A228465 Recurrence a(n) = a(n-1) + 2^n*a(n-2) with a(0)=0, a(1)=1.

0, 1, 1, 9, 25, 313, 1913, 41977, 531705, 22023929, 566489849, 45671496441, 2366013917945, 376506912762617, 39141278944373497, 12376519796349807353, 2577539376694811306745, 1624792742123856760679161, 677311275106408471956040441, 852536648457739021814912002809

Offset: 0

Author

Vaclav Kotesovec, Aug 22 2013

Keywords

• nonn

Comments

Generally (if p>0, q>1), recurrence a(n) = b*a(n-1) + (p*q^n+d)*a(n-2), a(n) is asymptotic to c*q^(n^2/4)*(p*q)^(n/2), where c is for fixed parameters b, p, d, q, a(0), a(1) constant, independent on n.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..100

Crossrefs

Cf. A006116, A006117, A006118, A006119, A006120, A006121, A006122.

Programs

Magma

[n le 2 select (n-1) else Self(n-1)+Self(n-2)*2^(n-1): n in [1..20]]; // Vincenzo Librandi, Aug 23 2013

Mathematica

RecurrenceTable[{a[n]==a[n-1]+2^n*a[n-2],a[0]==0,a[1]==1},a,{n,0,20}]
(* Alternative: *)
a[n_] := Sum[2^(k^2-1) QBinomial[n - k , k - 1, 2], {k, 1, n}];
Table[a[n], {n, 0, 19}] (* After Vladimir Kruchinin. Peter Luschny, Jan 20 2020 *)

SageMath

def a(n):
    return sum(2^(k^2 - 1)*q_binomial(n-k , k-1, 2) for k in (1..n))
print([a(n) for n in range(20)]) # Peter Luschny, Jan 20 2020

Formula

a(n) ~ c * 2^(n^2/4 + n/2), where c = 0.548441579870783378573455400152590154... if n is even and c = 0.800417244834941368929416800341853541... if n is odd.

a(n) = Sum_{k=1..floor(n/2+1/2)} qbinomial(n-k,k-1)*2^(k^2-1), where q-binomial is triangle A022166, that is, with q=2. - Vladimir Kruchinin, Jan 20 2020