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Start with the seed a(0)=2. The minimum number, different from 1, that multiplied by 2 (seed) produces a number with 2 as its rightmost digit is a(1)=6. Then 6*2=12. Again, the minimum number that multiplied by 12 produces 12 as its rightmost digits is a(2)=26 (12*26=312). And so on.
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%I #23 Aug 08 2023 11:58:56

%S 2,6,26,126,6251,62500001,6250000000000001,

%T 6250000000000000000000000000001,

%U 62500000000000000000000000000000000000000000000000000000000001

%N Start with the seed a(0)=2. The minimum number, different from 1, that multiplied by 2 (seed) produces a number with 2 as its rightmost digit is a(1)=6. Then 6*2=12. Again, the minimum number that multiplied by 12 produces 12 as its rightmost digits is a(2)=26 (12*26=312). And so on.

%D G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 99.

%e a(0)=2;

%e a(1)=6 because 2*6 = 12;

%e a(2)=26 because 12*26 = 312;

%e a(3)=126 because 312*126 = 39312;

%e a(4)=6251 because 39312*6251 = 245739312;

%e a(5)=62500001 because 245739312*62500001 = 15358707245739312.

%p P:=proc(q,h) local a,b,k,n; a:=h; b:=ilog10(a)+1; print(h);

%p for k from 1 to 10 do for n from 2 to q do

%p if ((a*n) mod 10^b)=a then print(n); a:=a*n; b:=ilog10(a)+1;

%p break; fi; od; od; end: P(10^9,2);

%o (Python)

%o import math

%o p,n = 2,0

%o while n<10:

%o ndigits, oldp = len(str(p)), p

%o p += math.lcm(p,10**ndigits)

%o print("a(%d) = %d"%(n:=n+1, p//oldp))

%o # _Bert Dobbelaere_, Aug 08 2023

%Y Cf. A123873, A123874, A123875.

%K nonn,base

%O 0,1

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Oct 16 2006

%E More terms from _Bert Dobbelaere_, Aug 08 2023