A129111 Sums of three consecutive heptagonal numbers.

8, 26, 59, 107, 170, 248, 341, 449, 572, 710, 863, 1031, 1214, 1412, 1625, 1853, 2096, 2354, 2627, 2915, 3218, 3536, 3869, 4217, 4580, 4958, 5351, 5759, 6182, 6620, 7073, 7541, 8024, 8522, 9035, 9563, 10106, 10664, 11237, 11825, 12428, 13046, 13679, 14327, 14990

Offset: 0

Author

Jonathan Vos Post, May 24 2007

Keywords

• nonn
• easy

Comments

Arises in heptagonal number analog to A129803 (Triangular numbers which are the sum of three consecutive triangular numbers).

What are the heptagonal numbers which are the sum of three consecutive heptagonal numbers?

Prime for a(2) = 59, a(3) = 107, a(7) = 449, a(10) = 863, a(11) = 1031, a(23) = 4217, a(26) = 5351, a(31) = 7541, a(42) = 13679, a(43) = 14327, a(46) = 16361, a(51) = 20051.

Examples

			a(0) = Hep(0) + Hep(1) + Hep(2) = 0 + 1 + 7 = 8 = (15/2)*0^2 + (21/2)*0 + 8.
a(1) = Hep(1) + Hep(2) + Hep(3) = 1 + 7 + 18 = 26 = (15/2)*1^2 + (21/2)*1 + 8.
a(2) = Hep(2) + Hep(3) + Hep(4) = 7 + 18 + 34 = 59 = (15/2)*2^2 + (21/2)*2 + 8.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000566, A007667, A034961, A129803, A129863.

Programs

Magma

I:=[8,26,59]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012

Mathematica

LinearRecurrence[{3,-3,1},{8,26,59},50] (* Vincenzo Librandi, Feb 12 2012 *)

PARI

a(n)=3*n*(5*n+7)/2+8 \\ Charles R Greathouse IV, Jun 17 2017

Python

def a(n): return 3*n*(5*n+7)//2 + 8
print([a(n) for n in range(44)]) # Michael S. Branicky, Aug 26 2021

Formula

a(n) = Hep(n) + Hep(n+1) + Hep(n+2) where Hep(n) = A000566(n) = n*(5*n-3)/2.

a(n) = (15/2)*n^2 + (21/2)*n + 8.

From Colin Barker, Feb 20 2012: (Start)

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

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

E.g.f.: exp(x)*(16 + 36*x + 15*x^2)/2. - Elmo R. Oliveira, Nov 16 2024