A084984 Numbers containing no prime digits.
0, 1, 4, 6, 8, 9, 10, 11, 14, 16, 18, 19, 40, 41, 44, 46, 48, 49, 60, 61, 64, 66, 68, 69, 80, 81, 84, 86, 88, 89, 90, 91, 94, 96, 98, 99, 100, 101, 104, 106, 108, 109, 110, 111, 114, 116, 118, 119, 140, 141, 144, 146, 148, 149, 160, 161, 164, 166, 168, 169
Offset: 1
Examples
166 has digits 1 and 6 and they are nonprime digits. a(1000) = 8686. a(10^4) = 118186 a(10^5) = 4090986. a(10^6) = 66466686.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
- Index entries for 10-automatic sequences.
Crossrefs
Programs
-
Haskell
a084984 n = a084984_list !! (n-1) a084984_list = filter (not . any (`elem` "2357") . show ) [0..] -- Reinhard Zumkeller, Jul 19 2011
-
Magma
[n: n in [0..169] | forall{d: d in [2,3,5,7] | d notin Set(Intseq(n))}]; // Bruno Berselli, Jul 19 2011
-
Mathematica
npdQ[n_]:=And@@Table[FreeQ[IntegerDigits[n],i],{i,{2,3,5,7}}]; Select[ Range[ 0,200],npdQ] (* Harvey P. Dale, Jul 22 2013 *)
-
PARI
is(n)=isprime(eval(Vec(Str(n))))==0 \\ Charles R Greathouse IV, Feb 20 2012
-
PARI
my(table=[0,1,4,6,8,9]); \ a(n) = fromdigits([table[d+1] |d<-digits(n-1,6)]); \\ Kevin Ryde, May 27 2025
Formula
A193238(a(n)) = 0. - Reinhard Zumkeller, Jul 19 2011
a(n) >> n^1.285. - Charles R Greathouse IV, Feb 20 2012
From Hieronymus Fischer, May 30 and Jun 25 2012: (Start)
a(n) = ((2*b_m(n)+1) mod 10 + floor((b_m(n)+4)/5) - floor((b_m(n)+1)/5))*10^m + sum_{j=0..m-1} ((2*b_j(n))) mod 12 + floor(b_j(n)/6) - floor((b_j(n)+1)/6) + floor((b_j(n)+4)/6) - floor((b_j(n)+5)/6)))*10^j, where n>1, b_j(n)) = floor((n-1-6^m)/6^j), m = floor(log_6(n-1)).
Special values:
a(1*6^n+1) = 1*10^n.
a(2*6^n+1) = 4*10^n.
a(3*6^n+1) = 6*10^n.
a(4*6^n+1) = 8*10^n.
a(5*6^n+1) = 9*10^n.
a(2*6^n) = 2*10^n - 1.
a(n) = 10^log_6(n-1) for n=6^k+1, k>0.
Inequalities:
a(n) < 10^log_6(n-1) for 6^k+10.
a(n) > 10^log_6(n-1) for 2*6^k=0.
a(n) <= 4*10^(log_6(n-1)-log_6(2)) = 1.641372618*10^(log_6(n-1)), equality holds for n=2*6^k+1, k>=0.
a(n) > 2*10^(log_6(n-1)-log_6(2)) = 0.820686309*10^(log_6(n-1)).
a(n) >= A202267(n), equality holds if the representation of n-1 as a base-6 number has only digits 0 or 1.
Lower and upper limits:
lim inf a(n)/10^log_6(n) = 2/10^log_6(2) = 0.820686309, for n --> inf.
lim sup a(n)/10^log_6(n) = 4/10^log_6(2) = 1.641372618, for n --> inf.
where 10^log_6(n) = n^1.2850972089...
G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^6^j * (1-x^6^j)*((1+x^6^j)^4 + 4(1+2x^6^j) * x^(3*6^j))/(1-x^6^(j+1)).
Also: g(x) = (x/(1-x))*(h_(6,1)(x) + 3*h_(6,2)(x) + 2*h_(6,3)(x) + 2*h_(6,4)(x) + h_(6,5)(x) - 9*h_(6,6)(x)), where h_(6,k)(x) = sum_{j>=0} 10^j*x^(k*6^j)/(1-x^6^(j+1)). (End)
Sum_{n>=2} 1/a(n) = 3.614028405471074989720026361356036456697082276983705341077940360653303099111... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024
Extensions
0 added by N. J. A. Sloane, Feb 02 2009
100 added by Arkadiusz Wesolowski, Mar 10 2011
Examples for n>=10^3 added by Hieronymus Fischer, May 30 2012
Comments