)* cos(3*Pi*(n+1)/4).
a(n) = 2^n*A009014(n). a(n) = -n!*A009116(n+1).
For x > -1/2 we have (2*x+1)/(2*x^2+2*x+1) = 2*int {t = 0..inf} exp(-t*(2*x+1))*cos(t). Using this we obtain a(n) = 2*(-2)^n*int {t = 0..inf} t^n*exp(-t)*cos(t). - Peter Bala, Oct 05 2011

Extended with signs by Olivier Gérard, Mar 15 1997

A133209 a(n) = 4a(n-1) - 6a(n-2) + 4a(n-3), n > 3; a(0) = 3, a(1) = 2, a(2) = a(3) = 0.

Original entry on oeis.org

3, 2, 0, 0, 8, 32, 80, 160, 288, 512, 960, 1920, 3968, 8192, 16640, 33280, 66048, 131072, 261120, 522240, 1046528, 2097152, 4198400, 8396800, 16785408, 33554432, 67092480, 134184960, 268402688, 536870912, 1073807360, 2147614720

Offset: 0

Paul Curtz, Oct 11 2007

Index entries for linear recurrences with constant coefficients, signature (4, -6, 4).

Cf. A131470, A009116, A099087.

a[0]:=3: a[1]:=2: a[2]:=0: a[3]:=0; for n from 4 to 27 do a[n]:=4*a[n-1]-6*a[n-2]+4*a[n-3] end do: seq(a[n],n=0..27); # Emeric Deutsch, Oct 14 2007

a = {3, 2, 0, 0}; Do[AppendTo[a, 4*a[[ -1]] - 6*a[[ -2]] + 4*a[[ -3]]], {30}]; a (* Stefan Steinerberger, Oct 14 2007 *)
LinearRecurrence[{4, -6, 4},{3, 2, 0},32] (* Ray Chandler, Sep 23 2015 *)

Sequence is identical to its fourth differences.
a(n) = 2^n + 2^[(n+3)/2]*cos((n+1)Pi/4); a(n)=2^n + (1+i)^(n+1) + (1-i)^(n+1), where i=sqrt(-1). - Emeric Deutsch, Oct 14 2007
G.f.: -(3-10*x+10*x^2)/(2*x-1)/(2*x^2-2*x+1). - R. J. Mathar, Nov 14 2007

More terms from Stefan Steinerberger and Emeric Deutsch, Oct 14 2007

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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A120743 a(n) = (1/2)*(1 + 3*i)^n + (1/2)*(1 - 3*i)^n where i = sqrt(-1).

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A100212 Expansion of 4*x^4*(2 + x)/(1 - 2*x + 2*x^2 - 4*x^4 + 8*x^5 - 8*x^6).

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A108086 Triangle, read by rows, where T(0,0) = 1, T(n,k) = (-1)^(n+k)*T(n-1,k) + T(n-1,k-1); a signed version of Pascal's triangle.

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A133212 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3; a(0) = 1, a(1) = 4, a(2) = 12, a(3) = 32.

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A134813 a(n) = b(n+1)-2b(n) where b() is A134812.

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

A120743 a(n) = (1/2)(1 + 3i)^n + (1/2)(1 - 3i)^n where i = sqrt(-1).

A100212 Expansion of 4x^4(2 + x)/(1 - 2x + 2x^2 - 4x^4 + 8x^5 - 8*x^6).

A133212 a(n) = 4a(n-1) - 6a(n-2) + 4*a(n-3), n > 3; a(0) = 1, a(1) = 4, a(2) = 12, a(3) = 32.