A106180 -id:A106180 - cp's OEIS frontend

A046854 Triangle read by rows: T(n, k) = binomial(floor((n+k)/2), k), n >= k >= 0.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 3, 4, 1, 1, 1, 3, 6, 4, 5, 1, 1, 1, 4, 6, 10, 5, 6, 1, 1, 1, 4, 10, 10, 15, 6, 7, 1, 1, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 1, 6, 21, 35, 70, 56, 84, 36, 45, 10, 11, 1, 1

Offset: 0

Wouter Meeussen

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004
Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005
As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008
From Emeric Deutsch, Jun 18 2010: (Start)
T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.
T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)
From L. Edson Jeffery, Mar 01 2011: (Start)
This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table
  1
  1
  1  1
  1  1
  1  2  1
  1  2  1
  1  3  3  1
  1  3  3  1
  ...,
then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with
  1;
  1, -1;
  1, -1, -1;
  1, -2, -1, 1;
  1, -2, -3, 1, 1;
  ...
(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.
For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)
Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011
Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012
Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014
From Jeremy Dover, Jun 07 2016: (Start)
T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.
T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

			Triangle begins:
  1;
  1 1;
  1 1 1;
  1 2 1 1;
  1 2 3 1 1;
  1 3 3 4 1 1;
  ...

J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1978. [Emeric Deutsch, Jun 18 2010]

Nathaniel Johnston, Rows 0..100, flattened
Robert G. Donnelly, Molly W. Dunkum, and Rachel McCoy, Olry Terquem's forgotten problem, arXiv:2303.05949 [math.HO], 2023.
Jeremy M. Dover, Some Notes on Pairs in Binary Strings, arXiv:1609.00980 [math.CO], 2016.
Dominique Foata and Guo-Niu Han, Multivariable Tangent and Secant q-derivative Polynomials, arXiv:1304.2486 [math.CO], 2013; see Fig. 10.1.
Index entries for triangles and arrays related to Pascal's triangle

Reflected version of A065941, which is considered the main entry. A deficient version is in A030111.
Cf. A066170, A097806, A134513, A168561, A187660.
Cf. A055244. - Gary W. Adamson, Dec 23 2008

Flat(List([0..16], n-> List([0..n], k-> Binomial(Int((n+k)/2), k) ))); # G. C. Greubel, Jul 13 2019

a046854 n k = a046854_tabl !! n !! k
a046854_row n = a046854_tabl !! n
a046854_tabl = [1] : f [1] [1,1] where
   f us vs = vs : f vs  (zipWith (+) (us ++ [0,0]) ([0] ++ vs))
-- Reinhard Zumkeller, Apr 24 2013

[Binomial(Floor((n+k)/2), k): k in [0..n], n in [0..16]]; // G. C. Greubel, Jul 13 2019

A046854:= proc(n,k): binomial(floor(n/2+k/2), k) end: seq(seq(A046854(n,k),k=0..n),n=0..16); # Nathaniel Johnston, Jun 30 2011

Table[Binomial[Floor[(n+k)/2], k], {n,0,16}, {k,0,n}]//Flatten
(* next program: Polynomials *)
z = 12; f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1;
t = Table[Factor[f[x, n]], {n, 1, z}]
u = Flatten[CoefficientList[Numerator[t], x]] (* this sequence *)
v = Flatten[CoefficientList[Denominator[t], x]]
(* Clark Kimberling, Oct 13 2014 *)

T(n,k) = binomial((n+k)\2, k); \\ G. C. Greubel, Jul 13 2019

[[binomial(floor((n+k)/2), k) for k in (0..n)] for n in (0..16)] # G. C. Greubel, Jul 13 2019

T(n,k) = binomial(floor((n+k)/2), k).
G.f.: (1+x)/(1-x*y-x^2). - Ralf Stephan, Feb 13 2005
Triangle = A097806 * A168561, as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007
T(n,k) = A065941(n,n-k) = abs(A130777(n,k)) = abs(A066170(n,k)) = abs(A187660(n,k)). - Johannes W. Meijer, Aug 08 2011
For n > 1: T(n, k) = T(n-1, k-1) + T(n-2, k), 0 < k < n-1. - Reinhard Zumkeller, Apr 24 2013
T(n,k) = A168561(n,k) + A168561(n-1,k). - R. J. Mathar, Feb 10 2024

A063886 Number of n-step walks on a line starting from the origin but not returning to it.

Original entry on oeis.org

1, 2, 2, 4, 6, 12, 20, 40, 70, 140, 252, 504, 924, 1848, 3432, 6864, 12870, 25740, 48620, 97240, 184756, 369512, 705432, 1410864, 2704156, 5408312, 10400600, 20801200, 40116600, 80233200, 155117520, 310235040, 601080390, 1202160780, 2333606220, 4667212440

Offset: 0

Henry Bottomley, Aug 28 2001

A Chebyshev transform of A007877(n+1). The g.f. is transformed to (1+x)/((1-x)(1+x^2)) under the mapping G(x)->(1/(1+x^2))G(1/(1+x^2)). - Paul Barry, Oct 12 2004
a(n-1) = 2*C(n-2, floor((n-2)/2)) is also the number of bit strings of length n in which the number of 00 substrings is equal to the number of 11 substrings. For example, when n = 4 we have 4 such bit strings: 0011, 0101, 1010, and 1100. - Angel Plaza, Apr 23 2009
Hankel transform is A120617. - Paul Barry, Aug 10 2009
The Hankel transform of a(n) is (-2)^C(n+1,2). The Hankel transform of (-1)^C(n+1,2)*a(n) is (-1)^C(n+1,2)*A164584(n). - Paul Barry, Aug 17 2009
For n > 1, a(n) is also the number of n-step walks starting from the origin and returning to it exactly once. - Geoffrey Critzer, Jan 24 2010
-a(n) is the Z-sequence for the Riordan array A130777. (See the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices). - Wolfdieter Lang, Jul 12 2011
Number of subsets of {1,...,n} in which the even elements appear as often at even positions as at odd positions. - Gus Wiseman, Mar 17 2018

			a(4) = 6 because there are six length four walks that do not return to the origin: {-1, -2, -3, -4}, {-1, -2, -3, -2}, {-1, -2, -1, -2}, {1, 2, 1, 2}, {1, 2, 3, 2}, {1, 2, 3, 4}. There are also six such walks that return exactly one time: {-1, -2, -1, 0}, {-1, 0, -1, -2}, {-1, 0, 1, 2}, {1, 0, -1, -2}, {1, 0, 1, 2}, {1, 2, 1, 0}. - _Geoffrey Critzer_, Jan 24 2010
The a(5) = 12 subsets in which the even elements appear as often at even positions as at odd positions: {}, {1}, {3}, {5}, {1,3}, {1,5}, {2,4}, {3,5}, {1,2,4}, {1,3,5}, {2,4,5}, {1,2,4,5}. - _Gus Wiseman_, Mar 17 2018

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Emeric Deutsch, Problem 11424, The American Mathematical Monthly, Vol. 116, No. 3 (March 2009), p. 277.
D. Perrin, A conjecture on rational sequences, pp. 267-274 of R. M. Capocelli, ed., Sequences, Springer-Verlag, NY 1990.

Apart from initial terms, same as A182027.
Cf. A000108, A000246, A000712, A000984, A001405, A001700, A002420, A006232, A007877, A026010, A028329, A045931, A047073, A089849, A097613, A106180, A120617, A130777, A130780, A138364, A164584, A171966, A239241, A300787, A300788, A300789.
Cf. A307768 (complementary event).

[1] cat [2*Binomial(n-1, Floor((n-1)/2)): n in [1..40]]; // G. C. Greubel, Jun 07 2023

seq(seq(binomial(2*j,j)*i, i=1..2),j=0..16); # Zerinvary Lajos, Apr 28 2007
# second Maple program:
a:= proc(n) option remember; `if`(n<2, n+1,
       4*a(n-2) +2*(a(n-1) -4*a(n-2))/n)
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Feb 10 2014
# third program:
A063886 := series(BesselI(0, 2*x)*(1 + x*2 + x*Pi*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x), x = 0, 34): seq(n!*coeff(A063886, x, n), n = 0 .. 33); # Mélika Tebni, Jun 17 2024

Table[Length[Select[Map[Accumulate, Strings[{-1, 1}, n]], Count[ #, 0] == 0 &]], {n, 0, 20}] (* Geoffrey Critzer, Jan 24 2010 *)
CoefficientList[Series[Sqrt[(1+2x)/(1-2x)],{x,0,40}],x] (* Harvey P. Dale, Apr 28 2016 *)

PARI
```
a(n)=(n==0)+2*binomial(n-1,(n-1)\2)
```

a(n) = 2^n*prod(k=0,n-1,(k/n+1/n)^((-1)^k)); \\ Michel Marcus, Dec 03 2013

from math import ceil
from sympy import binomial
def a(n):
    if n==0: return 1
    return 2*binomial(n-1,(n-1)//2)
print([a(n) for n in range(18)])
# David Nacin, Feb 29 2012

[2*binomial(n-1, (n-1)//2) + int(n==0) for n in range(41)] # G. C. Greubel, Jun 07 2023

G.f.: sqrt((1+2*x)/(1-2*x)).
a(n+1) = 2*C(n, floor(n/2)) = 2*A001405(n); a(2n) = C(2n, n) = A000984(n) = 4*a(2n-2)-|A002420(n)| = 4*a(2n-2)-2*A000108(n-1) = 2*A001700(n-1); a(2n+1) = 2*a(2n) = A028329(n).
2*a(n) = A047073(n+1).
a(n) = Sum_{k=0..n} abs(A106180(n,k)). - Philippe Deléham, Oct 06 2006
a(n) = Sum_{k=0..n} (k+1)binomial(n, (n-k)/2) ( 1-cos((k+1)*Pi/2) (1+(-1)^(n-k))/(n+k+2) ). - Paul Barry, Oct 12 2004
G.f.: 1/(1-2*x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+ ... (continued fraction). - Paul Barry, Aug 10 2009
G.f.: 1 + 2*x/(G(0)-x+x^2) where G(k)= 1 - 2*x^2 - x^4/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 10 2012
D-finite with recurrence: n*a(n) = 2*a(n-1) + 4*(n-2)*a(n-2). - R. J. Mathar, Dec 03 2012
From Sergei N. Gladkovskii, Jul 26 2013: (Start)
G.f.: 1/G(0), where G(k) = 1 - 2*x/(1 + 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: G(0), where G(k) = 1 + 2*x/(1 - 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: W(0)/2*(1+2*x), where W(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/W(k+1) )), abs(x) < 1/2; (continued fraction). (End)
a(n) = 2^n*Product_{k=0..n-1} (k/n + 1/n)^((-1)^k). - Peter Luschny, Dec 02 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k+1)/((2*k+1)*(1+2*x) - (2*k+1)*(4*k+3)*x*(1+2*x)/((4*k+3)*x + (k+1)*(1+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 19 2014
From Peter Bala, Mar 29 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-1)^(n+k)*binomial(1/2, k)*binomial(- 1/2, n-k) = 2^n * A000246(n)/n!.
a(n) = (1/2^n) * binomial(2*n, n) * hypergeom([-1/2, -n], [1/2 - n], -1). (End)
E.g.f.: BesselI(0, 2*x)*(1 + x*(2 + Pi)*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x). - Stefano Spezia, May 11 2024
a(n) = A089849(n) + A138364(n). - Mélika Tebni, Jun 17 2024
From Amiram Eldar, Aug 15 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 + Pi/(9*sqrt(3)). (End)

A105523 Expansion of 1-x*c(-x^2) where c(x) is the g.f. of A000108.

Original entry on oeis.org

1, -1, 0, 1, 0, -2, 0, 5, 0, -14, 0, 42, 0, -132, 0, 429, 0, -1430, 0, 4862, 0, -16796, 0, 58786, 0, -208012, 0, 742900, 0, -2674440, 0, 9694845, 0, -35357670, 0, 129644790, 0, -477638700, 0, 1767263190, 0

Offset: 0

Paul Barry, Apr 11 2005

Row sums of A105522. Row sums of inverse of A105438.

First column of number triangle A106180.

			G.f. = 1 - x + x^3 - 2*x^5 + 5*x^7 - 14*x^9 + 42*x^11 - 132*x^13 + 429*x^15 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, Aequat. Math., 80 (2010), 291-318. see p. 313.
R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, arXiv:1103.4936 [math.CO], 2011.

Cf. A000108, A097331, A090192, A090181, A105522, A105438.

m:=25; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1 + 2*x - Sqrt(1+4*x^2))/(2*x))); // G. C. Greubel, Sep 16 2018

A105523_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w]:=-a[w-1]+(-1)^w*add(a[j]*a[w-j-1],j=1..w-1) od; convert(a,list)end: A105523_list(40); # Peter Luschny, May 19 2011

a[n_?EvenQ] := 0; a[n_?OddQ] := 4^n*Gamma[n/2] / (Gamma[-n/2]*(n+1)!); a[0] = 1; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
CoefficientList[Series[(1 + 2 x - Sqrt[1 + 4 x^2])/(2 x), {x, 0, 50}], x] (* Vincenzo Librandi, Nov 01 2014 *)
a[ n_] := SeriesCoefficient[ (1 + 2 x - Sqrt[ 1 + 4 x^2]) / (2 x), {x, 0, n}]; (* Michael Somos, Jun 17 2015 *)
a[ n_] := If[ n < 1, Boole[n == 0], a[n] = -2 a[n - 1] + Sum[ a[j] a[n - j - 1], {j, 0, n - 1}]]; (* Michael Somos, Jun 17 2015 *)

{a(n) = local(A); if( n<0, 0, n++; A = vector(n); A[1] = 1; for( k=2, n, A[k] = -2 * A[k-1] + sum( j=1, k-1, A[j] * A[k-j])); A[n])}; /* Michael Somos, Jul 24 2011 */

def A105523(n):
    if is_even(n): return 0 if n>0 else 1
    return -(sqrt(pi)*2^(n-1))/(gamma(1-n/2)*gamma((n+3)/2))
[A105523(n) for n in (0..29)] # Peter Luschny, Oct 31 2014

G.f.: (1 + 2*x - sqrt(1+4*x^2))/(2*x).
a(n) = 0^n + sin(Pi*(n-2)/2)(C((n-1)/2)(1-(-1)^n)/2).
G.f.: 1/(1+x/(1-x/(1+x/(1-x/(1+x/(1-x.... (continued fraction). - Paul Barry, Jan 15 2009
a(n) = Sum{k = 0..n} A090181(n,k)*(-1)^k. - Philippe Deléham, Feb 02 2009
a(n) = (1/n)*sum_{i = 0..n-1} (-2)^i*binomial(n, i)*binomial(2*n-i-2, n-1). - Vladimir Kruchinin, Dec 26 2010
With offset 1, a(n) = -2 * a(n-1) + Sum_{k=1..n-1} a(k) * a(n-k), for n>1. - Michael Somos, Jul 25 2011
D-finite with recurrence: (n+3)*a(n+2) = -4*n*a(n), a(0)=1, a(1)=-1. - Fung Lam, Mar 18 2014
For nonzero terms, a(n) ~ (-1)^((n+1)/2)/sqrt(2*Pi)*2^(n+1)/(n+1)^(3/2). - Fung Lam, Mar 17 2014
a(n) = -(sqrt(Pi)*2^(n-1))/(Gamma(1-n/2)*Gamma((n+3)/2)) for n odd. - Peter Luschny, Oct 31 2014
From Peter Bala, Apr 20 2024: (Start)
a(n) = Sum_{k = 0..n} (-2)^(n-k)*binomial(n + k, 2*k)*Catalan(k), where Catalan(k) = A000108(k).
a(n) = (-2)^n * hypergeom([-n, n+1], [2], 1/2).
O.g.f.: A(x) = 1/x * series reversion of x*(1 - x)/(1 - 2*x).  Cf. A152681. (End)

Typo in definition corrected by Robert Israel, Oct 31 2014

A127872 Triangle formed by reading A039599 mod 2.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1

Offset: 0

Philippe Deléham, Apr 05 2007

Also triangle formed by reading triangles A061554, A106180, A110519, A124574, A124576, A126953, A127543 modulo 2.

			Triangle begins:
1;
1, 1;
0, 1, 1;
1, 1, 1, 1;
0, 0, 0, 1, 1;
0, 0, 1, 1, 1, 1;
0, 1, 1, 0, 0, 1, 1;
1, 1, 1, 1, 1, 1, 1, 1;
0, 0, 0, 0, 0, 0, 0, 1, 1;
0, 0, 0, 0, 0, 0, 1, 1, 1, 1;
0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1;
0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1;
0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1;
0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1;
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1; ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A061554, A106180, A110519, A124574, A124576, A126953, A127543.

Cf. A000007, A036987, A001316, A062878, A050614, A000045, A001519

T[0, 0] := 1; T[n_, k_] := Binomial[2*n - 1, n - k] - Binomial[2*n - 1, n - k - 2]; Table[Mod[T[n, k], 2], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Apr 18 2017 *)

Sum_{k=0..n} T(n,k)*x^k = A000007(n), A036987(n), A001316(n), A062878(n) for x=-1,0,1,2 respectively.

Sum_{k=0..n} T(n,k)*Fibonacci(2*k+1) = A050614(n), see A000045 and A001519. - Philippe Deléham, Aug 30 2007

A106181 Expansion of c(-x^2)(1+2x-sqrt(1+4x^2))/2, c(x) the g.f. of A000108.

Original entry on oeis.org

0, 1, -1, -1, 2, 2, -5, -5, 14, 14, -42, -42, 132, 132, -429, -429, 1430, 1430, -4862, -4862, 16796, 16796, -58786, -58786, 208012, 208012, -742900, -742900, 2674440, 2674440, -9694845, -9694845, 35357670, 35357670, -129644790, -129644790, 477638700, 477638700, -1767263190, -1767263190

Offset: 0

Paul Barry, Apr 24 2005

Second column of number triangle A106180.

Cf. A000108, A099363.

a(n) = sin(Pi*n/2)*(C((n-1)/2)*(1-(-1)^n)/2) + sin(Pi*(n+1)/2)*(C(n/2)*(1+(-1)^n)/2) - 0^n for n > 0.

Conjecture: (n+2)*a(n) + n*a(n-1) + 4*(n-1)*a(n-2) + 4*(n-3)*a(n-3) = 0. - R. J. Mathar, Nov 15 2011

A124448 Riordan array (sqrt(1+4x^2)-2x, (1+2x-sqrt(1+4x^2))/2).

Original entry on oeis.org

1, -2, 1, 2, -3, 1, 0, 4, -4, 1, -2, -1, 7, -5, 1, 0, -4, -4, 11, -6, 1, 4, 2, -6, -10, 16, -7, 1, 0, 8, 8, -6, -20, 22, -8, 1, -10, -5, 11, 19, -1, -35, 29, -9, 1, 0, -20, -20, 7, 34, 13, -56, 37, -10, 1, 28, 14, -26, -46, -12, 49, 41, -84

Offset: 0

Paul Barry, Nov 01 2006

Inverse of triangle A106195.
Row sums are A105523 (expansion of 1-xc(-x^2) where c(x) is the g.f. of A000108).
Product of A007318 and A124448 is inverse of A053538.
A124448*A007318 = A106180, as infinite lower triangular matrices. - Philippe Deléham, Oct 16 2007
Triangle T(n,k), read by rows, given by (-2,1,-1,1,-1,1,-1,1,-1,...) DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 09 2011

			Triangle begins
   1;
  -2,   1;
   2,  -3,   1;
   0,   4,  -4,   1;
  -2,  -1,   7,  -5,   1;
   0,  -4,  -4,  11,  -6,   1;
   4,   2,  -6, -10,  16,  -7,   1;
   0,   8,   8,  -6, -20,  22,  -8,   1;

Cf. A106195, A000045, A164948, A164942.

N=12;
T(n, k)=sum(i=0, n-k, binomial(k, i)*binomial(n-k, i)*2^(n-k-i));
M=matrix(N, N);
for(n=1, N, for(k=1, n, M[n, k]=T(n-1, k-1))); /* A106195 */
A=M^-1;  /* A124448 */
/* for (n=1, N, for(k=1, n, print1(M[n, k], ", "))); */ /* A106195 */
for (n=1, N, for(k=1, n, print1(A[n, k], ", "))); /* A124448 */
/* Joerg Arndt, May 14 2011 */

Edited by N. J. A. Sloane, Dec 29 2011

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A124448 Riordan array (sqrt(1+4x^2)-2x, (1+2x-sqrt(1+4x^2))/2).

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