A093852 a(n) = 10^(n-1) - 1 + n*floor(9*10^(n-1)/(n+1)).
4, 69, 774, 8199, 84999, 871425, 8874999, 89999999, 909999999, 9181818179, 92499999999, 930769230759, 9357142857140, 93999999999999, 943749999999999, 9470588235294111, 94999999999999999, 952631578947368403, 9549999999999999999, 95714285714285714279
Offset: 1
Examples
n-th row of the following triangle contains n uniformly located n-digit numbers. i.e. n terms of an arithmetic progression with 10^(n-1)-1 as the term preceding the first term and (n+1)-th term is the largest possible n-digit term. Given the triangle defined in A093850: ...4; ..39 69; .324 549 774; 2799 4599 6399 8199..... then this sequence is the leading diagonal.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..995
Programs
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Magma
[10^(n-1) -1 +n*Floor(9*10^(n-1)/(n+1)): n in [1..25]]; // G. C. Greubel, Mar 21 2019
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Maple
A093852 := proc(n) r := n ; 10^(n-1)-1+r*floor(9*10^(n-1)/(n+1)) ; end proc: seq(A093852(n),n=1..50) ; # R. J. Mathar, Oct 01 2011
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Mathematica
Table[10^(n-1) -1 +n*Floor[9*10^(n-1)/(n+1)], {n,25}] (* G. C. Greubel, Mar 21 2019 *)
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PARI
{a(n) = 10^(n-1) -1 +n*floor(9*10^(n-1)/(n+1))}; \\ G. C. Greubel, Mar 21 2019
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Sage
[10^(n-1) -1 +n*floor(9*10^(n-1)/(n+1)) for n in (1..25)] # G. C. Greubel, Mar 21 2019
Comments