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Smallest exponent m such that 2^m begins and ends with the same n digits, allowing any other digits in between.
0

%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