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 A129109 Sums of three consecutive hexagonal numbers. 1
 7, 22, 49, 88, 139, 202, 277, 364, 463, 574, 697, 832, 979, 1138, 1309, 1492, 1687, 1894, 2113, 2344, 2587, 2842, 3109, 3388, 3679, 3982, 4297, 4624, 4963, 5314, 5677, 6052, 6439, 6838, 7249, 7672, 8107, 8554, 9013, 9484, 9967, 10462, 10969, 11488 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Arises in hexagonal number analog to A129803 Triangular numbers which are the sum of three consecutive triangular numbers. What are the hexagonal numbers which are the sum of three consecutive hexagonal numbers? Prime for a(0) = 7, a(4) = 139, a(6) = 277, a(8) = 463, a(18) = 2113, a(22) = 3109, a(26) = 4297, a(38) = 9013, a(40) = 9967. 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) = H(n) + H(n+1) + H(n+2) where H(n) = A000384(n) = n(2n-1). a(n) = 6*n^2 + 9*n + 7. From Colin Barker, Feb 20 2012: (Start) a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). G.f.: (7+x+4*x^2)/(1-x)^3. (End) EXAMPLE a(0) = H(0) + H(1) + H(2) = 0 + 1 + 6 = 7 = 6*0^2 + 9*0 + 7. a(1) = H(1) + H(2) + H(3) = 1 + 6 + 15 = 22 = 6*1^2 + 9*1 + 7. a(2) = H(2) + H(3) + H(4) = 6 + 15 + 28 = 49 = 6*2^2 + 9*2 + 7. MATHEMATICA LinearRecurrence[{3, -3, 1}, {7, 22, 49}, 50] (* Vincenzo Librandi, Feb 20 2012 *) PROG (MAGMA) I:=[7, 22, 49]; [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)=6*n^2+9*n+7 \\ Charles R Greathouse IV, Feb 20 2012 CROSSREFS Cf. A000384, A007667, A034961, A129803, A129863. Sequence in context: A223833 A014073 A288114 * A224141 A002412 A211652 Adjacent sequences:  A129106 A129107 A129108 * A129110 A129111 A129112 KEYWORD easy,nonn AUTHOR Jonathan Vos Post, May 24 2007 STATUS approved

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Last modified October 15 12:31 EDT 2019. Contains 328026 sequences. (Running on oeis4.)