1,2

a(n) = A126890(n+1,n-2) for n>1. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 30 2006

a(n) = (3/2)n^2 + (1/2)n - 1. a(n+3), (n>3) = 3*a(n+2) - 3*a(n+1) + a(n). a(n+1) right term in M^n * [1 1 1].

sum (n+j-1,j=0..n). - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Sep 12 2006

Row sums of triangle A131414.

Equals binomial transform of (1, 5, 3, 0, 0, 0,...). Equals A051340 * (1, 2, 3,...).

G.f.: x*(-1-3*x+x^2)/(-1+x)^3 = 1-3/(-1+x)^3-4/(-1+x)^2 . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 19 2007

1. a(4) = 25 = A005449(4) - 1

2. a(5) = 39 = f(n), n=5 in (3/2)n^2 + (1/2)n - 1.

3. a(7) = 76 = 3*56 - 3*39 + 25

4. a(5) = 39 = right term of M^4 * [1 1 1] = [1 5 39].

A005449 := proc(n) RETURN(n*(3*n+1)/2) ; end: A095794 := proc(n) RETURN(A005449(n)-1) ; end: for n from 1 to 100 do printf("%a, ", A095794(n)) ; od: - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 23 2006

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]-a[n-2]-3 od: seq(-a[n], n=2..50); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 18 2008

a[n_]:=Sum[i+n-3, {i, 1, n}]; [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 04 2008]

s = 1; lst = {s}; Do[s += n + 4; AppendTo[lst, s], {n, 1, 200, 3}]; lst [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 11 2009]

Table[Sum[i + n - 3, {i, 1, n}], {n, 2, 50}] [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 11 2009]

Cf. A005449, A051340, A131414.

A000217 [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 11 2009]

Adjacent sequences: A095791 A095792 A095793 this_sequence A095795 A095796 A095797

Sequence in context: A028557 A083657 A010740 this_sequence A119867 A026055 A165986

nonn

Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 06 2004, Jul 08 2007

Corrected and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 23 2006

Comment corrected by Jason Bandlow (jbandlow(AT)math.upenn.edu), Feb 28 2009