A254315 Number of distinct digits in the prime factorization of n (counting terms of the form p^1 as p).

1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 3, 3, 2, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 2, 2, 2, 3, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3

Offset

2,5

Comments

Write n as product of primes raised to powers; then a(n) is the total number of distinct digits in product representation (number of distinct digits in all the primes and number of distinct digits in all the exponents that are greater than 1).

a(n)<=10. The least n such that a(n)=10 is n = 41701690 = 2*5*47*83*1069.

Property: a(p) = A043537(p), for p prime.

From Michel Marcus, Feb 21 2015: (Start)

For p in A038604, a(p^2) = A043537(p) + 1.

For p in A038611, a(p^3) = A043537(p) + 1.

For p in A038612, a(p^4) = A043537(p) + 1.

For p in A038613, a(p^5) = A043537(p) + 1.

For p in A038614, a(p^6) = A043537(p) + 1.

For p in A038615, a(p^7) = A043537(p) + 1.

For p in A038616, a(p^8) = A043537(p) + 1.

For p in A038617, a(p^9) = A043537(p) + 1.

(End)

Links

Michel Lagneau, Table of n, a(n) for n = 2..10000

Example

a(36)=2 because 36 = 2^2 * 3^2 => 2 distinct digits.
a(414)=2 because 414 = 2 * 3^2 * 23 => 2 distinct digits.

Maple

with(ListTools):
nn:=100:
  for n from 2 to nn do:
    n0:=length(n):lst:={}:x0:=ifactors(n):
    y:=Flatten(x0[2]):z:=convert(y, set):
    z1:=z minus {1}:nn0:=nops(z1):
     for k from 1 to nn0 do :
      t1:=convert(z1[k], base, 10):z2:=convert(t1, set):
      lst:=lst union z2:
     od:
     nn1:=nops(lst):printf(`%d, `, nn1):
     od :

Mathematica

f[n_] := Block[{pf = FactorInteger@ n, i}, Length@ DeleteDuplicates@ Flatten@ IntegerDigits@ Rest@ Flatten@ Reap@ Do[If[Last[pf[[i]]] == 1, Sow@ First@ pf[[i]], Sow@ FromDigits@ Flatten[IntegerDigits /@ pf[[i]]]], {i, Length@ pf}]]; Array[f, 100] (* Michael De Vlieger, Jan 29 2015 *)

Prog

(PARI) print1(1, ", "); for(k=2, 100, s=[]; F=factor(k); for(i=1, #F[, 1], s=concat(s, digits(F[i, 1])); if(F[i, 2]>1, s=concat(s, digits(F[i, 2])))); print1(#vecsort(s, , 8), ", ")) \\ Derek Orr, Jan 30 2015
(Python)
from sympy import factorint
def A254315(n):
....return len(set([x for l in [[d for d in str(p)]+[d for d in str(e) if d != '1'] for p, e in factorint(n).items()] for x in l]))
# Chai Wah Wu, Feb 24 2015

Crossrefs

Cf. A027748, A043537, A050252, A254317.

Sequence in context: A160242 A043529 A201219 * A080942 A099812 A246600

Adjacent sequences: A254312 A254313 A254314 * A254316 A254317 A254318

Keyword

nonn,base,easy

Author

Michel Lagneau, Jan 28 2015

Status

approved