A202268 Numbers in which all digits are nonprimes (1, 4, 6, 8, 9).

1, 4, 6, 8, 9, 11, 14, 16, 18, 19, 41, 44, 46, 48, 49, 61, 64, 66, 68, 69, 81, 84, 86, 88, 89, 91, 94, 96, 98, 99, 111, 114, 116, 118, 119, 141, 144, 146, 148, 149, 161, 164, 166, 168, 169, 181, 184, 186, 188, 189, 191, 194, 196, 198, 199, 411, 414, 416, 418, 419

Offset: 1

Author

Jaroslav Krizek, Dec 25 2011

Keywords

• nonn
• base
• easy

Comments

Supersequence of A029581.

Subsequence of A084984.

If n-1 is represented as a zerofree base-5 number (see A084545) according to n-1=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=1,4,6,8,9 for k=1..5. - Hieronymus Fischer, May 30 2012

Examples

			From _Hieronymus Fischer_, May 30 2012: (Start)
a(1000) = 14889.
a(10^4) = 498889
a(10^5) = 11188889.
a(10^6) = 446888889. (End)
		

Links

• Hieronymus Fischer, 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

Cf. A046034 (numbers in which all digits are primes), A001742 (numbers in which all digits are noncomposites excluding 0), A202267 (numbers in which all digits are noncomposites), A084984 (numbers in which all digits are nonprimes), A029581 (numbers in which all digits are composites).

Cf. A084545, A001743, A001744, A014261, A014263.

Programs

Magma

[n: n in [1..500] | Set(Intseq(n)) subset [1, 4, 6, 8, 9]]; // Vincenzo Librandi, Dec 17 2018

Mathematica

Table[FromDigits/@Tuples[{1, 4, 6, 8, 9}, n], {n, 3}] // Flatten (* Vincenzo Librandi, Dec 17 2018 *)

Formula

From Hieronymus Fischer, May 30 2012: (Start)

a(n) = Sum_{j=0..m-1} ((2*b_j(n)+1) mod 10 + floor((b_j(n)+4)/5) - floor((b_j(n)+1)/5))*10^j, where b_j(n))=floor((4*n+1-5^m)/(4*5^j)), m=floor(log_5(4*n+1)).

a(1*(5^n-1)/4) = 1*(10^n-1)/9.

a(2*(5^n-1)/4) = 4*(10^n-1)/9.

a(3*(5^n-1)/4) = 6*(10^n-1)/9.

a(4*(5^n-1)/4) = 8*(10^n-1)/9.

a(5*(5^n-1)/4) = 10^n-1.

a(n) = (10^log_5(4*n+1)-1)/9 for n=(5^k-1)/4, k>0.

a(n) <= 36/(9*2^log_5(9)-1)*(10^log_5(4*n+1)-1)/9 for n>0, equality holds for n=2.

a(n) > 0.776*10^log_5(4*n+1)-1)/9 for n>0.

a(n) >= A001742(n), equality holds for n=(5^k-1)/4, k>0.

a(n) = A084545(n) iff all digits of A084545(n) are 1, a(n)>A084545(n), else.

G.f.: g(x) = (x^(1/4)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(5/4)*(1-z(j))*(1 + 4z(j) + 6*z(j)^2 + 8*z(j)^3 + 9*z(j)^4)/(1-z(j)^5), where z(j)=x^5^j.

Also: g(x) = (1/(1-x))*(h_(5,0)(x) + 3h_(5,1)(x) + 2h_(5,2)(x) + 2h_(5,3)(x) + h_(5,4)(x) - 9*h_(5,5)(x)), where h_(5,k)(x) = Sum_{j>=0} 10^j*x^((5^(j+1)-1)/4)*(x^5^j)^k/(1-(x^5^j)^5). (End)

Sum_{n>=1} 1/a(n) = 2.897648425695540438556738520657902585305276107220152307051361916356295164643... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024