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A091634
Number of primes less than 10^n which do not contain the digit 0.
30
4, 25, 153, 1010, 7122, 52313, 397866, 3103348, 24649318, 198536215, 1616808581, 13287264748, 110033428309, 917072930187
OFFSET
1,1
FORMULA
Number of primes less than 10^n after removing any primes with at least one digit 0.
a(n) <= A052386(n) = 9*(9^n-1)/8. - Charles R Greathouse IV, Sep 13 2016
a(n) <= (9^n-1)/2 = A052386(n)*4/9 since the last digit of a prime of n digits can only be one of 4 numbers, (2,3,5,7) when n = 1 and (1,3,7,9) when n > 1. - Chai Wah Wu, Mar 18 2018
EXAMPLE
a(3) = 153 because there are 168 primes less than 10^3, 15 primes have at least one zero; 168 - 15 = 153.
MATHEMATICA
NextPrim[n_] := Block[{k = n + 1}, While[ !PrimeQ[k], k++ ]; k]; c = 0; p = 1; Do[ While[ p = NextPrim[p]; p < 10^n, If[ Position[ IntegerDigits[p], 0] == {}, c++ ]]; Print[c]; p--, {n, 1, 8}] (* Robert G. Wilson v, Feb 02 2004 *)
Table[PrimePi[10^n]-Total[Boole[DigitCount[#, 10, 0]>0]&/@ Prime[ Range[ PrimePi[ 10^n]]]], {n, 8}] (* The program generates the first 8 terms of the sequence. To generate more, increase the digit 8 but the program may take a long time to run. *) (* Harvey P. Dale, Aug 26 2021 *)
PROG
(Python)
from sympy import sieve # use primerange for larger terms
def nodigs0(n): return '0' not in str(n)
def aupton(terms):
ps, alst = 0, []
for n in range(1, terms+1):
ps += sum(nodigs0(p) for p in sieve.primerange(10**(n-1), 10**n))
alst.append(ps)
return alst
print(aupton(7)) # Michael S. Branicky, Apr 25 2021
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Enoch Haga, Jan 30 2004
EXTENSIONS
Edited and extended by Robert G. Wilson v, Feb 02 2004
a(9)-a(12) from Donovan Johnson, Feb 14 2008
a(13) from Robert Price, Nov 08 2013
a(14) from Giovanni Resta, Mar 20 2017
STATUS
approved