A054106 -id:A054106 - cp's OEIS frontend

A054109 a(n) = T(2*n+1, n), array T as in A054106.

1, 2, 8, 27, 99, 363, 1353, 5082, 19228, 73150, 279566, 1072512, 4127788, 15930512, 61628248, 238911947, 927891163, 3609676487, 14062955413, 54860308997, 214268628223, 837780853637, 3278934510163, 12844867331387

Offset: 0

Clark Kimberling

nonn

Hankel transform of A054109. - Paul Barry, Nov 04 2009
From Paul Barry, Mar 29 2010: (Start)
Hankel transform is A167478 (correction of previous entry).
The aerated sequence 0,0,1,0,2,0,8,0,... has e.g.f. Integral_{t=0..x} cos(x-t)*Bessel_I(1,2t). (End)
Hankel transform of 0,1,2,8,27,... is -F(2n). - Paul Barry, Jan 17 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A054106, A000108, A000984.

a := n -> abs(add(binomial(-j-1, -2*j-2), j=0..n)):
seq(a(n), n=0..23); # Zerinvary Lajos, Oct 03 2007
gf := ((1 - 4*x)^(-1/2) - 1)/(2*x*(x + 1)): ser := series(gf, x, 32):
seq(coeff(ser, x, n), n=0..23); # Peter Luschny, Jan 18 2020

Table[FullSimplify[1/2*(-1)^(1+n) * (-1+1/Sqrt[5]-(-1)^n*Binomial[2*(2+n), 2+n] * Hypergeometric2F1[1, 5/2+n, 3+n, -4])],{n,0,20}] (* Vaclav Kotesovec, Feb 12 2014 *)
Table[1/2*(-1)^(n+1)*Sum[(-1)^k*Binomial[2*k, k],{k,1,n+1}],{n,0,20}] (* Vaclav Kotesovec, Feb 12 2014 *)

a(n)=(1/2)*(-1)^(n+1)*sum(k=1,n+1,(-1)^k*binomial(2*k,k))

a(n-1) = (1/2)*(-1)^n*Sum_{k=1..n} (-1)^k*binomial(2k, k). - Benoit Cloitre, Nov 07 2002
Conjecture: (n+1)*a(n) + (-3*n-1)*a(n-1) + 2*(-2*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 24 2012
a(n) ~ 2^(2*n+3) / (5*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(2k+1,k+1). - Paul Barry, Jan 17 2020
G.f.: c(x)B(x)/(1+x), c(x) g.f. of A000108, B(x) g.f. of A000984. - Paul Barry, Jan 17 2020
a(n) = binomial(2*n+3, n+2)*hypergeom([1, n+5/2], [n+3], -4) + (-1)^n*(5 - sqrt(5)) /10. - Peter Luschny, Jan 18 2020

A054107 T(n,n-3), array T as in A054106.

Original entry on oeis.org

1, 3, 8, 15, 27, 44, 67, 97, 135, 182, 239, 307, 387, 480, 587, 709, 847, 1002, 1175, 1367, 1579, 1812, 2067, 2345, 2647, 2974, 3327, 3707, 4115, 4552, 5019, 5517, 6047, 6610, 7207, 7839, 8507, 9212, 9955, 10737, 11559

Offset: 3

Clark Kimberling

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

LinearRecurrence[{4,-6,4,-1},{1,3,8,15,27,44,67},50] (* Harvey P. Dale, Apr 24 2020 *)

For n>5, a(n) = 1/6 * (n^3 - 6n^2 + 23n - 48).

G.f.: x^3*(1-x+x^2)*(1+x^2-2*x^3+x^4)/(1-x)^4. [Colin Barker, Sep 09 2012]

A013580 Triangle formed in same way as Pascal's triangle (A007318) except 1 is added to central element in even-numbered rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 5, 9, 5, 1, 1, 6, 14, 14, 6, 1, 1, 7, 20, 29, 20, 7, 1, 1, 8, 27, 49, 49, 27, 8, 1, 1, 9, 35, 76, 99, 76, 35, 9, 1, 1, 10, 44, 111, 175, 175, 111, 44, 10, 1, 1, 11, 54, 155, 286, 351, 286, 155, 54, 11, 1, 1, 12, 65, 209, 441, 637, 637, 441, 209, 65

Offset: 0

Martin Hecko (bigusm(AT)interramp.com)

From Gus Wiseman, Apr 19 2023: (Start)
Appears to be the number of nonempty subsets of {1,...,n} with median k, where the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). For example, row n = 5 counts the following subsets:
  {1}  {2}      {3}          {4}      {5}
       {1,3}    {1,5}        {3,5}
       {1,2,3}  {2,4}        {1,4,5}
       {1,2,4}  {1,3,4}      {2,4,5}
       {1,2,5}  {1,3,5}      {3,4,5}
                {2,3,4}
                {2,3,5}
                {1,2,4,5}
                {1,2,3,4,5}
Including half-steps gives A231147.
For mean instead of median we have A327481.
(End)

			Triangle begins:
   1
   1   1
   1   3   1
   1   4   4   1
   1   5   9   5   1
   1   6  14  14   6   1
   1   7  20  29  20   7   1
   1   8  27  49  49  27   8   1
   1   9  35  76  99  76  35   9   1
   1  10  44 111 175 175 111  44  10   1
   1  11  54 155 286 351 286 155  54  11   1
   1  12  65 209 441 637 637 441 209  65  12   1

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

Row sums give A000975, A054106.
Central diagonal T(2n+1,n+1) appears to be A006134.
Central diagonal T(2n,n) appears to be A079309.
For partitions instead of subsets we have A359901, row sums A325347.
A000975 counts subsets with integer median.
A007318 counts subsets by length, A359893 by twice median.
Cf. A000984, A024718, A057552, A231147, A327475, A327481, A361654.

CoefficientList[CoefficientList[Series[1/(1 - (1 + y)*x)/(1 - y*x^2), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Oct 10 2017 *)

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

More terms from James Sellers

A054108 a(n) = (-1)^(n+1)Sum_{k=0..n+1}(-1)^kbinomial(2*k,k).

Original entry on oeis.org

1, 5, 15, 55, 197, 727, 2705, 10165, 38455, 146301, 559131, 2145025, 8255575, 31861025, 123256495, 477823895, 1855782325, 7219352975, 28125910825, 109720617995, 428537256445, 1675561707275, 6557869020325, 25689734662775

Offset: 0

Clark Kimberling

nonn

p divides a((p-3)/2) for p in A045468 (primes congruent to {1, 4} mod 5). - Alexander Adamchuk, Jul 05 2006
The sequence 1,1,5,15,55,... has general term sum{k=0..n, (-1)^(n-k)*C(2k,k)}. Its Hankel transform is A082761. - Paul Barry, Apr 10 2007
From Paul Barry, Mar 29 2010: (Start)
The sequence 1,1,5,15,... has g.f. 1/((1+x)*sqrt(1-4x)).
The doubled sequence 1,1,1,1,5,5,... has e.g.f. dif(int((sin(x-t)+cos(x-t))*Bessel_I(0,2t),t,0,x),x). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

T(2n, n), array T as in A054106.

Cf. A066796, A000984, A054109, A006134, A045468.

Table[Sum[(-1)^(k+n)*((2k)!/(k!)^2),{k,0,n}], {n,1,50}] (* Alexander Adamchuk, Jul 05 2006 *)
CoefficientList[Series[(1/Sqrt[1-4*x]/(1+x)-1)/x, {x, 0, 20}], x]
(* or *)
Table[(-1)^(n+1)*Sum[(-1)^k*Binomial[2*k, k], {k, 0, n+1}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 06 2012 *)
Round@Table[Binomial[2 (n + 2), n + 2] Hypergeometric2F1[1, n + 5/2, n + 3, -4] - (-1)^n/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

a(n)=(-1)^(n+1)*sum(k=0,n+1,(-1)^k*binomial(2*k,k))

from math import comb
def A054108(n): return (1 if n % 2 else -1)*sum((-1 if k % 2 else 1)*comb(2*k,k) for k in range(n+2)) # Chai Wah Wu, Jan 19 2022

a(n) = C(2n, n) - a(n-1) with a(0)=1. - Labos Elemer, Apr 26 2003
C(2n,n) - C(2n-2,n-1) + ... +(-1)^(k+n)*C(2k,k)+ ... + (-1)^(1+n)*C(2,1) + (-1)^n*C(0,0), where C(2k,k)=(2k)!/(k!)^2 - central binomial coefficients A000984[k]. - Alexander Adamchuk, Jul 05 2006
a(n) = Sum_{k=0..n} (-1)^(k+n)*((2k)!/(k!)^2). - Alexander Adamchuk, Jul 05 2006
G.f.: (1/sqrt(1-4*x)/(1+x)-1)/x = (-1 + 2/(U(0)-2*x))/(1+x) where U(k)= 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 27 2012
a(n) ~ 2^(2*n+4)/(5*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012
Recurrence: (n+1)*a(n) = (3*n+1)*a(n-1) + 2*(2*n+1)*a(n-2). - Vaclav Kotesovec, Nov 06 2012

Formula from Benoit Cloitre, Sep 29 2002

Definition corrected by Vaclav Kotesovec, Nov 06 2012

cp's OEIS Frontend

A054109 a(n) = T(2*n+1, n), array T as in A054106.

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A054107 T(n,n-3), array T as in A054106.

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A013580 Triangle formed in same way as Pascal's triangle (A007318) except 1 is added to central element in even-numbered rows.

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A054108 a(n) = (-1)^(n+1)*Sum_{k=0..n+1}(-1)^k*binomial(2*k,k).

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A054108 a(n) = (-1)^(n+1)Sum_{k=0..n+1}(-1)^kbinomial(2*k,k).