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%I #15 Sep 08 2022 08:46:11
%S 16,24,458,18341,16641,1116830,10971256,105969200,193295556,
%T 48221950695,72585186028
%N Smallest exponent m such that 2^m begins and ends with the same n digits, allowing any other digits in between.
%e 2^24 = 16777216 starts and ends with the same two digits.
%e Solution details:
%e A is the substring of beginning and ending digits
%e B is the total number of digits in 2^a(n)
%e n.......a(n)..........A..........B
%e 1.........16..........6..........5
%e 2.........24.........16..........8
%e 3........458........744........138
%e 4......18341.......1552.......5522
%e 5......16641......27552.......5010
%e 6....1116830.....213824.....336200
%e 7...10971256....1399936....3302678
%e 8..105969200...65541376...31899908 - _Lars Blomberg_, May 07 2015
%o (Magma) lst:=[]; c:=Log(10, 2); for n in [1..5] do k:=Ceiling(2*n/c); t:=0; while t eq 0 do m:=2^k mod 10^n; if #Intseq(m) eq n then if Truncate(2^k/10^(Floor(k*c+1)-n)) eq m then Append(~lst, k); t:=1; end if; end if; k+:=1; end while; end for; lst;
%K nonn,base,more
%O 1,1
%A _Arkadiusz Wesolowski_, Apr 01 2015
%E a(6)-a(8) from _Lars Blomberg_, May 07 2015
%E a(9)-a(11) from _Hiroaki Yamanouchi_, Jun 03 2015