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

Author

Michel Lagneau, Jan 28 2015

Keywords

• nonn
• base
• easy

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)

Examples

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

Links

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

Crossrefs

Cf. A027748, A043537, A050252, A254317.

Programs

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 *)

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