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%I #65 Mar 31 2023 10:16:33
%S 0,120,528,139128,609960,160554240,703893960,185279454480,
%T 812293020528,213812329916328,937385441796000,246739243443988680,
%U 1081741987539564120,284736873122033021040,1248329316235215199128,328586104843582662292128,1440570949193450800230240,379188080252621270252095320
%N Triangular numbers that are eight times another triangular number.
%C The triangular numbers T(t) that are eight times another triangular number T(u) : T(t) = 8*T(u). The t's are in A336625, the T(u)'s are in A336624 and the u's are in A336623.
%C Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.
%H Vladimir Pletser, <a href="/A336626/b336626.txt">Table of n, a(n) for n = 1..653</a>
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2101.00998">Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers</a>, arXiv:2101.00998 [math.NT], 2021.
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.12392">Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2102.12392 [math.GM], 2021.
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.13494">Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations</a>, arXiv:2102.13494 [math.NT], 2021.
%H Vladimir Pletser, <a href="https://www.researchgate.net/profile/Vladimir-Pletser/publication/359808848_USING_PELL_EQUATION_SOLUTIONS_TO_FIND_ALL_TRIANGULAR_NUMBERS_MULTIPLE_OF_OTHER_TRIANGULAR_NUMBERS/">Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers</a>, 2022.
%H V. Pletser, <a href="https://doi.org/10.1007/s13226-021-00172-y">Recurrent relations for triangular multiples of other triangular numbers</a>, Indian J. Pure Appl. Math. 53 (2022) 782-791
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1154,-1154,-1,1).
%F a(n) = 8*A336624(n).
%F a(n) = 1154*a(n-2) - a(n-4) + 648, for n>=2 with a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
%F a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=528, a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
%F a(n) = ((10*sqrt(2))/17 + 15/17)*(17 + 12*sqrt(2))^n + (-(10*sqrt(2))/17 + 15/17)*(17 - 12*sqrt(2))^n + (-15/17 - (45*sqrt(2))/68)*(-17 - 12*sqrt(2))^n + (-15/17 + (45*sqrt(2))/68)*(-17 + 12*sqrt(2))^n - 27*(-4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 + 12*sqrt(2)))^n/(1088*(-17 + 12*sqrt(2))) - 27*(4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 - 12*sqrt(2)))^n/(1088*(-17 - 12*sqrt(2))) - 9/16 - 9*(-3 + 2*sqrt(2))*sqrt(2)*(-1/(17 - 12*sqrt(2)))^n/(272*(17 - 12*sqrt(2))) - 9*(3 + 2*sqrt(2))*sqrt(2)*(-1/(17 + 12*sqrt(2)))^n/(272*(17 + 12*sqrt(2))).
%F Let b(n) be A336625(n). Then a(n) = b(n)*(b(n)+1)/2.
%F G.f.: 24*x^2*(5 + 17*x + 5*x^2)/(1 - x - 1154*x^2 + 1154*x^3 + x^4 - x^5). - _Stefano Spezia_, Oct 05 2020
%F From _Vladimir Pletser_, Feb 21 2021: (Start)
%F a(n) = ((11*(1 + sqrt(2))^2 - (-1)^n*6*(4 + 3*sqrt(2)))*(1 + sqrt(2))^(4n) + (11*(1 - sqrt(2))^2 - (-1)^n*6*(4 - 3*sqrt(2)))*(1 - sqrt(2))^(4n))/32 - 9/16.
%F a(n) = ((1 + 2*sqrt(2))^2*(1 + sqrt(2))^(4n) + (1 - 2*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for even n.
%F a(n) = ((5 + 4*sqrt(2))^2*(1 + sqrt(2))^(4n) + (5 - 4*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for odd n. (End)
%e a(2) = 120 is a term because it is triangular and 120/8 = 15 is also triangular.
%e a(3) = 1154*a(1) - a(-1) + 648 = 0 - 120 + 648 = 528;
%e a(4) = 1154*a(2) - a(0) + 648 = 1154*120 - 0 + 648 = 139128, etc.
%e .
%e From _Peter Luschny_, Oct 19 2020: (Start)
%e Related sequences in context, as computed by the Julia function:
%e n [A336623, A336624, A336625, A336626 ]
%e [0] [0, 0, 0, 0 ]
%e [1] [5, 15, 15, 120 ]
%e [2] [11, 66, 32, 528 ]
%e [3] [186, 17391, 527, 139128 ]
%e [4] [390, 76245, 1104, 609960 ]
%e [5] [6335, 20069280, 17919, 160554240 ]
%e [6] [13265, 87986745, 37520, 703893960 ]
%e [7] [215220, 23159931810, 608735, 185279454480 ]
%e [8] [450636, 101536627566, 1274592, 812293020528 ]
%e [9] [7311161, 26726541239541, 20679087, 213812329916328] (End)
%p f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 648, a(2) = 120, a(1) = 0, a(0) = 0, a(-1) = 120}, a(n), remember); map(f, [$ (1 .. 1000)])[]; #
%t LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 120, 528, 139128, 609960}, 18]
%o (Julia)
%o function omnibus()
%o println("[A336623, A336624, A336625, A336626]")
%o println([0, 0, 0, 0])
%o t, h = 1, 1
%o for n in 1:999999999
%o d, r = divrem(t, 8)
%o if r == 0
%o d2 = 2*d
%o s = isqrt(d2)
%o d2 == s * (s + 1) && println([s, d, n, t])
%o end
%o t, h = t + h + 1, h + 1
%o end
%o end
%o omnibus() # _Peter Luschny_, Oct 19 2020
%Y Subsequence of A000217.
%Y Cf. A336623, A336624, A336625.
%Y Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077260, A077261, A077262, A077288, A077289, A077290, A077291, A077398, A077399, A077400, A077401.
%K easy,nonn
%O 1,2
%A _Vladimir Pletser_, Oct 04 2020
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