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Numbers in which all digits are nonprimes (1, 4, 6, 8, 9).
18

%I #33 Feb 15 2024 01:57:38

%S 1,4,6,8,9,11,14,16,18,19,41,44,46,48,49,61,64,66,68,69,81,84,86,88,

%T 89,91,94,96,98,99,111,114,116,118,119,141,144,146,148,149,161,164,

%U 166,168,169,181,184,186,188,189,191,194,196,198,199,411,414,416,418,419

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

%C Supersequence of A029581.

%C Subsequence of A084984.

%C 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

%H Hieronymus Fischer, <a href="/A202268/b202268.txt">Table of n, a(n) for n = 1..10000</a>

%H Robert Baillie and Thomas Schmelzer, <a href="https://library.wolfram.com/infocenter/MathSource/7166/">Summing Kempner's Curious (Slowly-Convergent) Series</a>, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.

%H <a href="/index/Ar#10-automatic">Index entries for 10-automatic sequences</a>.

%F From _Hieronymus Fischer_, May 30 2012: (Start)

%F 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)).

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

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

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

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

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

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

%F 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.

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

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

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

%F 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.

%F 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)

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

%e From _Hieronymus Fischer_, May 30 2012: (Start)

%e a(1000) = 14889.

%e a(10^4) = 498889

%e a(10^5) = 11188889.

%e a(10^6) = 446888889. (End)

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

%o (Magma) [n: n in [1..500] | Set(Intseq(n)) subset [1, 4, 6, 8, 9]]; // _Vincenzo Librandi_, Dec 17 2018

%Y 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).

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

%K nonn,base,easy

%O 1,2

%A _Jaroslav Krizek_, Dec 25 2011