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A211222
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Minimum integer m for which n·m = n'·m', where n' and m' are the arithmetic derivatives of n and m. Zero if m does not exist.
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1
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16, 64, 4, 1024, 20, 16384, 9, 8, 1372, 4194304, 0, 67108864, 0, 0, 2, 17179869184, 77948684, 274877906944, 6, 1000, 42417997492, 70368744177664, 1771561, 32, 300, 3125, 0, 288230376151711744, 550618520345910837374536871905139185678862401, 4611686018427387904
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OFFSET
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2,1
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COMMENTS
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The entries of the sequence are the solution of the differential equation m'=n/n'*m.
If n=n' then m'=m (A051674) and m=p^p, with p prime. Taking the minimum prime p=2, m=4.
If n'|n then m'=b*m, where b=n/n', and m=2^(2*b).
In general n and n’ can have same common factor. Let us consider the fraction k/h obtained by reduction of n/n' [GCD(k,h)=1]. If h=p, p prime, then if k>h m=4*p^(k-h) else if k<h m=p^k.
If h is composite let us make the variable substitution m=h*t. Now, m'=h'*t+h*t'=(k/h)*h*t=k*t that leads to t'=1/h*(k-h')*t. Let us reduce the fraction (k-h')/h, if necessary. If we get a prime denominator we can solve the equation by the previous formulas. Otherwise we must iterate the variable substitutions until we get:
1) a prime denominator.
2) a differential equation of the type y'=0, whose solution is y=1.
3) a differential equation of the type y'=x*y, with x<0. In this case there is no solution. This corresponds to the zeros in the sequence (12, 14, 15, 28, 35, 39, 44, 46, 50, 51, 55, 65, etc.).
In cases 1) and 2), moving backward through all the substitutions, we reach the final solution.
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LINKS
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EXAMPLE
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n=18; n'=11. m'=18/11*m. Here n>n', n'=p and therefore m=4*11^7.
n=24; n'=44. m'=24/44*m. Simplifying the fraction we have m'=6/11*m. Here k<k' e k'=p and therefore m=116.
n=44; n'=48. m'=44/48*m. Simplifying the fraction we have m'=11/12*m. Let m=12t, m’=16t+12t'=11t. We have t'=-5/12*t that has no solutions. Therefore a(44)=0.
n=62; n'=33. m'=62/33*m. Let m=33*t, m'=14*t+33*t'=62*t. We have t’=48/33*t=16/11*t. therefore t=4*11^5 and m=33*4*11^5.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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