A254317 -id:A254317 - cp's OEIS frontend

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

Michel Lagneau, Jan 28 2015

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)

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

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

Cf. A027748, A043537, A050252, A254317.

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 :

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 *)

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

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

A254318 Hyper equidigital numbers.

Original entry on oeis.org

2, 3, 4, 5, 7, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 28, 29, 31, 32, 35, 36, 37, 39, 41, 43, 46, 47, 49, 50, 53, 54, 58, 59, 61, 64, 67, 69, 71, 72, 73, 79, 81, 83, 89, 92, 93, 97, 98, 100, 101, 103, 104, 105, 106, 107, 109, 113, 116, 119

Offset: 1

Michel Lagneau, Jan 28 2015

The distinction between the equidigital numbers (A046758) is that only the distinct digits are counted instead of all digits. Hence the definition:

Write n as product of primes raised to powers, let D(n) = 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) and nbd(n) = A043537(n) = number of distinct digits in n; sequence gives n such that D(n) = nbd(n).

			116 is in the sequence because 116 = 2^2*29 => D(116)= A043537(116)=2.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A043537, A046760, A046758, A046759, A254315, A254317, A254319, A254321.

Cases[Range[400], n_ /; Length[Union[Flatten[IntegerDigits[FactorInteger[n] /. 1 -> Sequence[]]]]]==Length[Union[Flatten[IntegerDigits[n]]]]]

for(n=1,100,s=[];F=factor(n);for(i=1,#F[,1],s=concat(s,digits(F[i,1]));if(F[i,2]>1,s=concat(s,digits(F[i,2]))));if(#vecsort(digits(n),,8)==#vecsort(s,,8),print1(n,", "))) \\ Derek Orr, Jan 30 2015

cp's OEIS Frontend

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

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