

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
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OFFSET

1,1


COMMENTS

Is it difficult to prove that the sequence continues in the expected way?


LINKS



FORMULA

a(n) = 6*10^(n1)  1.
a(n) = 11*a(n1)  10*a(n2) 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^(r1)  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



EXTENSIONS



STATUS

approved



