A265676 a(n) is the total number of petals of the Flower of Life at the n-th iteration.

0, 1, 7, 19, 43, 67, 97, 139, 181, 229, 289, 349, 415, 493, 571, 655, 751, 847, 949, 1063, 1177, 1297, 1429, 1561, 1699, 1849, 1999, 2155, 2323, 2491, 2665, 2851, 3037, 3229, 3433, 3637, 3847, 4069, 4291, 4519, 4759, 4999, 5245, 5503, 5761, 6025, 6301, 6577

Offset: 0

Author

Kival Ngaokrajang, Dec 13 2015

Keywords

• nonn
• easy

Comments

Inspired by A264788, but counting on petals of the Flower of Life instead of circles. For n >= 3, the second differences seem to be cyclic of 6, 12, 0.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Crystalinks, Flower of Life
• Kival Ngaokrajang, Illustration of initial terms, For n = 11
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A264788.

Programs

Magma

I:=[0,1,7,19,43,67]; [n le 6 select I[n] else 2*Self(n-1)-Self(n-2)+Self(n-3)-2*Self(n-4)+Self(n-5): n in [1..60]]; // Vincenzo Librandi, Dec 14 2015

Mathematica

CoefficientList[Series[x (1 + 5 x + 6 x^2 + 11 x^3 - 5 x^4)/((1 - x)^3 (1 + x + x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 14 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{0,1,7,19,43,67},50] (* Harvey P. Dale, Aug 26 2024 *)

PARI

{ a = 7; d1 = 6; print1("0, 1, ", a, ", "); for(n = 3, 100, if (Mod(n,3) == 0, d2 = 6); if (Mod(n,3) == 1, d2 = 12); if (Mod(n,3) == 2, d2 = 0); d1 = d1 + d2; a = a + d1; print1(a, ", "))}

PARI

concat(0, Vec(x*(1+5*x+6*x^2+11*x^3-5*x^4) / ((1-x)^3*(1+x+x^2)) + O(x^100))) \\ Colin Barker, Dec 13 2015

Formula

From Colin Barker, Dec 13 2015: (Start)

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n>5.

G.f.: x*(1+5*x+6*x^2+11*x^3-5*x^4) / ((1-x)^3*(1+x+x^2)).

(End)