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%I #7 May 21 2015 21:25:42
%S 1,2,1,2,1,2,1,2,1,2,2,2,1,3,1,2,1,2,1,2,1,3,1,2,1,3,1,3,2,2,1,2,2,2,
%T 2,2,1,3,1,2,1,3,1,3,1,2,2,2,1,3,1,3,1,2,3,3,1,3,1,2,1,3,1,2,1,3,2,2,
%U 1,4,1,2,1,2,1,3,2,3,2,2,2,2,3,3,2,2,2,3,1,2,2,2,1,3,2,2,1,3,3,3,1,2,2,3,2
%N Number of solutions of the equation (n+k)' = n' + k', with 1 <= k <= 2n, where n' denotes the arithmetic derivative of n.
%C Observe that when n and c*n have the same parity, a(c*n) >= a(n) for all integers c. For even n, there are always at least two solutions, k=n/2 and k=2n. For odd n, k=2n is always a solution.
%C a(A258138(n)) = n and a(m) != n for m < A258138(n). - _Reinhard Zumkeller_, May 21 2015
%D See A003415
%H Reinhard Zumkeller, <a href="/A099305/b099305.txt">Table of n, a(n) for n = 1..10000</a>
%t dn[0]=0; dn[1]=0; dn[n_]:=Module[{f=Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Plus@@(n*f[[2]]/f[[1]])]]; Table[lst={}; k=0; While[k<2n, k++; While[k<=2n && dn[n]+dn[k] != dn[n+k], k++ ]; If[dn[n]+dn[k]==dn[n+k], AppendTo[lst, k]]]; Length[lst], {n, 100}]
%o (Haskell)
%o a099305 n = a099305_list !! (n-1)
%o a099305_list = f 1 $ h 1 empty where
%o f x ad = y : f (x + 1) (h (3 * x + 1) ad) where
%o y = length [() | k <- [1 .. 2 * x],
%o let x' = ad ! x, ad ! (x + k) == x' + ad ! k]
%o h z = insert z (a003415 z) .
%o insert (z+1) (a003415 (z+1)) . insert (z+2) (a003415 (z+2))
%o -- _Reinhard Zumkeller_, May 21 2015
%Y Cf. A003415 (arithmetic derivative of n), A099304 (least k > 0 such that (n+k)' = n' + k').
%Y Cf. A258138.
%K nonn
%O 1,2
%A _T. D. Noe_, Oct 12 2004
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