

A129111


Sums of three consecutive heptagonal numbers.


1



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
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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(5n3)/2.
a(n) = (15/2)*n^2 + (21/2)*n + 8.
G.f. (8+2*x+5*x^2)/(1x)^3; a(n) = 3*a(n1)3*a(n2)+a(n3).  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(n1)3*Self(n2)+1*Self(n3): 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


CROSSREFS

Cf. A000566, A007667, A034961, A129803, A129863.
Sequence in context: A005897 A215097 A111694 * A002413 A218325 A252870
Adjacent sequences: A129108 A129109 A129110 * A129112 A129113 A129114


KEYWORD

nonn,easy


AUTHOR

Jonathan Vos Post, May 24 2007


STATUS

approved



