A091634 Number of primes less than 10^n which do not contain the digit 0.
4, 25, 153, 1010, 7122, 52313, 397866, 3103348, 24649318, 198536215, 1616808581, 13287264748, 110033428309, 917072930187
Offset: 1
Examples
a(3) = 153 because there are 168 primes less than 10^3, 15 primes have at least one zero; 168 - 15 = 153.
Crossrefs
Programs
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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 *) -
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
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
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