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%I #11 Jun 29 2023 18:50:40
%S 5,59,599,5999,59999,599999,5999999,59999999,599999999,5999999999,
%T 59999999999,599999999999,5999999999999,59999999999999,
%U 599999999999999,5999999999999999,59999999999999999,599999999999999999,5999999999999999999,59999999999999999999,599999999999999999999
%N Positive integers a such that f(3a)+f(a)=concatenation of 3a and a, where f(k)=k(k+3)/2 (A000096).
%C Is it difficult to prove that the sequence continues in the expected way?
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11, -10).
%F From _Chai Wah Wu_, Jun 15 2020: (Start)
%F a(n) = 6*10^(n-1) - 1.
%F a(n) = 11*a(n-1) - 10*a(n-2) for n > 2.
%F G.f.: x*(4*x + 5)/((x - 1)*(10*x - 1)).
%F Proof: let m be a term and r be the number of decimal digits of m. Then m satisfies the equation 3m(3m+3)/2 + m(m+3)/2 = 3m*10^r + m = m(3*10^r+1). Solving for m we get m = 6*10^(r-1) - 1 and it is clear that m indeed has r decimal digits.
%F (End)
%e 599 is in the sequence because (3*599)(3*599+3)/2 + 599(602)/2 = 1797*1800/2 + 599*602/2 = 1797599.
%Y Cf. A096032, A099148, A099149, A099150.
%K nonn,base,easy
%O 1,1
%A _John W. Layman_, Sep 30 2004
%E Edited by _Charles R Greathouse IV_, Apr 29 2010
%E More terms from _Chai Wah Wu_, Jun 15 2020
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