OFFSET
0,1
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.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).
FORMULA
a(n) = Hep(n) + Hep(n+1) + Hep(n+2) where Hep(n) = A000566(n) = n(5n-3)/2.
a(n) = (15/2)*n^2 + (21/2)*n + 8.
G.f. (8+2*x+5*x^2)/(1-x)^3; a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Colin Barker, Feb 20 2012
EXAMPLE
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.
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {8, 26, 59}, 50] (* Vincenzo Librandi, Feb 12 2012 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, May 24 2007
STATUS
approved