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A307019
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Squares which can be expressed as the product of a number and its reversal in exactly three different ways.
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6
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6350400, 43560000, 635040000, 768398400, 4356000000, 42033200400, 55847142400, 63504000000, 64780430400, 72694944400, 76839840000, 78243278400, 234101145600, 435600000000, 4203320040000, 5086017248400, 5584714240000, 6350400000000, 6363107150400, 6478043040000, 6757504230400
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OFFSET
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1,1
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COMMENTS
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1) Why do all these terms end with an even number of zeros?
1.1) Is it possible to find a term that does not end with zeros? If such a term m exists, this number must satisfy the Diophantine equation m^2 = a*rev(a) = b*rev(b) = c*rev(c). No solution (m,a,b,c) with m that does not end with zeros is known.
1.2) Consider now the Diophantine equation: m^2 = a*rev(a) = b*rev(b) where a is a palindrome and b is not a palindrome. For each solution (m,a,b), we generate terms (10*m)^2 of this sequence and we get: (10*m)^2 = 100 * m^2 = (100*a)*(rev(100*a) = (100*b)*(rev(100*b)) = (100*rev(b)) * (rev(100*rev(b))).
Example: with a(1) = 63504 = 252^2 = 252 * 252 = 144 * 441, so (m,a,b) = (63504,252,144), we obtain the 3 following ways: 6350400 = 25200 * 252 = 14400 * 441 = 44100 * 144.
2) When can square numbers be expressed in this way in more than three different ways?
If the Diophantine equation: m^2 = a*rev(a) = b*rev(b), with a <> b and a and b not palindromes has a solution, then it is possible to get integers equal to (10*m)^2 which can be expressed as the product of a number and its reversal in exactly four different ways.
We don't know if such a solution (m,a,b) exists.
David A. Corneth has found 70 terms < 6*10^15 belonging to this sequence (see links in A083408), but no square has four solutions for m^2 = k * rev(k) until 6*10^15.
There is no square less than 10^24 with 4 or more different ways. - Chai Wah Wu, Apr 12 2019
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LINKS
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EXAMPLE
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6350400 = 2520^2 = 25200 * 252 = 14400 * 441 = 44100 * 144.
43560000 = 6600^2 = 660000 * 66 = 52800 * 825 = 82500 * 528.
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PROG
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(PARI) is(n) = {if(!issquare(n), return(0)); my(d = divisors(n), t = 0); forstep(i = #d, #d \ 2 + 1, -1, revd = fromdigits(Vecrev(digits(d[i]))); if(revd * d[i] == n, t++; if(t > 3, return(0)); ) ); t==3 } \\ David A. Corneth, Mar 20 2019
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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