This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A211222 #20 Jun 16 2016 23:27:45 %S A211222 16,64,4,1024,20,16384,9,8,1372,4194304,0,67108864,0,0,2,17179869184, %T A211222 77948684,274877906944,6,1000,42417997492,70368744177664,1771561,32, %U A211222 300,3125,0,288230376151711744,550618520345910837374536871905139185678862401,4611686018427387904 %N 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. %C A211222 The entries of the sequence are the solution of the differential equation m'=n/n'*m. %C A211222 If n=n' then m'=m (A051674) and m=p^p, with p prime. Taking the minimum prime p=2, m=4. %C A211222 If n'|n then m'=b*m, where b=n/n', and m=2^(2*b). %C A211222 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 A211222 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 A211222 1) a prime denominator. %C A211222 2) a differential equation of the type y'=0, whose solution is y=1. %C A211222 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 A211222 In cases 1) and 2), moving backward through all the substitutions, we reach the final solution. %H A211222 Paolo P. Lava, <a href="/A211222/b211222.txt">Table of n, a(n) for n = 2..100</a> %e A211222 n=18; n'=11. m'=18/11*m. Here n>n', n'=p and therefore m=4*11^7. %e A211222 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 A211222 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 A211222 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 A211222 Cf. A003415, A051674, A165558. %K A211222 nonn %O A211222 2,1 %A A211222 _Paolo P. Lava_, Apr 20 2012