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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
OFFSET
1,1
COMMENTS
Is it difficult to prove that the sequence continues in the expected way?
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
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