A007910 -id:A007910 - cp's OEIS frontend

A137505 Inverse binomial transform of A007910.

1, 1, 0, 2, 0, 0, 4, -4, 4, 4, -12, 20, -12, -12, 52, -76, 52, 52, -204, 308, -204, -204, 820, -1228, 820, 820, -3276, 4916, -3276, -3276, 13108, -19660, 13108, 13108, -52428, 78644, -52428, -52428, 209716, -314572, 209716, 209716, -838860, 1258292, -838860, -838860, 3355444, -5033164, 3355444

Offset: 0

Paul Curtz, Apr 23 2008

sign

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,0,2).

LinearRecurrence[{-1,0,2},{1,1,0},50] (* Harvey P. Dale, Sep 17 2012 *)

Recurrence: a(n) = -a(n-1) + 2a(n-3), starting 1,1,0.
O.g.f.: (1+x)^2/((1-x)(1+2x+2x^2)). - R. J. Mathar, Jun 12 2008
a(4n) = a(4n+1) = (-1)^n*A109499(n). - Paul Curtz, Nov 01 2009
a(n) = (1/5) * (A137429(n-1) + 4) = A077973(n-2) + 2*A077973(n-1) + A077973(n). - Ralf Stephan, Aug 18 2013

More terms from R. J. Mathar, Jun 12 2008

A137500 Binomial transform of b(n) = (0, 0, A007910).

Original entry on oeis.org

0, 0, 1, 5, 17, 51, 149, 439, 1309, 3927, 11797, 35423, 106301, 318903, 956645, 2869807, 8609293, 25827879, 77483893, 232452191, 697357085, 2092071255, 6276212741, 18828636175, 56485906477, 169457719431, 508373162389, 1525119495359, 4575358494269, 13726075482807

Offset: 0

Paul Curtz, Apr 27 2008

b(n) is binomial transform of (0, 0, A077973).

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-8,6).

Cf. A007910, A009545.

LinearRecurrence[{5,-8,6},{0,0,1},40] (* Harvey P. Dale, Sep 27 2020 *)

concat([0,0], Vec(1/((1 - 3*x)*(1 - 2*x + 2*x^2)) + O(x^40))) \\ Andrew Howroyd, Jan 03 2020

a(n) = 3*a(n-1) + A009545(n-1) for n > 0.
From Andrew Howroyd, Jan 03 2020: (Start)
a(n) = Sum_{k=0..n-2} binomial(n, k+2)*A007910(k).
a(n) = 5*a(n-1) - 8*a(n-2) + 6*a(n-3) for n >= 3.
G.f.: x*2/((1 - 3*x)*(1 - 2*x + 2*x^2)). (End)

Terms a(11) and beyond from Andrew Howroyd, Jan 03 2020

A381255 Positive integers not of the form round(2^(k+2)/5). Complement of A007910.

Original entry on oeis.org

4, 5, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74

Offset: 1

Chai Wah Wu, Apr 21 2025

nonn

Cf. A007910.

def A381255(n): return (m:=n-2+(k:=(5*n+3).bit_length()))+(m>=((1<

a(n) = m+1 if m>=floor((2^k+2)/5) otherwise a(n) = m where k = floor(log_2(5*n+3))+1 and m = n-2+k.

A113405 Expansion of x^3/(1 - 2x + x^3 - 2x^4) = x^3/( (1-2x)(1+x)*(1-x+x^2) ).

Original entry on oeis.org

0, 0, 0, 1, 2, 4, 7, 14, 28, 57, 114, 228, 455, 910, 1820, 3641, 7282, 14564, 29127, 58254, 116508, 233017, 466034, 932068, 1864135, 3728270, 7456540, 14913081, 29826162, 59652324, 119304647, 238609294, 477218588, 954437177, 1908874354, 3817748708

Offset: 0

Paul Barry, Oct 28 2005

A transform of the Jacobsthal numbers. A059633 is the equivalent transform of the Fibonacci numbers.
Paul Curtz, Aug 05 2007, observes that the inverse binomial transform of 0,0,0,1,2,4,7,14,28,57,114,228,455,910,1820,... gives the same sequence up to signs. That is, the extended sequence is an eigensequence for the inverse binomial transform (an autosequence).
The round() function enables the closed (non-recurrence) formula to take a very simple form: see Formula section. This can be generalized without loss of simplicity to a(n) = round(b^n/c), where b and c are very small, incommensurate integers (c may also be an integer fraction). Particular choices of small integers for b and c produce a number of well-known sequences which are usually defined by a recurrence - see Cross Reference. - Ross Drewe, Sep 03 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, arXiv:math/0205301 [math.CO], 2002. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
N. J. A. Sloane, Transforms
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,2).

From Ross Drewe, Sep 03 2009: (Start)
Other sequences a(n) = round(b^n / c), where b and c are very small integers:
A001045 b = 2; c = 3
A007910 b = 2; c = 5
A016029 b = 2; c = 5/3
A077947 b = 2; c = 7
abs(A078043) b = 2; c = 7/3
A007051 b = 3; c = 2
A015518 b = 3; c = 4
A034478 b = 5; c = 2
A003463 b = 5; c = 4
A015531 b = 5; c = 6
(End)

[Round(2^n/9): n in [0..40]]; // Vincenzo Librandi, Aug 11 2011

A010892 := proc(n) op((n mod 6)+1,[1,1,0,-1,-1,0]) ; end proc:
A113405 := proc(n) (2^n-(-1)^n)/9 -A010892(n-1)/3; end proc: # R. J. Mathar, Dec 17 2010

CoefficientList[Series[x^3/(1-2x+x^3-2x^4),{x,0,40}],x] (* or *) LinearRecurrence[{2,0,-1,2},{0,0,0,1},40] (* Harvey P. Dale, Apr 30 2011 *)

a(n)=2^n\/9 \\ Charles R Greathouse IV, Jun 05 2011

def A113405(n): return ((1<Chai Wah Wu, Apr 17 2025

a(n) = 2a(n-1) - a(n-3) + 2a(n-4).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k)*A001045(k).
a(n) = Sum_{k=0..n} binomial((n+k)/2,k)*A001045((n-k)/2)*(1+(-1)^(n-k))/2.
a(3n) = A015565(n), a(3n+1) = 2*A015565(n), a(3n+2) = 4*A015565(n). - Paul Curtz, Nov 30 2007
From Paul Curtz, Dec 16 2007: (Start)
a(n+1) - 2a(n) = A131531(n).
a(n) + a(n+3) = 2^n. (End)
a(n) = round(2^n/9). - Ross Drewe, Sep 03 2009
9*a(n) = 2^n + (-1)^n - 3*A010892(n). - R. J. Mathar, Mar 24 2018

Edited by N. J. A. Sloane, Dec 13 2007

A077854 Expansion of 1/((1-x)(1-2x)*(1+x^2)).

Original entry on oeis.org

1, 3, 6, 12, 25, 51, 102, 204, 409, 819, 1638, 3276, 6553, 13107, 26214, 52428, 104857, 209715, 419430, 838860, 1677721, 3355443, 6710886, 13421772, 26843545, 53687091, 107374182, 214748364, 429496729, 858993459, 1717986918, 3435973836, 6871947673

Offset: 0

N. J. A. Sloane, Nov 17 2002

Partial sums of A007910. - Mircea Merca, Dec 27 2010
This is the decimal representation of the middle column of "Rule 54" elementary cellular automaton. - Karl V. Keller, Jr., Sep 26 2021
This same sequence (except that the offset is changed to 4) is 2^n with the final digit chopped off. - J. Lowell, May 11 2022

			The sequence in hexadecimal shows the pattern
1, 3, 6, c,
19, 33, 66, cc,
199, 333, 666, ccc,
1999, 3333, 6666, cccc,
19999, 33333, 66666, ccccc,
199999, 333333, 666666, cccccc,
1999999, 3333333, 6666666, ccccccc,
19999999, 33333333, 66666666, cccccccc,
... - _Armands Strazds_, Oct 09 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,3,-2).

Equals A007909(n+3) - [n congruent 2, 3 mod 4].

Cf. A130306, A043291 (subsequence); A000975, A007910, A133872, A259661 (binary).

import Data.Bits (xor)
a077854 n = a077854_list !! n
a077854_list = scanl1 xor $ tail a000975_list :: [Integer]
-- Reinhard Zumkeller, Jan 04 2013

[Round((2^(n+4)-5)/10): n in [0..40]]; // Vincenzo Librandi, Jun 25 2011

a := proc(n) option remember; if n=0 then RETURN(1); fi; if n=1 then RETURN(3); fi; if n=2 then RETURN(6); fi; if n=3 then RETURN(12); fi; 3*a(n-1)-3*a(n-2)+3*a(n-3)-2*a(n-4); end;
seq(iquo(2^n,5),n=3..35); # Zerinvary Lajos, Apr 20 2008

CoefficientList[Series[1/((1 - x) (1 - 2 x) (1 + x^2)), {x, 0, 32}], x] (* Michael De Vlieger, Mar 29 2016 *)
LinearRecurrence[{3,-3,3,-2},{1,3,6,12},40] (* Harvey P. Dale, Feb 06 2019 *)

a(n)=(16<Charles R Greathouse IV, Sep 23 2012

Vec(1/(1-3*x+3*x^2-3*x^3+2*x^4)+O(x^99)) \\ Derek Orr, Oct 26 2014

print([2**(n+3)//5 for n in range(50)]) # Karl V. Keller, Jr., Sep 26 2021

a(n) = 3*a(n-1) - 3*a(n-2) + 3*a(n-3) - 2*a(n-4), with initial values a(0) = 1, a(1) = 3, a(2) = 6, a(3) = 12.
a(n) = (1/10)*(2^(n+4) + (-1)^floor(n/2) - 2*(-1)^floor((n+1)/2) - 5).
Row sums of A130306. - Gary W. Adamson, May 20 2007
a(n) = floor(2^(n+3)/5). - Gary Detlefs, Sep 06 2010
a(n) = round((2^(n+4)-5)/10) = floor((2^(n+3)-1)/5) = ceiling((2^(n+3)-4)/5) = round((2^(n+3)-2)/5); a(n) = a(n-4) + 3*2^(n-1), n > 3. - Mircea Merca, Dec 27 2010
a(n) = 2^(n+1) - 1 - a(n-2); a(n) = a(n-1)/2 for n == 2, 3 (mod 4); a(n) = (a(n-1)-1)/2 for n == 0, 1 (mod 4). - Arie Bos, Apr 06 2013
a(n) = floor(A000975(n+2)*3/5). - Armands Strazds, Oct 18 2014
a(n) = Sum_{k=1..n+3} floor(1 + sin(k*Pi/2 + 3*Pi/4))*2^(n-k+3). - Andres Cicuttin, Mar 28 2016
a(n) = (-15 + 3*2^(3+n) + 2^(1 + n - 4*floor((1+n)/4)) + 2^(2 + n - 4*floor((2+n)/4)))/15. - Andres Cicuttin, Mar 28 2016
a(n) = (16*2^n+(-1)^((2*n-1+(-1)^n)/4)-2*(-1)^((2*n+1-(-1)^n)/4)-5)/10. - Wesley Ivan Hurt, Apr 01 2016

A007909 Expansion of (1-x)/(1-2x+x^2-2x^3).

Original entry on oeis.org

1, 1, 1, 3, 7, 13, 25, 51, 103, 205, 409, 819, 1639, 3277, 6553, 13107, 26215, 52429, 104857, 209715, 419431, 838861, 1677721, 3355443, 6710887, 13421773, 26843545, 53687091, 107374183, 214748365, 429496729, 858993459, 1717986919, 3435973837, 6871947673

Offset: 0

Mogens Esrom Larsen (mel(AT)math.ku.dk)

Equals INVERT transform of (1, 0, 2, 2, 2, ...). - Gary W. Adamson, Apr 28 2009

a(n) is the number of compositions (ordered partitions) of n into parts 1 (one kind), and parts >= 3 of three kinds (no parts 2). - Joerg Arndt, Apr 22 2025

Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
M. E. Larsen, Summa Summarum, A. K. Peters, Wellesley, MA, 2007; see p. 38.

M. F. Hasler, Table of n, a(n) for n = 0..1000 (in replacement of a(0..999) indexed 1..1000 from Vincenzo Librandi).
Charles K. Cook and Michael R. Bacon, Some identities for Jacobsthal and Jacobsthal-Lucas numbers satisfying higher order recurrence relations, Annales Mathematicae et Informaticae, 41 (2013) pp. 27-39.
Shanzhen Gao, Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
I. Gessel, Problem 10424, Amer. Math. Monthly, 102 (1995), 70.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 444
Yüksel Soykan, Properties of Generalized (r, s, t, u)-Numbers, Earthline J. of Math. Sci. (2021) Vol. 5, No. 2, 297-327.
Index entries for linear recurrences with constant coefficients, signature (2,-1,2).

Cf. A005251, A007679, A007910.

I:=[1, 1, 1]; [n le 3 select I[n] else 2*Self(n-1)-Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 17 2012

U:=n->(1/5)*(2^(n+1)+3*cos(n*Pi/2)+sin(n*Pi/2)); [seq(U(n),n=0..50)];

CoefficientList[Series[(1-x)/(1-2*x+x^2-2*x^3),{x,0,40}],x] (* Vincenzo Librandi, Jun 17 2012 *)
LinearRecurrence[{2,-1,2},{1,1,1},40] (* Harvey P. Dale, Jul 26 2016 *)

a(n)=2^(n+1)\5+(n%4<2) \\ M. F. Hasler, Feb 22 2018

def A007909(n): return (2<Chai Wah Wu, Apr 22 2025

G.f.: (1-x)/(1-2*x+x^2-2*x^3).
a(n) = (1/5)*(2^(n+1)+3*cos(n*Pi/2)+sin(n*Pi/2)).
a(n) = Sum_{k=0..floor((n-1)/3)} binomial(n-k-1, 2*k)*2^k. - Paul Barry, Sep 16 2004
a(n) = (1/5)*(2^(n+1) + (-1)^[(n+1)/2] + 2*(-1)^[n/2]). - Ralf Stephan, Jun 09 2005
a(n) = 2*a(n-1)-a(n-2)+2*a(n-3). Sequence is identical to its half second differences from the second term; a(n)+a(n+2)=2^(n+1). - Paul Curtz, Dec 17 2007
a(n+1) = (2^n)*Sum_{k=1..n} (-1)^floor(k/2)/2^k. - Yalcin Aktar, Jul 20 2008

Offset corrected by M. F. Hasler, Feb 22 2018

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957

Offset: 0

M. F. Hasler, Feb 22 2018

nonn

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);

LinearRecurrence[{17, -16}, {1, 13}, 20] (* Jean-François Alcover, Feb 22 2018 *)

PARI
```
A299960(n)=4^(2*n+1)\5+1
```

def A299960(n): return ((1<<(n<<2)+2)+1)//5 # Chai Wah Wu, Jul 29 2022

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
  (53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
  (107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
  (173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
  (733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
  (997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
  (41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
  (509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
  (1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
  (773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A054106 Alternating sums of vertically aligned numbers in Pascal's triangle: T(n,k) = C(n,k) - C(n-2,k-1) + C(n-4,k-2) - ... +- C(n-2[n/2],m).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 1, 1, 4, 8, 8, 4, 1, 1, 5, 12, 15, 12, 5, 1, 1, 6, 17, 27, 27, 17, 6, 1, 1, 7, 23, 44, 55, 44, 23, 7, 1, 1, 8, 30, 67, 99, 99, 67, 30, 8, 1, 1, 9, 38, 97, 166, 197, 166, 97, 38, 9, 1, 1, 10, 47, 135, 263, 363, 363

Offset: 0

Clark Kimberling

			Rows: {1}; {1,1}; {1,1,1}; {1,2,2,1}; {1,3,5,3,1} ...

For (nonalternating) vertically aligned sums, see A013580.

Row sums of this array: A007910.

G.f.: 1/(1-(1+y)*x)/(1+y*x^2). - Vladeta Jovovic, Oct 12 2003

A100088 Expansion of (1-x^2)/((1-2x)(1+x^2)).

Original entry on oeis.org

1, 2, 2, 4, 10, 20, 38, 76, 154, 308, 614, 1228, 2458, 4916, 9830, 19660, 39322, 78644, 157286, 314572, 629146, 1258292, 2516582, 5033164, 10066330, 20132660, 40265318, 80530636, 161061274, 322122548, 644245094, 1288490188, 2576980378

Offset: 0

Paul Barry, Nov 03 2004

A Chebyshev transform of A100087, under the mapping A(x) -> ((1-x^2)/(1+x^2)) * A(x/(1+x^2)).

A176742(n+2) = A084099(n+2) = period 4:repeat 0, -2, 0, 2.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,2).

Cf. A007910, A084099, A100087, A122117, A176742.

[n le 3 select Floor((n+2)/2) else 2*Self(n-1) - Self(n-2) +2*Self(n-3): n in [1..41]]; // G. C. Greubel, Jul 08 2022

CoefficientList[Series[(1-x^2)/((1-2x)(1+x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{2,-1,2},{1,2,2},40] (* Harvey P. Dale, May 12 2011 *)

def A100088(n): return ((4<Chai Wah Wu, Apr 22 2025

def b(n): return (2/5)*(3*2^(2*n-1) + (-1)^n) # b=A122117
def A100088(n): return b(n/2) if (n%2==0) else 2*b((n-1)/2)
[A100088(n) for n in (0..60)]  # G. C. Greubel, Jul 08 2022

a(n) = (3*2^n + 2*cos(Pi*n/2) + 4*sin(Pi*n/2))/5.
a(n) = n*Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*A100087(n-2*k)/(n-k).
a(n) = 2*a(n-1) + period 4:repeat 0, -2, 0, 2, with a(0) = 1.
a(n) = A007910(n+1) - A007910(n-1).
a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3).
a(n) = (1/5)*(3*2^n + i^n*(1+(-1)^n) - 2*i^(n+1)*(1-(-1)^n)). - G. C. Greubel, Jul 08 2022
a(n) = A122117(n/2) if (n mod 2 = 0) otherwise 2*A122117((n-1)/2). - G. C. Greubel, Jul 21 2022

A118404 Triangle T, read by rows, where all columns of T are different and yet all columns of the matrix square T^2 (A118407) are equal; also equals the matrix inverse of triangle A118400.

Original entry on oeis.org

1, 1, -1, -1, 0, 1, -1, 1, -1, -1, 1, 0, 0, 2, 1, 1, -1, 0, -2, -3, -1, -1, 0, 1, 2, 5, 4, 1, -1, 1, -1, -3, -7, -9, -5, -1, 1, 0, 0, 4, 10, 16, 14, 6, 1, 1, -1, 0, -4, -14, -26, -30, -20, -7, -1, -1, 0, 1, 4, 18, 40, 56, 50, 27, 8, 1, -1, 1, -1, -5, -22, -58, -96, -106, -77, -35, -9, -1, 1, 0, 0, 6, 27, 80, 154, 202, 183, 112, 44, 10, 1, 1, -1, 0, -6, -33, -107, -234, -356, -385, -295, -156, -54, -11, -1, -1, 0, 1, 6, 39, 140, 341, 590, 741, 680, 451, 210, 65, 12, 1, -1, 1, -1, -7, -45, -179, -481, -931, -1331, -1421, -1131, -661, -275, -77, -13, -1, 1, 0, 0, 8, 52, 224, 660, 1412, 2262, 2752, 2552, 1792, 936, 352, 90, 14, 1

Offset: 0

Paul D. Hanna, Apr 27 2006

Appears to coincide with triangle (5.2) in Lee-Oh (2016), although there is no obvious connection! - N. J. A. Sloane, Dec 07 2016

			Triangle begins:
1;
1,-1;
-1, 0, 1;
-1, 1,-1,-1;
1, 0, 0, 2, 1;
1,-1, 0,-2,-3,-1;
-1, 0, 1, 2, 5, 4, 1;
-1, 1,-1,-3,-7,-9,-5,-1;
1, 0, 0, 4, 10, 16, 14, 6, 1;
1,-1, 0,-4,-14,-26,-30,-20,-7,-1;
-1, 0, 1, 4, 18, 40, 56, 50, 27, 8, 1;
-1, 1,-1,-5,-22,-58,-96,-106,-77,-35,-9,-1;
1, 0, 0, 6, 27, 80, 154, 202, 183, 112, 44, 10, 1;
1, -1, 0, -6, -33, -107, -234, -356, -385, -295, -156, -54, -11, -1;
-1, 0, 1, 6, 39, 140, 341, 590, 741, 680, 451, 210, 65, 12, 1;
-1, 1, -1, -7, -45, -179, -481, -931, -1331, -1421, -1131, -661, -275, -77, -13, -1;
1, 0, 0, 8, 52, 224, 660, 1412, 2262, 2752, 2552, 1792, 936, 352, 90, 14, 1;
1, -1, 0, -8, -60, -276, -884, -2072, -3674, -5014, -5304, -4344, -2728, -1288, -442, -104, -15, -1;
-1, 0, 1, 8, 68, 336, 1160, 2956, 5746, 8688, 10318, 9648, 7072, 4016, 1730, 546, 119, 16, 1; ...
The matrix square is A118407:
1;
0, 1;
-2, 0, 1;
2,-2, 0, 1;
0, 2,-2, 0, 1;
-2, 0, 2,-2, 0, 1;
4,-2, 0, 2,-2, 0, 1;
-6, 4,-2, 0, 2,-2, 0, 1;
4,-6, 4,-2, 0, 2,-2, 0, 1;
6, 4,-6, 4,-2, 0, 2,-2, 0, 1; ...
in which all columns are equal.

Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.

Cf. A118405 (row sums), A118406 (unsigned row sums), A118407 (matrix square), A118400 (matrix inverse).

Columns or diagonals (modulo offsets): A219977, A011848, A212342, A007598, A005581, A007910.

T[n_, k_] := SeriesCoefficient[(-1)^k/((1+x^2)(1+x)^(k-1)), {x, 0, n-k}];
Table[T[n, k], {n, 0, 16}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2018 *)

{T(n,k)=polcoeff(polcoeff((1+x)^2/(1+x^2)/(1+x+x*y +x*O(x^n)),n,x)+y*O(y^k),k,y)}
for(n=0, 16, for(k=0, n, print1(T(n, k), ", ")); print(""))

G.f.: A(x,y) = (1+x)^2 / ( (1+x^2) * (1+x + x*y) ).

G.f. of column k: (-1)^k / ( (1+x^2) * (1+x)^(k-1) ) for k>=0.

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