A014994 -id:A014994 - cp's OEIS frontend

A014983 a(n) = (1 - (-3)^n)/4.

0, 1, -2, 7, -20, 61, -182, 547, -1640, 4921, -14762, 44287, -132860, 398581, -1195742, 3587227, -10761680, 32285041, -96855122, 290565367, -871696100, 2615088301, -7845264902, 23535794707, -70607384120, 211822152361, -635466457082, 1906399371247

Offset: 0

Olivier Gérard

q-integers for q=-3.
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-3, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^n*charpoly(A,0). - Milan Janjic, Jan 27 2010
Pisano period lengths:  1, 2, 1, 4, 4, 2, 3, 8, 1, 4, 10, 4, 6, 6, 4, 16, 16, 2, 9, 4, ... - R. J. Mathar, Aug 10 2012

T. D. Noe, Table of n, a(n) for n = 0..200
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 927
László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
R. A. Sulanke, Moments of generalized Motzkin paths, J. Integer Sequences, Vol. 3 (2000), #00.1.
Index entries for linear recurrences with constant coefficients, signature (-2,3)

Cf. A077925, A014985, A014986, A014987, A014989, A014990, A014991, A014992, A014993, A014994. - Zerinvary Lajos, Dec 16 2008

[(1-(-3)^n)/4: n in [0..30]]; // G. C. Greubel, May 26 2018

a:=n->sum ((-3)^j, j=0..n): seq(a(n), n=-1..25); # Zerinvary Lajos, Dec 16 2008

nn = 25; CoefficientList[Series[x/((1 - x)*(1 + 3*x)), {x, 0, nn}], x] (* T. D. Noe, Jun 21 2012 *)
Table[(1 - (-3)^n)/4, {n, 0, 27}] (* Michael De Vlieger, Nov 23 2016 *)

PARI
```
a(n)=(1-(-3)^n)/4
```

[gaussian_binomial(n,1,-3) for n in range(0,27)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + (-3)^(n-1).
G.f.: x/((1-x)*(1+3*x)).
a(n) = -(-1)^n*A015518(n).
a(n) = the (1, 2)-th element of M^n, where M = ((1, 1, 1, -2), (1, 1, -2, 1), (1, -2, 1, 1), (-2, 1, 1, 1)). - Simone Severini, Nov 25 2004
a(0)=0, a(1)=1, a(n) = -2*a(n-1) + 3*a(n-2) for n>1. - Philippe Deléham, Sep 19 2009
From Sergei N. Gladkovskii, Apr 29 2012: (Start)
G.f. A(x) = G(0)/4; G(k) = 1 - 1/(3^(2*k) - 3*x*3^(4*k)/(3*x*3^(2*k) + 1/(1 + 1/(3*3^(2*k) - 3^(3)*x*3^(4*k)/(3^2*x*3^(2*k) - 1/G(k+1)))))); (continued fraction, 3rd kind, 6-step).
E.g.f. E(x) = G(0)/4; G(k) = 1 - 1/(9^k - 3*x*81^k/(3*x*9^k + (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k - (2*k+2)/G(k+1)))))); (continued fraction, 3rd kind, 6-step). (End)
a(n) = A084222(n) - 1. - Filip Zaludek, Nov 19 2016
E.g.f.: sinh(x)*cosh(x)*exp(-x). - Ilya Gutkovskiy, Nov 20 2016

A014985 a(n) = (1 - (-4)^n)/5.

Original entry on oeis.org

1, -3, 13, -51, 205, -819, 3277, -13107, 52429, -209715, 838861, -3355443, 13421773, -53687091, 214748365, -858993459, 3435973837, -13743895347, 54975581389, -219902325555, 879609302221, -3518437208883, 14073748835533

Offset: 1

Olivier Gérard

q-integers for q=-4.
In Penrose's book, presented as partial sums of the series for 1/(1-x^2) evaluated at x=2. - Olivier Gérard, May 22 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-3, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Jan 27 2010

Roger Penrose, "The Road to Reality, A complete guide to the Laws of the Universe", Jonathan Cape, London, 2004, pages 79-80. - Olivier Gérard, May 22 2009

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-3,4).

Cf. A077925, A014983, A014986, A014987, A014989-A014994.

I:=[1, -3]; [n le 2 select I[n] else -3*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 21 2012

a:=n->sum ((-4)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

LinearRecurrence[{-3, 4}, {1, -3}, 50] (* or *) CoefficientList[ Series[ 1/((1-x)*(1+4*x)), {x,0,30}], x] (* Vincenzo Librandi, Oct 21 2012 *)
(1-(-4)^Range[30])/5 (* Harvey P. Dale, Oct 06 2024 *)

a(n)=(1-(-4)^n)/5 \\ Charles R Greathouse IV, Sep 24 2015

[gaussian_binomial(n,1,-4) for n in range(1,24)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}, with q=-4.
From Paul Barry, Jan 12 2007: (Start)
G.f.: x/(1+3*x-4*x^2).
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*4^k*(-3)^(n-2k). (End)
a(n) = -3*a(n-1) +4*a(n-2). - Vincenzo Librandi, Oct 21 2012

G.f. adapted to the offset by Vincenzo Librandi, Oct 21 2012

Better name from Ralf Stephan, Jul 14 2013

A014987 a(n) = (1 - (-6)^n)/7.

Original entry on oeis.org

1, -5, 31, -185, 1111, -6665, 39991, -239945, 1439671, -8638025, 51828151, -310968905, 1865813431, -11194880585, 67169283511, -403015701065, 2418094206391, -14508565238345, 87051391430071, -522308348580425, 3133850091482551

Offset: 1

Olivier Gérard

q-integers for q=-6.

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,2). - Milan Janjic, Jan 27 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (-5,6).

Absolute values are in A015540.

Cf. A077925, A014983, A014985, A014986, A014989-A014994.

I:=[1,-5]; [n le 2 select I[n] else -5*Self(n-1)+6*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-6)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

LinearRecurrence[{-5, 6}, {1, -5}, 30] (* Vincenzo Librandi Oct 22 2012 *)

a(n)=(1-(-6)^n)/7 \\ Charles R Greathouse IV, Sep 24 2015

[gaussian_binomial(n,1,-6) for n in range(1,22)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
G.f.: x/((1+6*x)*(1-x)).
a(n) = -5*a(n-1) + 6*a(n-2). - Vincenzo Librandi Oct 22 2012
E.g.f.: (exp(x) - exp(-6*x))/7. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

A015125 Triangle of q-binomial coefficients for q=-12.

Original entry on oeis.org

1, 1, 1, 1, -11, 1, 1, 133, 133, 1, 1, -1595, 19285, -1595, 1, 1, 19141, 2775445, 2775445, 19141, 1, 1, -229691, 399683221, -4793193515, 399683221, -229691, 1, 1, 2756293, 57554154133, 8283038077141, 8283038077141, 57554154133, 2756293, 1, 1

Offset: 0

Olivier Gérard

May be read as a symmetric triangular (T[n,k]=T[n,n-k]; k=0,...,n; n=0,1,...) or square array (A[n,r]=A[r,n]=T[n+r,r], read by antidiagonals). The diagonals of the former, or rows/columns of the latter, are, for k=0,...,12: A000012, A014994, A015264, A015281, A015302, A015319, A015336, A015354, A015369, A015384, A015401, A015421, A015436. - M. F. Hasler, Nov 04 2012

Index entries related to Gaussian binomial coefficients.

Cf. analog triangles for other negative q=-2,...,-15: A015109 (q=-2), A015110 (q=-3), A015112 (q=-4), A015113 (q=-5), A015116 (q=-6), A015117 (q=-7), A015118 (q=-8), A015121 (q=-9), A015123 (q=-10), A015124 (q=-11), A015129 (q=-13), A015132 (q=-14), A015133 (q=-15). - M. F. Hasler, Nov 04 2012

Cf. analog triangles for positive q=2,...,24: A022166 (q=2), A022167 (q=3), A022168, A022169, A022170, A022171, A022172, A022173, A022174 (q=10), A022175, A022176, A022177, A022178, A022179, A022180, A022181, A022182, A022183, A022184 (q=20), A022185, A022186, A022187, A022188. - M. F. Hasler, Nov 05 2012

T015125(n, k, q=-12)=prod(i=1, k, (q^(1+n-i)-1)/(q^i-1)) \\ (Indexing is that of the triangular array: 0 <= k <= n = 0,1,2,...) - M. F. Hasler, Nov 04 2012

A014986 a(n) = (1 - (-5)^n)/6.

Original entry on oeis.org

1, -4, 21, -104, 521, -2604, 13021, -65104, 325521, -1627604, 8138021, -40690104, 203450521, -1017252604, 5086263021, -25431315104, 127156575521, -635782877604, 3178914388021, -15894571940104, 79472859700521

Offset: 1

Olivier Gérard

q-integers for q = -5.

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 27 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-4,5).

Cf. A077925, A014983, A014985, A014987, A014989, A014990, A014991, A014992, A014993, A014994.

I:=[1, -4]; [n le 2 select I[n] else -4*Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 19 2012

a:=n->sum ((-5)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

LinearRecurrence[{-4,5},{1,-4},30] (* Vincenzo Librandi, Jun 19 2012 *)

a(n)=(1-(-5)^n)/6 \\ Charles R Greathouse IV, Dec 07 2011

[gaussian_binomial(n,1,-5) for n in range(1,22)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
G.f.: x/((1-x)*(1+5*x)). - Bruno Berselli, Dec 07 2011
a(n) = -4*a(n-1) + 5*a(n-2). - Vincenzo Librandi, Jun 19 2012
E.g.f.: (exp(x) - exp(-5*x))/6. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

A014989 a(n) = (1 - (-7)^n)/8.

Original entry on oeis.org

1, -6, 43, -300, 2101, -14706, 102943, -720600, 5044201, -35309406, 247165843, -1730160900, 12111126301, -84777884106, 593445188743, -4154116321200, 29078814248401, -203551699738806, 1424861898171643, -9974033287201500

Offset: 1

Olivier Gérard

q-integers for q = -7.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (-6,7).

Cf. A077925, A014983, A014985-A014987, A014990-A014994.

I:=[1,-6]; [n le 2 select I[n] else -6*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-7)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

LinearRecurrence[{-6, 7}, {1, -6}, 30] (* Vincenzo Librandi, Oct 22 2012 *)

x='x+O('x^30); Vec(x/((1-x)*(1+7*x))) \\ G. C. Greubel, May 26 2018

[gaussian_binomial(n,1,-7) for n in range(1,21)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
a(n) = -6*a(n-1) + 7*a(n-2). - Vincenzo Librandi, Oct 22 2012
From G. C. Greubel, May 26 2018: (Start)
G.f.: x/((1-x)*(1+7*x)).
E.g.f.: (exp(x) - exp(-7*x))/8. (End)

Better name from Ralf Stephan, Jul 14 2013

A014990 a(n) = (1 - (-8)^n)/9.

Original entry on oeis.org

1, -7, 57, -455, 3641, -29127, 233017, -1864135, 14913081, -119304647, 954437177, -7635497415, 61083979321, -488671834567, 3909374676537, -31274997412295, 250199979298361, -2001599834386887, 16012798675095097

Offset: 1

Olivier Gérard

q-integers for q=-8.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-7,8).

Cf. A015565, A077925, A014983, A014985, A014986, A014987, A014989, A014991, A014992, A014993, A014994.

I:=[1, -7]; [n le 2 select I[n] else -7*Self(n-1) +8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-8)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

QBinomial[Range[20],1,-8] (* or *) LinearRecurrence[{-7,8},{1,-7},20] (* Harvey P. Dale, Dec 19 2011 *)

a(n)=(1-(-8)^n)/9 \\ Charles R Greathouse IV, Oct 07 2015

[gaussian_binomial(n,1,-8) for n in range(1,20)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}
From Philippe Deléham, Feb 13 2007: (Start)
a(1)=1, a(2)=-7, a(n) = -7*a(n-1) + 8*a(n-2) for n > 2.
a(n) = (-1)^(n+1)*A015565(n).
G.f.: x/(1 + 7*x - 8*x^2). (End)
E.g.f.: (exp(x) - exp(-8*x))/9. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

A014992 a(n) = (1 - (-10)^n)/11.

Original entry on oeis.org

1, -9, 91, -909, 9091, -90909, 909091, -9090909, 90909091, -909090909, 9090909091, -90909090909, 909090909091, -9090909090909, 90909090909091, -909090909090909, 9090909090909091, -90909090909090909

Offset: 1

Olivier Gérard

q-integers for q = -10.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-9,10).

Cf. A077925, A014983, A014985, A014986, A014987, A014989, A014990, A014991, A014993, A014994. - Zerinvary Lajos, Dec 16 2008

I:=[1, -9]; [n le 2 select I[n] else -9*Self(n-1) +10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-10)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

CoefficientList[Series[1/((1 - x)*(1 + 10*x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 22 2012 *)

for(n=1, 30, print1((1-(-10)^n)/11, ", ")) \\ G. C. Greubel, May 26 2018

[gaussian_binomial(n,1,-10) for n in range(1,19)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
G.f.: x/((1 - x)*(1 + 10*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -9*a(n-1) + 10*a(n-2). - Vincenzo Librandi, Oct 22 2012
a(n) = (-1)^(n+1)*A015585(n). - R. J. Mathar, Oct 26 2015
E.g.f.: (exp(x) - exp(-10*x))/11. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

A014991 a(n) = (1 - (-9)^n)/10.

Original entry on oeis.org

1, -8, 73, -656, 5905, -53144, 478297, -4304672, 38742049, -348678440, 3138105961, -28242953648, 254186582833, -2287679245496, 20589113209465, -185302018885184, 1667718169966657, -15009463529699912

Offset: 1

Olivier Gérard

q-integers for q = -9.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-8,9).

Cf. A077925, A014983, A014985-A014987, A014989-A014994. - Zerinvary Lajos, Dec 16 2008

I:=[1,-8]; [n le 2 select I[n] else -8*Self(n-1)+9*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-9)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

((-9)^Range[30]-1)/-10 (* or *) LinearRecurrence[{-8,9},{1,-8},30] (* Harvey P. Dale, Aug 08 2011 *)
CoefficientList[Series[1/((1 - x)*(1 + 9*x)), {x, 0, 30}], x]; (* Vincenzo Librandi, Oct 22 2012 *)

for(n=1,30, print1((1-(-9)^n)/10, ", ")) \\ G. C. Greubel, May 26 2018

[gaussian_binomial(n,1,-9) for n in range(1,19)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
a(0)=1, a(1)=-8, a(n) = -8*a(n-1) + 9*a(n-2). - Harvey P. Dale, Aug 08 2011
G.f.: x/((1 - x)*(1 + 9*x)). - Vincenzo Librandi, Oct 22 2012
E.g.f.: (exp(x) - exp(-9*x))/10. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

A014993 a(n) = (1 - (-11)^n)/12.

Original entry on oeis.org

1, -10, 111, -1220, 13421, -147630, 1623931, -17863240, 196495641, -2161452050, 23775972551, -261535698060, 2876892678661, -31645819465270, 348104014117971, -3829144155297680, 42120585708274481

Offset: 1

Olivier Gérard

q-integers for q = -11.

Vincenzo Librandi, Table of n, a(n) for n = 1..900
Index entries for linear recurrences with constant coefficients, signature (-10,11).

Cf. A077925, A014983, A014985, A014986, A014987, A014989, A014990, A014991, A014992, A014994. - Zerinvary Lajos, Dec 16 2008

I:=[1, -10]; [n le 2 select I[n] else -10*Self(n-1) +11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-11)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

LinearRecurrence[{-10, 11}, {1, -10}, 40] (* Vincenzo Librandi, Oct 22 2012 *)

for(n=1,30, print1((1-(-11)^n)/12, ", ")) \\ G. C. Greubel, May 26 2018

[gaussian_binomial(n,1,-11) for n in range(1,18)] # Zerinvary Lajos, May 28 2009

a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}.
G.f.: x/((1 - x)*(1 + 11*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -10*a(n-1) + 11*a(n-2). - Vincenzo Librandi, Oct 22 2012
E.g.f.: (exp(x) - exp(-11*x))/12. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

cp's OEIS Frontend

A014983 a(n) = (1 - (-3)^n)/4.

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A014985 a(n) = (1 - (-4)^n)/5.

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A014987 a(n) = (1 - (-6)^n)/7.

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A015125 Triangle of q-binomial coefficients for q=-12.

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A014986 a(n) = (1 - (-5)^n)/6.

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Maple

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A014989 a(n) = (1 - (-7)^n)/8.

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Maple