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Triangular numbers that are the sum of three consecutive triangular numbers.
19

%I #36 Oct 31 2024 13:17:23

%S 10,136,1891,26335,366796,5108806,71156485,991081981,13803991246,

%T 192264795460,2677903145191,37298379237211,519499406175760,

%U 7235693307223426,100780206894952201,1403687203222107385,19550840638214551186,272308081731781609216

%N Triangular numbers that are the sum of three consecutive triangular numbers.

%C Indices m: 4, 16, 61, 229, 856, 3196, 11929, with recurrence m(i) = 5(m(i-1) - m(i-2)) + m(i-3) (see A133161).

%C If first term is omitted, same sequence as A128862. - _R. J. Mathar_, Jun 13 2008

%H Colin Barker, <a href="/A129803/b129803.txt">Table of n, a(n) for n = 1..850</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15, -15, 1).

%F a(n) = tr(m) = tr(k) + tr(k+1) + tr(k+2), where tr(k) = k(k+1)/2 = A000217(k).

%F From _Richard Choulet_, Oct 06 2007: (Start)

%F a(n+2) = 14*a(n+1) - a(n) - 3.

%F a(n+1) = 7*a(n) - 3/2 + 1/2*sqrt(192*a(n)^2 - 96*a(n) - 15).

%F G.f.: x*(10-14*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)

%F a(n) = (4-3*(7-4*sqrt(3))^n*(-2+sqrt(3))+3*(2+sqrt(3))*(7+4*sqrt(3))^n)/16. - _Colin Barker_, Mar 05 2016

%e With tr(k) = k(k+1)/2 = A000217(k):

%e 10 = tr(4) = tr(1) + tr(2) + tr(3) = 1 + 3 + 6,

%e 136 = tr(16) = tr(8) + tr(9) + tr(10) = 36 + 45 + 55,

%e 1891 = tr(61) = tr(34) + tr(35) + tr(36) = 595 + 630 + 666,

%e 26335 = tr(229) = tr(131) + tr(132) + tr(133) = 8646 + 8778 + 8911,

%e 366796 = tr(856) = tr(493) + tr(494) + tr(495) = 121771 + 122265 + 122760.

%t LinearRecurrence[{15,-15,1},{10,136,1891},20] (* _Harvey P. Dale_, Oct 31 2024 *)

%o (PARI) Vec((10*z - 14*z^2 + z^3)/((1-z)*(1 - 14*z + z^2)) + O(z^30)) \\ _Michel Marcus_, Sep 16 2015

%Y Cf. A000217, A128862, A133161.

%K nonn,easy

%O 1,1

%A _Zak Seidov_, May 18 2007