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)
(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
KEYWORD
nonn,base,easy
AUTHOR
Michel Lagneau, Jan 28 2015
STATUS
approved