A129152 The n-th arithmetic derivative of 5^6.

15625, 18750, 34375, 37500, 87500, 187500, 475000, 1212500, 2437500, 6362500, 12737500, 25487500, 50987500, 101987500, 206975000, 530037500, 1060087500, 3890025000, 15175012500, 45525375000, 177026512500, 596222100000, 2708984250000, 12765250350000

Offset

0,1

Comments

In general, the trajectory of p^(p+1) under A003415 is equal to p^p times the trajectory of p under A129283: n -> n + n'. Here we have the case p = 5 (see A129286 for a(n)/5^5), see A129150 and A129151 for p = 2 and 3. - M. F. Hasler, Nov 28 2019

Links

Table of n, a(n) for n=0..23.

Formula

a(n+1) = A003415(a(n)), a(0) = 5^6 = 15625.

a(n) = A129286(n)*5^5; A129251(a(n)) > 0. - Reinhard Zumkeller, Apr 07 2007

Mathematica

dn[0] = 0; dn[1] = 0; dn[n_?Negative] := -dn[-n]; dn[n_] := Module[{f = Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Total[n*f[[2]]/f[[1]]]]]; s = 5^6; Join[{s}, Table[s = dn[s], {18}]] (* T. D. Noe, Mar 07 2013 *)

Prog

(Haskell)
a129152 n = a129152_list !! n
a129152_list = iterate a003415 15625  -- Reinhard Zumkeller, Apr 29 2012
(PARI) A129152_upto(N)=vector(N, n, N=if(n>1, A003415(N), 5^6)) \\ gives a(0..N-1). To get a(1..N) put A003415() around if() instead inside.  M. F. Hasler, Nov 28 2019

Crossrefs

Cf. A129150, A129151, A068327.

Cf. A099309, A051674, A100716.

Sequence in context: A206183 A205365 A254029 * A223185 A097219 A016818

Adjacent sequences: A129149 A129150 A129151 * A129153 A129154 A129155

Keyword

nonn

Author

Reinhard Zumkeller, Apr 01 2007

Status

approved