Alec Mihailovs

Dr. Aleksandrs Mihailovs

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19 years, 276 days
Mihailovs, Inc.
Owner, President, and CEO
Tyngsboro, Massachusetts, United States

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I received my Ph.D. from the University of Pennsylvania in 1998 and I have been teaching since then at SUNY Oneonta for 1 year, at Shepherd University for 5 years, at Tennessee Tech for 2 years, at Lane College for 1 year, and this year I taught at the University of Massachusetts Lowell. My research interests include Representation Theory and Combinatorics.

MaplePrimes Activity

These are replies submitted by Alec Mihailovs

Correspondingly, here is the improved version of a1,
local A,i,j,k,arem,rem,dig;
arem:=proc() rem:=irem(b*j,i); dig:=convert(j,base,b); 
          k = 1 .. iquo(b-1+rem,i)) end;
A[1] := [$1..b-1];
for i from 2 do
     A[i] := [seq](arem(), j = A[i-1]);
     if A[i] = [] then break fi od;
i-1 = A[i-1]
For decimal system, it works very fast,
tt:=time(): a1(10); time()-tt;

                           9 = [381654729]

In addition to the values in the worksheet, I calculated
                   18 = [509504810627259318940020]


                   19 = [5213608783930183587520297]


  21 = [3017921136920002608782486643, 3202513921881602294966195322,

        10912277070078955588719411381, 11506928512609494037599575154,

        19136629049078922486640055262, 19563154864313291881366262910,

        20927442077211608842020972981, 22095893110934309880723905997,

        24479055281091281744429147151, 26034588683899462487295851544,

        27944901128718982841182304058, 28236074834439966722409861288,

        30327538677373615144106864301, 30327538677373615144106864322,

        33286709815522787435710859625, 33637644479370238638607793874,

s:=proc(L) local n,a,b,i;

For small b, say from 2 to 20, or 30, K(b) ~ (b-1)*e gives a good approximation. More precise asymptotics that is better for large b, such as b close to 1,000,000 and larger, seems to be K(b) ~ b*e - ln(b)*γ
Looking at my user number, 135, Joe Riel's 84, and Thomas Richard's 50, I found out that we have something in common. gcd(50,84)=2 gcd(50,135)=5 gcd(84,135)=3
The same calculation for even bases shows that the maximal number of digits cannot be b -2 for even bases - either it is b -1, or ≤b -3, because for a (b -2)-digit number with all digits different and non-zero, adding the remaining non-zero digit at the end automatically produces a number divisible by b -1. P.S. I called the previous comment Even Bases so I didn't have other choice as to call this Odd Bases. They are odd as in the well-known phrase: "All prime numbers are odd and 2 is the oddest."
Thank you, Joe. Both things are good. The next improvement can be achieved by separating the loop in 2 parts,
local A,i,j,k,arem,brem,rem;
arem:=proc() rem:= irem(b*j,i); 
     seq(b*j+k*i-rem,k = `if`(rem=0,0,1) .. iquo(b-1+rem,i)) end;
brem:=proc() rem:=irem(b*j,i);
     if rem=0 then b*j elif b-1+rem<i then NULL else b*j+i-rem fi end;
A[1] := [$1..b-1];
for i from 2 to b-1 do
     A[i] := [seq](arem(), j = A[i-1]) od;
for i from b do 
     A[i] := [seq](brem(), j = A[i-1]);
     if A[i] = [] then break fi od;
i-1 = A[i-1]
Meanwhile, I calculated a(16): 39 = [18872900738885736149574055538327802527212537551, 60753927368683934227793588395570842550542338031, 89515749136034833729775437005460258167590093634] Converted to hex, the numbers look like 34E4A468166CD8604EC0F8106AB4326098286CF AA44CE207C78FC30003C3CC0D8382E2078D07EF FAE06678C2E884607EB8B4E0B0A0F0603420342
Yes, there is a simple proof. A number written in base b is divisible by b -1 if and only if the sum of digits is divisible by b -1 because a[0]*b^k + a[1]*b^(k-1) + a[k] mod (b-1) = a[0] + a[1]+ a[k] mod (b-1). If all b -1 digits are different and non-zero, then their sum is 1+2+...+(b -1) = b*(b -1)/2. This number, divided by b -1 is b/2 that is not integer for odd b. Thus, there are no (b -1)-digit numbers in base b with all digits different and non-zero, divisible by b -1, q.e.d.
The probability that a K-digit number is divisible by K is approximately 1/K. The probability that first (K-1) digit of it is divisible by (K-1) is approximately 1/(K-1), etc... That gives a probability that a K-digit number satisfies the problem conditions, approximately 1/K! Now, there are (b-1)*b^(K-1) K-digit numbers in base b. Thus, the expected number of them, satisfying the problem conditions, is (b-1)*b^(K-1)/K! Approximating factorial by Stirling formula, we get the following asymptotics of K with expected number 1, K(b) ~ (b-1)*e that is pretty close to the calculated maximal possible values of K.
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