A261954 Start with a single equilateral triangle for n=0; for the odd n-th generation add a triangle at each expandable side of the triangles of the (n-1)-th generation (this is the "side to side" version); for the even n-th generation use the "side to vertex" version; a(n) is the number of triangles added in the n-th generation.

1, 3, 3, 6, 12, 15, 21, 18, 30, 27, 39, 30, 48, 39, 57, 42, 66, 51, 75, 54, 84, 63, 93, 66, 102, 75, 111, 78, 120, 87, 129, 90, 138, 99, 147, 102, 156, 111, 165, 114, 174, 123, 183, 126, 192, 135, 201, 138, 210, 147, 219

Offset: 0

Author

Kival Ngaokrajang, Sep 06 2015

Keywords

• nonn
• easy

Comments

See a comment on V-V and V-S at A249246.

There are a total of 16 combinations as shown in the table below:

+-------------------------------------------------------+

| Even n-th version V-V S-V V-S S-S |

+-------------------------------------------------------+

| Odd n-th version |

| V-V A008486 A248969 A261951 A261952 |

| S-V A261950 A008486 A008486 A261956 |

| V-S A249246 A008486 A008486 A261957 |

| S-S A261953 a(n) A261955 A008486 |

+-------------------------------------------------------+

Note: V-V = vertex to vertex, S-V = side to vertex,

V-S = vertex to side, S-S = side to side.

Links

• Kival Ngaokrajang, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,1,0,-1).

Crossrefs

Cf. A008486, A017197, A248969, A249246.

Programs

PARI

a=3; print1("1, ", a, ", "); for (n=2, 100, if (Mod(n,4)==0||Mod(n,4)==2, print1(9*(n/2-1)+3, ", "), if (Mod(n,4)==1, a=a+9, a=a+3); print1(a, ", ")))

Formula

a(0) = 1, a(1) = 3; for even n >= 2, a(n) = 9*(n/2-1) + 3 or a(n) = A017197(n/2-1); for odd n >= 3, a(n) = a(n-2) + 9, if mod(n,4) = 1 otherwise a(n) = a(n-2) + 3.

Conjectures from Colin Barker, Sep 10 2015: (Start)

a(n) = a(n-2)+a(n-4)-a(n-6) for n>6.

G.f.: (7*x^6+6*x^5+8*x^4+3*x^3+2*x^2+3*x+1) / ((x-1)^2*(x+1)^2*(x^2+1)).

(End)