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