A046034 Numbers whose digits are primes.

2, 3, 5, 7, 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77, 222, 223, 225, 227, 232, 233, 235, 237, 252, 253, 255, 257, 272, 273, 275, 277, 322, 323, 325, 327, 332, 333, 335, 337, 352, 353, 355, 357, 372, 373, 375, 377, 522, 523, 525, 527, 532

Offset

1,1

Comments

If n is represented as a zerofree base-4 number (see A084544) according to n=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)=2,3,5,7 for k=1..4. - Hieronymus Fischer, May 30 2012

According to A153025, it seems that 5, 235 and 72335 are the only terms whose square is also a term, i.e., which are also in the sequence A275971 of square roots of the terms which are squares, listed in A191486. - M. F. Hasler, Sep 16 2016

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.

Eric Weisstein's World of Mathematics, Smarandache Sequences.

Index entries for 10-automatic sequences.

Formula

A055642(a(n)) = A193238(a(n)). - Reinhard Zumkeller, Jul 19 2011

From Hieronymus Fischer, Apr 20, May 30 and Jun 25 2012: (Start)

a(n) = Sum_{j=0..m-1} ((2*b(j)+1) mod 8 + floor(b(j)/4) - floor((b(j)-1)/4))*10^j, where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).

a(n) = Sum_{j=0..m-1} (A010877(A005408(b(j)) + A002265(b(j)) - A002265(b(j)-1))*10^j.

Special values:

a(1*(4^n-1)/3) = 2*(10^n-1)/9.

a(2*(4^n-1)/3) = 1*(10^n-1)/3.

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

a(4*(4^n-1)/3) = 7*(10^n-1)/9.

Inequalities:

a(n) <= 2*(10^log_4(3*n+1)-1)/9, equality holds for n = (4^k-1)/3, k>0.

a(n) <= 2*A084544(n), equality holds iff all digits of A084544(n) are 1.

a(n) > A084544(n).

Lower and upper limits:

lim inf a(n)/10^log_4(n) = (7/90)*10^log_4(3) = 0.48232167706987..., for n -> oo.

lim sup a(n)/10^log_4(n) = (2/9)*10^log_4(3) = 1.378061934485343..., for n -> oo.

where 10^log_4(n) = n^1.66096404744...

G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(2 + z(j) + 2*z(j)^2 + 2*z(j)^3 - 7*z(j)^4)/(1-z(j)^4), where z(j) = x^4^j.

Also g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1-z(j))*(2 + 3*z(j) + 5*z(j)^2 + 7*z(j)^3)/(1-z(j)^4), where z(j)=x^4^j.

Also: g(x) = (1/(1-x))*(2*h_(4,0)(x) + h_(4,1)(x) + 2*h_(4,2)(x) + 2*h_(4,3)(x) - 7*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3)*x^(k*4^j)/(1-x^4^(j+1)). (End)

Sum_{n>=1} 1/a(n) = 1.857333779940977502574887651449435985318556794733869779170825138954093657197... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Example

a(100)   = 2277,
a(10^3)  = 55327,
a(9881)  = 3233232,
a(10^4)  = 3235757,
a(10922) = 3333333,
a(10^5)  = 227233257.

Mathematica

Table[FromDigits /@ Tuples[{2, 3, 5, 7}, n], {n, 3}] // Flatten (* Michael De Vlieger, Sep 19 2016 *)

Prog

(PARI) is_A046034(n)=Set(isprime(digits(n)))==[1] \\ M. F. Hasler, Oct 12 2013
(Haskell)
a046034 n = a046034_list !! (n-1)
a046034_list = filter (all (`elem` "2357") . show ) [0..]
-- Reinhard Zumkeller, Jul 19 2011
(Magma) [n: n in [2..532] | Set(Intseq(n)) subset [2, 3, 5, 7]];  // Bruno Berselli, Jul 19 2011
(Python)
def A046034(n):
    m = (3*n+1).bit_length()-1>>1
    return int(''.join(('2357'[(3*n+1-(1<<(m<<1)))//(3<<((m-1-j)<<1))&3] for j in range(m)))) # Chai Wah Wu, Feb 08 2023

Crossrefs

Cf. A046035, A118950, A019546 (primes), A203263, A035232, A039996, A085823, A052382, A084544, A084984, A017042, A001743, A001744, A014261, A014263, A153025, A191486, A193238, A202267, A202268, A211681, A365471 (complement).

Sequence in context: A340112 A085557 A125665 * A062087 A162457 A084983

Adjacent sequences: A046031 A046032 A046033 * A046035 A046036 A046037

Keyword

nonn,base,easy

Author

Eric W. Weisstein

Extensions

More terms from Cino Hilliard, Aug 06 2006

Typo in second formula corrected by Hieronymus Fischer, May 12 2012

Two typos in example section corrected by Hieronymus Fischer, May 30 2012

Status

approved