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%I #24 Sep 08 2022 08:45:30
%S 8,26,59,107,170,248,341,449,572,710,863,1031,1214,1412,1625,1853,
%T 2096,2354,2627,2915,3218,3536,3869,4217,4580,4958,5351,5759,6182,
%U 6620,7073,7541,8024,8522,9035,9563,10106,10664,11237,11825,12428,13046,13679,14327
%N Sums of three consecutive heptagonal numbers.
%C Arises in heptagonal number analog to A129803 (Triangular numbers which are the sum of three consecutive triangular numbers).
%C What are the heptagonal numbers which are the sum of three consecutive heptagonal numbers?
%C 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.
%H Vincenzo Librandi, <a href="/A129111/b129111.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) = Hep(n) + Hep(n+1) + Hep(n+2) where Hep(n) = A000566(n) = n(5n-3)/2.
%F a(n) = (15/2)*n^2 + (21/2)*n + 8.
%F 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
%e a(0) = Hep(0) + Hep(1) + Hep(2) = 0 + 1 + 7 = 8 = (15/2)*0^2 + (21/2)*0 + 8.
%e a(1) = Hep(1) + Hep(2) + Hep(3) = 1 + 7 + 18 = 26 = (15/2)*1^2 + (21/2)*1 + 8.
%e a(2) = Hep(2) + Hep(3) + Hep(4) = 7 + 18 + 34 = 59 = (15/2)*2^2 + (21/2)*2 + 8.
%t LinearRecurrence[{3,-3,1},{8,26,59},50] (* _Vincenzo Librandi_, Feb 12 2012 *)
%o (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
%o (PARI) a(n)=3*n*(5*n+7)/2+8 \\ _Charles R Greathouse IV_, Jun 17 2017
%o (Python)
%o def a(n): return 3*n*(5*n+7)//2 + 8
%o print([a(n) for n in range(44)]) # _Michael S. Branicky_, Aug 26 2021
%Y Cf. A000566, A007667, A034961, A129803, A129863.
%K nonn,easy
%O 0,1
%A _Jonathan Vos Post_, May 24 2007
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