A140229 Binomial transform of [1, 3, 3, 1, -2, 3, -4, 5, ...].

1, 4, 10, 20, 33, 49, 68, 90, 115, 143, 174, 208, 245, 285, 328, 374, 423, 475, 530, 588, 649, 713, 780, 850, 923, 999, 1078, 1160, 1245, 1333, 1424, 1518, 1615, 1715, 1818, 1924, 2033, 2145, 2260, 2378, 2499, 2623, 2750, 2880, 3013, 3149, 3288, 3430, 3575

Offset: 1

Author

Gary W. Adamson, May 13 2008

Keywords

• nonn
• easy

Comments

The falling diagonal starting with T(1,4) in A049777 (as a square array) gives the terms of this sequence for n >=3. - Bob Selcoe, Oct 27 2014

From Fatma E. Kurnaz, Salih E. Ariturk, and Arzu Caglar, Aug 08 2024: (Start)

Suppose a regular hexagon is divided into equilateral triangles. When we number the corners of the triangles in a spiral manner starting from the center, if the numbers increase counterclockwise, let's define them as "right triangle" and the remaining ones as "left triangle".

The number of downward-pointing left triangles, constitute this sequence for n>=2.

For n=3 a(3)=20, A hexagon with a side of 3 units consists of 54 equilateral triangles. The number of upward right triangles is 12, the number of upward left triangles is 15, the number of downward right triangles is 7. The number of downward left triangles is equal to 20. (End)

Examples

			a(5) = 33 = (1, 4, 6, 4, 1) dot (1, 3, 3, 1, -2) = (1 + 12 + 18 + 4 - 2).
		

Links

• İ. H. Karakaş and İ. Aliyev, Problem 8.36, Analiz ve Cebirde İlginç OLİMPİYAT PROBLEMLERİ VE ÇÖZÜMLERİ, TÜBİTAK Publications, 5 (2003), p. 298.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A049777

Programs

Magma

[1,4] cat [(n+1)*(3*n-4)/2: n in [3..50]]; // Vincenzo Librandi, Oct 27 2014

Maple

1,4,seq((1/2)*(n+1)*(3*n-4), n=3..40); # Emeric Deutsch, May 18 2008

Mathematica

s=-2;lst={1, 4};Do[s+=n+1;If[n>3, AppendTo[lst, s]], {n, 0, 6!, 3}];lst (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)

Formula

A007318 * [1, 3, 3, 1, -2, 3, -4, 5,...].

a(n) = (n+1)(3n-4)/2, for n>=3. - Emeric Deutsch, May 18 2008

G.f.: x(1+x+x^2+x^3-x^4)/(1-x)^3. a(n) = 3*a(n-1) -3*a(n-2) + a(n-3), n>5. a(n+1)-a(n) = A016777(n), n>3. - R. J. Mathar, Nov 25 2008

Extensions

More terms from Emeric Deutsch, May 18 2008

More terms from Vladimir Joseph Stephan Orlovsky, Oct 25 2008