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 A099151 Positive integers a such that f(3a)+f(a)=concatenation of 3a and a, where f(k)=k(k+3)/2 (A000096). 0
 5, 59, 599, 5999, 59999, 599999, 5999999, 59999999, 599999999, 5999999999, 59999999999, 599999999999, 5999999999999, 59999999999999, 599999999999999, 5999999999999999, 59999999999999999, 599999999999999999, 5999999999999999999, 59999999999999999999, 599999999999999999999 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Is it difficult to prove that the sequence continues in the expected way? LINKS Table of n, a(n) for n=1..21. Index entries for linear recurrences with constant coefficients, signature (11, -10). FORMULA From Chai Wah Wu, Jun 15 2020: (Start) a(n) = 6*10^(n-1) - 1. a(n) = 11*a(n-1) - 10*a(n-2) for n > 2. G.f.: x*(4*x + 5)/((x - 1)*(10*x - 1)). 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. (End) EXAMPLE 599 is in the sequence because (3*599)(3*599+3)/2 + 599(602)/2 = 1797*1800/2 + 599*602/2 = 1797599. CROSSREFS Cf. A096032, A099148, A099149, A099150. Sequence in context: A106105 A178003 A036947 * A243219 A113055 A020468 Adjacent sequences: A099148 A099149 A099150 * A099152 A099153 A099154 KEYWORD nonn,base,easy AUTHOR John W. Layman, Sep 30 2004 EXTENSIONS Edited by Charles R Greathouse IV, Apr 29 2010 More terms from Chai Wah Wu, Jun 15 2020 STATUS approved

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Last modified February 20 21:52 EST 2024. Contains 370219 sequences. (Running on oeis4.)