A099305 Number of solutions of the equation (n+k)' = n' + k', with 1 <= k <= 2n, where n' denotes the arithmetic derivative of n.

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, 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, 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

Offset: 1

Author

T. D. Noe, Oct 12 2004

Keywords

• nonn

Comments

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.

a(A258138(n)) = n and a(m) != n for m < A258138(n). - Reinhard Zumkeller, May 21 2015

References

• See A003415

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003415 (arithmetic derivative of n), A099304 (least k > 0 such that (n+k)' = n' + k').

Cf. A258138.

Programs

Haskell

a099305 n = a099305_list !! (n-1)
a099305_list = f 1 $ h 1 empty where
   f x ad = y : f (x + 1) (h (3 * x + 1) ad)  where
            y = length [() | k <- [1 .. 2 * x],
                             let x' = ad ! x, ad ! (x + k) == x' + ad ! k]
   h z = insert z (a003415 z) .
          insert (z+1) (a003415 (z+1)) . insert (z+2) (a003415 (z+2))
-- Reinhard Zumkeller, May 21 2015

Mathematica

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}]