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A200993
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Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3*T(m) = 2*T(k) for some k.
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9
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0, 10, 990, 97020, 9506980, 931587030, 91286021970, 8945098566040, 876528373449960, 85890835499530050, 8416425350580494950, 824723793521388975060, 80814515339745539060940, 7918997779501541438997070, 775980967875811315482651930, 76038215854050007375860892080
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OFFSET
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0,2
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COMMENTS
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For n>1, a(n) = 98*a(n-1) - a (n-2) + 10. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1)= A014105(m) and for n>1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m).
Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).
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LINKS
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FORMULA
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G.f. 10*x / ((1-x)*(x^2-98*x+1)). - R. J. Mathar, Dec 20 2011
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3) for n>2. - Colin Barker, Mar 02 2016
a(n) = (-10+(5-2*sqrt(6))*(49+20*sqrt(6))^(-n)+(5+2*sqrt(6))*(49+20*sqrt(6))^n)/96. - Colin Barker, Mar 07 2016
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EXAMPLE
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3*0 = 2*0.
3*10 = 2*15.
3*990 = 2*1485.
3*97020 = 2*145530.
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MATHEMATICA
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LinearRecurrence[{99, -99, 1}, {0, 10, 990}, 20] (* Harvey P. Dale, Feb 25 2018 *)
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PROG
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(PARI) concat(0, Vec(10*x/((1-x)*(1-98*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 02 2016
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(10*x/((1-x)*(x^2-98*x+1)))); // G. C. Greubel, Jul 15 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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