0,3

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000,100,110,111,200,210,211 decrease, so a(2) = 12-7 = 5).

Number of subgroups of C_(2^n) X C_(2^n) (see A060724).

Starting with "1" = row sums of triangle A054582. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 23 2008

Row sums of A125165: (1, 5, 15, 37...). Binomial transform of [1, 4, 6, 6, 6...] = [1, 5, 15, 37,...]. 4-th diagonal from the right of A126777 = (1, 5, 15,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 23 2006

a(n) = 2*a(n-1) + (2n-1); e.g. a(4) = 37 = 2*15 + 7. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 30 2007

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Feb 20 2009: (Start)

a(n) = 4a(n-1)-5a(n-2)+2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.

G.f.: z*(1+z)/((1-z)^2*(1-2*z))

(End)

s=0; lst={s}; Do[s+=n+=s-1; AppendTo[lst, s], {n, 2, 5!, 2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 07 2008]

A050487(2^m-1).

Equals (1/2) A051667.

Cf. A126277, A125165.

Cf. A054852.

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Feb 20 2009: (Start)

Cf. A156925

A050488(n+1) equals A156920(n+1,1)

A050488(n+1) equals A156919(n+1,1)/2^n

A050488(n+1) equals A142963(n+2,1)/2

(End)

Sequence in context: A109818 A146797 A005491 this_sequence A142964 A014316 A075717

Adjacent sequences: A050485 A050486 A050487 this_sequence A050489 A050490 A050491

nonn

James A. Sellers (sellersj(AT)math.psu.edu), Dec 26, 1999.