A014987 -id:A014987 - 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

A015116 Triangle of q-binomial coefficients for q=-6.

Original entry on oeis.org

1, 1, 1, 1, -5, 1, 1, 31, 31, 1, 1, -185, 1147, -185, 1, 1, 1111, 41107, 41107, 1111, 1, 1, -6665, 1480963, -8838005, 1480963, -6665, 1, 1, 39991, 53308003, 1910490043, 1910490043, 53308003, 39991, 1, 1, -239945, 1919128099, -412612541285

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 A000012 (k=0), A014987 (k=1), A015257 (k=2), A015273, A015292, A015310, A015328, A015345, A015361, A015378, A015392 (k=10), A015410, A015429,... - M. F. Hasler, Nov 04 2012

Cf. analog triangles for other q: A015109 (q=-2), A015110 (q=-3), A015112 (q=-4), A015113 (q=-5), A015117 (q=-7), A015118 (q=-8), A015121 (q=-9), A015123 (q=-10), A015124 (q=-11), A015125 (q=-12), A015129 (q=-13), A015132 (q=-14), A015133 (q=-15); 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 04 2012

Table[QBinomial[n, k, -6], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 09 2016 *)

T015116(n, k, q=-6)=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

A014994 a(n) = (1 - (-12)^n)/13.

Original entry on oeis.org

1, -11, 133, -1595, 19141, -229691, 2756293, -33075515, 396906181, -4762874171, 57154490053, -685853880635, 8230246567621, -98762958811451, 1185155505737413, -14221866068848955, 170662392826187461

Offset: 1

Olivier Gérard

q-integers for q=-12.

Vincenzo Librandi, Table of n, a(n) for n = 1..900
Index entries for linear recurrences with constant coefficients, signature (-11,12).

Cf. A077925, A014983, A014985-A014987, A014989-A014993.

I:=[1,-11]; [n le 2 select I[n] else -11*Self(n-1)+12*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012

a:=n->sum ((-12)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008

LinearRecurrence[{-11, 12}, {1, -11}, 30] (* Vincenzo Librandi, Oct 22 2012 *)

a(n)=(1-(-12)^n)/13 \\ Charles R Greathouse IV, Sep 24 2015

[gaussian_binomial(n,1,-12) 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 + 12*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -11*a(n-1) + 12*a(n-2). - Vincenzo Librandi, Oct 22 2012
E.g.f.: (exp(x) - exp(-12*x))/13. - G. C. Greubel, May 26 2018

Better name from Ralf Stephan, Jul 14 2013

A015540 a(n) = 5a(n-1) + 6a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 5, 31, 185, 1111, 6665, 39991, 239945, 1439671, 8638025, 51828151, 310968905, 1865813431, 11194880585, 67169283511, 403015701065, 2418094206391, 14508565238345, 87051391430071, 522308348580425, 3133850091482551, 18803100548895305, 112818603293371831

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct vertices of the complete graph K_7. Example: a(2)=5 because the walks of length 2 between the vertices A and B of the complete graph ABCDEFG are ACB, ADB, AEB, AFB and AGB. - Emeric Deutsch, Apr 01 2004
General form: k=6^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500. -Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Pisano period lengths: 1, 1, 2, 2, 2, 2, 14, 2, 2, 2, 10, 2, 12, 14, 2, 2, 16, 2, 18, 2, ... - R. J. Mathar, Aug 10 2012
Sum_{i=0..m} (-1)^(m+i)*6^i, for m >= 0, gives all terms after 0. - Bruno Berselli, Aug 28 2013
The ratio a(n+1)/a(n) converges to 6 as n approaches infinity. Also A053524, A080424, A051958. - Felix P. Muga II, Mar 09 2014

			G.f. = x + 5*x^2 + 31*x^3 + 185*x^4 + 1111*x^5 + 6665*x^6 + 39991*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
Index entries for linear recurrences with constant coefficients, signature (5,6).

Partial sums are in A033116. Cf. A014987.

[Floor(6^n/7-(-1)^n/7): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

seq(round(6^n/7),n=0..25); # Mircea Merca, Dec 28 2010

k=0; lst={k}; Do[k = 6^n-k; AppendTo[lst, k], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
CoefficientList[Series[x / ((1 - 6 x) (1 + x)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)
LinearRecurrence[{5,6},{0,1},30] (* Harvey P. Dale, May 12 2015 *)

my(x='x+O('x^30)); concat([0], Vec(x/((1-6*x)*(1+x)))) \\ G. C. Greubel, Jan 24 2018

a(n) = round(6^n/7); \\ Altug Alkan, Sep 05 2018

[lucas_number1(n,5,-6) for n in range(21)] # Zerinvary Lajos, Apr 24 2009

a(n) = 5*a(n-1) + 6*a(n-2).
From Paul Barry, Apr 20 2003: (Start)
a(n) = (6^n - (-1)^n)/7.
G.f.: x/((1-6*x)*(1+x)).
E.g.f.: (exp(6*x) - exp(-x))/7. (End)
a(n) = 6^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
a(n+1) = Sum_{k=0..n} binomial(n-k, k)*5^(n-2*k)*6^k. - Paul Barry, Jul 29 2004
a(n) = round(6^n/7). - Mircea Merca, Dec 28 2010
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-7)^k = (6^n - (-1)^n)/7 = (-1)^(n-1)*Sum_{k=0..n-1} (-6)^k. Equals (-1)^(n-1)*Phi(n,-6), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014

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

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1

Offset: 0

Tom Copeland, Mar 19 2014

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J. Adams, On the groups J(x)-II, Topology, Vol. 3, p. 137-171, Pergamon Press, (1965).
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mathoverflow, Inequality for Laguerre polynomials

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.
Cf. A049310, A112468, A341091.

[[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018

A239473 := proc(n,k)
    add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
end proc; # R. J. Mathar, Jul 21 2016

Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)

for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018

Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
     = Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
     = Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
     = Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
     = Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
     = Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) -  (x / (e^x-1)) ] / x =  Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Inverse array added by Tom Copeland, Mar 26 2014

Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

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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A015116 Triangle of q-binomial coefficients for q=-6.

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

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

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A014994 a(n) = (1 - (-12)^n)/13.

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A015540 a(n) = 5a(n-1) + 6a(n-2), a(0) = 0, a(1) = 1.