A014986 -id:A014986 - cp's OEIS frontend

A077925 Expansion of 1/((1-x)(1+2x)).

1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, -1365, 2731, -5461, 10923, -21845, 43691, -87381, 174763, -349525, 699051, -1398101, 2796203, -5592405, 11184811, -22369621, 44739243, -89478485, 178956971, -357913941, 715827883, -1431655765, 2863311531, -5726623061

Offset: 0

N. J. A. Sloane, Nov 17 2002

a(n+1) is the reflection of a(n) through a(n-1) on the numberline. - Floor van Lamoen, Aug 31 2004
If a zero is added as the (new) a(0) in front, the sequence represents the inverse binomial transform of A001045. Partial sums are in A077898. - R. J. Mathar, Aug 30 2008
a(n) = A077953(2*n+3). - Reinhard Zumkeller, Oct 07 2008
Related to the Fibonacci sequence by an INVERT transform: if A(x) = 1+x^2*g(x) is the generating function of the a(n) prefixed with 1, 0, then 1/A(x) = 2+(x+1)/(x^2-x+1) is the generating function of 1, 0, -1, 1, -2, 3, ..., the signed Fibonacci sequence A000045 prefixed with 1. - Gary W. Adamson, Jan 07 2011
Also: Gaussian binomial coefficients [n+1,1], or q-integers, for q=-2, diagonal k=1 in the triangular (or column r=1 in the square) array A015109. - M. F. Hasler, Nov 04 2012
With a leading zero, 0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, ... we obtain the Lucas U(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
Let m = a(n). Then 18*m^2 - 12*m + 1 = A000225(2n+3). - Roderick MacPhee, Jan 17 2013

			G.f. = 1 - x + 3*x^2 - 5*x^3 + 11*x^4 - 21*x^5 + 43*x^6 - 85*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (-1,2)
Index entries related to Gaussian binomial coefficients.
Index entries for Lucas sequences

Cf. A001045 (unsigned version).
Cf. A014983, A014985, A014986. - Zerinvary Lajos, Dec 16 2008
Cf. A232600, A232601, A232602.

[(1-(-2)^(n+1))/3: n in [0..40]]; // Vincenzo Librandi, Jun 21 2011

a:=n->sum ((-2)^j, j=0..n): seq(a(n), n=0..35); # Zerinvary Lajos, Dec 16 2008

CoefficientList[Series[(1 - x)^(-1)/(1 + 2 x), {x, 0, 50}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 20 2011 *)

a(n)=(1+(-2)^n*2)/3 \\ Charles R Greathouse IV, Jun 21 2011

[gaussian_binomial(n,1,-2) for n in range(1,35)] # Zerinvary Lajos, May 28 2009

G.f.: 1/(1+x-2*x^2).
a(n) = (1-(-2)^(n+1))/3. - Vladeta Jovovic, Apr 17 2003
a(n) = Sum_{k=0..n} (-2)^k. - Paul Barry, May 26 2003
a(n+1) - a(n) = A122803(n). - R. J. Mathar, Aug 30 2008
a(n) = Sum_{k=0..n} A112555(n,k)*(-2)^k. - Philippe Deléham, Sep 11 2009
a(n) = A082247(n+1) - 1. - Philippe Deléham, Oct 07 2009
G.f.: Q(0)/(3*x), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k + 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k-1 + 2*x)/( x*(4*k+1 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
E.g.f.: (2*exp(-2*x) + exp(x))/3. - Ilya Gutkovskiy, Nov 12 2016
a(n) = A086893(n+2) - A061547(n+3), n >= 0. - Yosu Yurramendi, Jan 16 2017
a(n) = (-1)^n*A001045(n+1). - M. F. Hasler, Feb 13 2020
a(n) - a(n-1) = a(n-1) - a(n+1) = (-2)^n, a(n+1) = - a(n) + 2*a(n-1) = 1 - 2*a(n). - Michael Somos, Feb 22 2023

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

Original entry on oeis.org

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

A015113 Triangle of q-binomial coefficients for q=-5.

Original entry on oeis.org

1, 1, 1, 1, -4, 1, 1, 21, 21, 1, 1, -104, 546, -104, 1, 1, 521, 13546, 13546, 521, 1, 1, -2604, 339171, -1679704, 339171, -2604, 1, 1, 13021, 8476671, 210302171, 210302171, 8476671, 13021, 1, 1, -65104, 211929796, -26279294704, 131649159046

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), A014986 (k=1), A015255 (k=2), A015272, A015291, A015309, A015327, A015344, A015360, A015377, A015391 (k=10), A015409, A015427,... - M. F. Hasler, Nov 04 2012

Cf. analog triangles for other q: A015109 (q=-2), A015110 (q=-3), A015112 (q=-4), A015116 (q=-6), 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, -5], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 09 2016 *)

T015113(n, k, q=-5)=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

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

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

A179897 a(n) = (n^(2*n+1) + 1) / (n+1).

Original entry on oeis.org

1, 1, 11, 547, 52429, 8138021, 1865813431, 593445188743, 250199979298361, 135085171767299209, 90909090909090909091, 74619186937936447687211, 73381705110822317661638341, 85180949465178001182799643437, 115244915978498073437814463065839, 179766618030828831251710653305053711

Offset: 0

Martin Saturka (martin(AT)saturka.net), Jul 31 2010

a(n) is the arithmetic mean of the multiset consisting of n lots of 1/n and one lot of n^(2*n+1). This multiset also has an integer valued geometric mean which is equal to n for n > 0.

According to search at OEIS for particular sequence members, a(n) is also: (1+2*n)-th q-integer for q=-n, (2*(n+1))-th cyclotomic polynomial at q=-n, Gaussian binomial coefficient [2*n+1, 2*n] for q=-n, number of walks of length 1+2*n between any two distinct vertices of the complete graph K_(n+1).

			For n = 2, a(2) = 11 which is the arithmetic mean of {1/2, 1/2, 2^5} = 33 / 3 = 11. The geometric mean is 8^(1/3) = 2, i.e. both are integral.

Andrew Howroyd, Table of n, a(n) for n = 0..100
Google Groups, Integer-valued arithmetic and geometric means of sequences with non-integer numbers

Main diagonal of A362783.

Values for n = 5, 6 via other ways. Q-integers: A014986, A014987, K_n paths: A015531, A015540, Cyclotomic polynomials: A020504, A020505, Gaussian binomial coefficients: A015391, A015429.

a(n) = (n^(2*n + 1) + 1)/(n + 1) \\ Andrew Howroyd, May 03 2023

[(n**(2*n+1)+1)//(n+1) for n in range(1,11)]

a(n) = Sum_{i=0..2*n} (-n)^i.

Edited, a(0)=1 prepended and more terms from Andrew Howroyd, May 03 2023

cp's OEIS Frontend

A077925 Expansion of 1/((1-x)*(1+2*x)).

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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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A015113 Triangle of q-binomial coefficients for q=-5.

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

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A077925 Expansion of 1/((1-x)(1+2x)).