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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: A215097 A331242 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 November 25 14:33 EST 2020. Contains 338624 sequences. (Running on oeis4.)