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A211222 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. 1

%I #20 Jun 16 2016 23:27:45

%S 16,64,4,1024,20,16384,9,8,1372,4194304,0,67108864,0,0,2,17179869184,

%T 77948684,274877906944,6,1000,42417997492,70368744177664,1771561,32,

%U 300,3125,0,288230376151711744,550618520345910837374536871905139185678862401,4611686018427387904

%N 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.

%C The entries of the sequence are the solution of the differential equation m'=n/n'*m.

%C If n=n' then m'=m (A051674) and m=p^p, with p prime. Taking the minimum prime p=2, m=4.

%C If n'|n then m'=b*m, where b=n/n', and m=2^(2*b).

%C 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.

%C 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:

%C 1) a prime denominator.

%C 2) a differential equation of the type y'=0, whose solution is y=1.

%C 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.).

%C In cases 1) and 2), moving backward through all the substitutions, we reach the final solution.

%H Paolo P. Lava, <a href="/A211222/b211222.txt">Table of n, a(n) for n = 2..100</a>

%e n=18; n'=11. m'=18/11*m. Here n>n', n'=p and therefore m=4*11^7.

%e 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.

%e 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.

%e 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.

%Y Cf. A003415, A051674, A165558.

%K nonn

%O 2,1

%A _Paolo P. Lava_, Apr 20 2012

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