A084077 -id:A084077 - cp's OEIS frontend

A084076 Length of list created by n substitutions k -> Range[-1-abs(k), abs(k)+1] starting with {1}.

1, 5, 27, 157, 963, 6141, 40323, 270845, 1852419, 12857341, 90337283, 641286141, 4592533507, 33139654653, 240723001347, 1758796578813, 12916805074947, 95300512382973, 706044251602947, 5250379998560253, 39176121681444867

Offset: 0

Wouter Meeussen, May 11 2003

nonn

2*a(n-1) is the second diagonal of the triangle A115195.

Row sums of A167432. Hankel transform is A167435. - Paul Barry, Nov 03 2009

			{1}
{-2,-1,0,1,2}
{-3,-2,-1,0,1,2,3,-2,-1,0,1,2,-1,0,1,-2,-1,0,1,2,-3,-2,-1,0,1,2,3}

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Third column (m=2) of triangle A115193, called C(1, 2).
Cf. A000108, A115139, A115195, A167432, A167435.
Cf. A084075, A084077, A084078.

R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( (1-5*x -(1- x)*Sqrt(1-8*x))/(4*x^2*(1+x)) )); // G. C. Greubel, Nov 23 2022

Rest@CoefficientList[InverseSeries[Series[ -((1+5*n+2*n^2-(1+2*n)*Sqrt[1+6*n+n^2] )/(4*n^2)), {n, 0, 28}]], n] or Length/@Flatten/@NestList[ # /. k_Integer :> Range[ -1-Abs[k], Abs[k]+1]&, {1}, 8]
Flatten[{1,RecurrenceTable[{(n+2)*(7*n-5)*a[n] == (7*n-2)*(7*n-1)*a[n-1] + 4*(2*n-1)*(7*n+2)*a[n-2],a[1]==5,a[2]==27},a,{n,20}]}] (* Vaclav Kotesovec, Oct 14 2012 *)

{a(n) = my(L); L = [1]; if(n < 0, 0, for(i = 1, n, L = concat([ vector(3 + 2*abs(k), i, i - abs(k) - 2) | k <- L])); #L)}; /* Michael Somos, Nov 23 2022 */

def A084076_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-5*x -(1-x)*sqrt(1-8*x))/(4*x^2*(1+x)) ).list()
A084076_list(40) # G. C. Greubel, Nov 23 2022

G.f. is the series reversion of -((1 + 5*x + 2*x^2 - (1 + 2*x)*sqrt(1 + 6*x + x^2))/(4*x^2)).
G.f.: 2*((c(2*x))^3)/(1+c(2*x)) with the o.g.f. c(x) of A000108 (Catalan numbers).
a(n) = Sum_{j=1..n+1} A115195(n, j), n >= 0.
G.f.: (-1 + (1-x)*c(2*x))/(x*(1+x)); cf. A115139. - Wolfdieter Lang, Feb 23 2006
D-finite with recurrence: (n+2)*(7*n-5)*a(n) = (7*n-2)*(7*n-1)*a(n-1) + 4*(2*n-1)*(7*n+2)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 7*2^(3n+3)/(9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
D-finite with recurrence (n+2)*a(n) = 2*(4*n+1)*a(n-1) + (n+16)*a(n-2) - 4*(2*n-3)*a(n-3). - R. J. Mathar, Mar 10 2022
a(n) = ( (7*n-1)*(7*n-2)*a(n-1) + 4*(2*n-1)*(7*n+2)*a(n-2) )/((n+2)*(7*n-5)), with a(0) = 1, a(1) = 5. - G. C. Greubel, Nov 23 2022

A084075 Length of list created by n substitutions k -> Range( -abs(k+1), abs(k-1), 2) starting with {1}.

Original entry on oeis.org

1, 2, 5, 12, 33, 86, 249, 680, 2033, 5722, 17485, 50260, 156033, 455534, 1431281, 4228752, 13412193, 40003058, 127840085, 384232156, 1235575201, 3737280582, 12080678505, 36736735672, 119276490193, 364372758986, 1187542872989

Offset: 0

Wouter Meeussen, May 11 2003

nonn

			{1}, {-2,0}, {-1,1,3,-1,1}, {0,2,-2,0,-4,-2,0,2,0,2,-2,0}

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A084076, A084077, A084078.

Cf. A027307, A215067, A034015 (even bisection).

I:=[1,2,5,12]; [n le 4 select I[n] else (6*(35*n^2-55*n-76)*Self(n-1) + (275*n^4-770*n^3-203*n^2+1736*n-912)*Self(n-2) -6*(5*n^2+5*n-28)*Self(n-3) + (n-4)*(n-2)*(25*n^2+5*n-48)*Self(n-4))/(n*(n+2)*(25*n^2-45*n-28)): n in [1..41]]; // G. C. Greubel, Nov 24 2022

Rest@CoefficientList[InverseSeries[Series[ (-1-6n-8n^2+(1+2n)^2 Sqrt[1+4n])/( 2(n+4n^2+4n^3)), {n, 0, 40}]], n]
Length/@Flatten/@NestList[ #/.k_Integer:>Range[-Abs[k+1], Abs[k-1], 2] &, {1}, 8]

# replace iterates lists as described in Example.
def replace(L):
    return [i for k in L for i in range(-abs(k + 1), 1 + abs(k - 1), 2)]
def a(n):
  L = [1]
  for k in range(n): L=replace(L)
  return len(L)
print([a(n) for n in range(12)]) # F. Chapoton, Nov 15 2024

@CachedFunction
def a(n): # a = A084075
    if n < 4: return (1, 2, 5, 12)[n]
    else: return (6*(35*n^2 +15*n -96)*a(n-1) +(275*n^4+330*n^3-863*n^2+120*n+126)*a(n-2) -6*(5*n^2+15*n-18)*a(n-3) +(n-3)*(n-1)*(25*n^2+55*n-18)*a(n-4))/((n+1)*(n+3)*(25*n^2+5*n-48))
[a(n) for n in range(41)] # G. C. Greubel, Nov 24 2022

G.f. is the series reversion of (-1 -6*x -8*x^2 + (1+2*x)^2 * sqrt(1+4*x))/(2*(x +4*x^2 +4*x^3)).
a(2*n) = A027307(n)/2, n >= 1.
a(n) = ( 6*(35*n^2 +15*n -96)*a(n-1) + (275*n^4 +330*n^3 -863*n^2 +120*n +126)*a(n-2) - 6*(5*n^2 +15*n -18)*a(n-3) + (n-3)*(n-1)*(25*n^2 +55*n -18)*a(n-4) )/((n+1)*(n+3)*(25*n^2 +5*n -48)), n >= 4. - G. C. Greubel, Nov 24 2022

cp's OEIS Frontend

A084076 Length of list created by n substitutions k -> Range[-1-abs(k), abs(k)+1] starting with {1}.

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