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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 (list; graph; refs; listen; history; text; internal format)
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

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

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Last modified December 16 01:36 EST 2017. Contains 296063 sequences.