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A129109 Sums of three consecutive hexagonal numbers. 1

%I

%S 7,22,49,88,139,202,277,364,463,574,697,832,979,1138,1309,1492,1687,

%T 1894,2113,2344,2587,2842,3109,3388,3679,3982,4297,4624,4963,5314,

%U 5677,6052,6439,6838,7249,7672,8107,8554,9013,9484,9967,10462,10969,11488

%N Sums of three consecutive hexagonal numbers.

%C 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.

%H Vincenzo Librandi, <a href="/A129109/b129109.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F 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.

%F From _Colin Barker_, Feb 20 2012: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: (7+x+4*x^2)/(1-x)^3. (End)

%e a(0) = H(0) + H(1) + H(2) = 0 + 1 + 6 = 7 = 6*0^2 + 9*0 + 7.

%e a(1) = H(1) + H(2) + H(3) = 1 + 6 + 15 = 22 = 6*1^2 + 9*1 + 7.

%e a(2) = H(2) + H(3) + H(4) = 6 + 15 + 28 = 49 = 6*2^2 + 9*2 + 7.

%t LinearRecurrence[{3,-3,1},{7,22,49},50] (* _Vincenzo Librandi_, Feb 20 2012 *)

%t Total/@Partition[PolygonalNumber[6,Range[0,50]],3,1] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Mar 14 2020 *)

%o (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

%o (PARI) a(n)=6*n^2+9*n+7 \\ _Charles R Greathouse IV_, Feb 20 2012

%Y Cf. A000384, A007667, A034961, A129803, A129863.

%K easy,nonn

%O 0,1

%A _Jonathan Vos Post_, May 24 2007

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Last modified November 27 15:33 EST 2020. Contains 338683 sequences. (Running on oeis4.)