A236333 The (n-2)-th (n>=3) triple of terms gives coefficients of double trinomial P_n(x) = ((n-2)^2*x^2 + n*x + 2)/2 (see comment).

1, 3, 2, 4, 4, 2, 9, 5, 2, 16, 6, 2, 25, 7, 2, 36, 8, 2, 49, 9, 2, 64, 10, 2, 81, 11, 2, 100, 12, 2, 121, 13, 2, 144, 14, 2, 169, 15, 2, 196, 16, 2, 225, 17, 2, 256, 18, 2, 289, 19, 2, 324, 20, 2, 361, 21, 2, 400, 22, 2, 441, 23, 2, 484, 24, 2, 529, 25, 2, 576, 26, 2, 625, 27, 2, 676, 28, 2, 729, 29, 2

Offset

3,2

Comments

Let {G_n(k)}_(k>=0) be sequence of n-gonal numbers. Then G_n(P_n(k)) = G_n(P_n(k)-1) + G_n((n-2)*k+1).

Links

Vincenzo Librandi, Table of n, a(n) for n = 3..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Formula

If n==0 (mod 3), then a(n) = n^2/9;

if n==1 (mod 3), then a(n) = (n+5)/3;

if n==2 (mod 3), then a(n) = 2.

G.f.: -x^3*(2*x^8+2*x^7-4*x^5-5*x^4+x^3+2*x^2+3*x+1) / ((x-1)^3*(x^2+x+1)^3). - Colin Barker, Jan 23 2014

Example

Let n=5, k=4. Then G_5(k)=k*(3*k-1)/2 (Cf. A000326) and the double trinomial 2*P_5(x)= 9*x^2+5*x+2, P_5(4)=(9*4^2+5*4+2)/2=83,
Thus, we have G_5(83)=G_5(82)+G_5(13), or 83*124 = 41*245 + 13*19 = 10292.

Mathematica

a[n_]:=Which[Mod[n, 3]==0, n^2/9, Mod[n, 3]==1, (n+5)/3, True, 2]; Map[a, Range[3, 103]]
CoefficientList[Series[(-1-3 x-2 x^2-x^3+5 x^4+4 x^5-2 x^7-2 x^8)/((-1+x)^3 (1+x+x^2)^3), {x, 0, 100}], x]

Prog

(PARI) Vec(-x^3*(2*x^8+2*x^7-4*x^5-5*x^4+x^3+2*x^2+3*x+1)/((x-1)^3*(x^2+x+1)^3) + O(x^100)) \\ Colin Barker, Jan 23 2014
(Magma) I:=[1, 3, 2, 4, 4, 2, 9, 5, 2]; [n le 9 select I[n] else 3*Self(n-3)-3*Self(n-6)+Self(n-9): n in [1..90]]; // Vincenzo Librandi, Feb 02 2014

Crossrefs

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A001106, A001107, A051682, A051624, A051865-A051876.

Sequence in context: A230499 A023630 A110550 * A128220 A258242 A287869

Adjacent sequences: A236330 A236331 A236332 * A236334 A236335 A236336

Keyword

nonn,easy

Author

Vladimir Shevelev, Jan 22 2014

Status

approved