A084544 -id:A084544 - cp's OEIS frontend

A052382 Numbers without 0 in the decimal expansion, colloquial 'zeroless numbers'.

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 111, 112, 113

Offset: 1

Henry Bottomley, Mar 13 2000

The entries 1 to 79 match the corresponding subsequence of A043095, but then 81, 91-98, 100, 102, etc. are only in one of the two sequences. - R. J. Mathar, Oct 13 2008
Complement of A011540; A168046(a(n)) = 1; A054054(a(n)) > 0; A007602, A038186, A038618, A052041, A052043, and A052045 are subsequences. - Reinhard Zumkeller, Apr 25 2012, Apr 07 2011, Dec 01 2009
a(n) = n written in base 9 where zeros are not allowed but nines are. The nine distinct digits used are 1, 2, 3, ..., 9 instead of 0, 1, 2, ..., 8. To obtain this sequence from the "canonical" base 9 sequence with zeros allowed, just replace any 0 with a 9 and then subtract one from the group of digits situated on the left. For example, 9^3 = 729 (10) (in base 10) = 1000 (9) (in base 9) = 889 (9-{0}) (in base 9 without zeros) because 100 (9) = [9-1]9 = 89 (9-{0}) and thus 1000 (9) = [89-1]9 = 889 (9-{0}). - Robin Garcia, Jan 15 2014
From Hieronymus Fischer, May 28 2014: (Start)
Inversion: Given a term m, the index n such that a(n) = m can be calculated by A052382_inverse(m) = m - sum_{1<=j<=k} floor(m/10^j)*9^(j-1), where k := floor(log_10(m)) [see Prog section for an implementation in Smalltalk].
Example 1: A052382_inverse(137) = 137 - (floor(137/10) + floor(137/100)*9) = 137 - (13*1 + 1*9) = 137 - 22 = 115.
Example 2: A052382_inverse(4321) = 4321 - (floor(4321/10) + floor(4321/100)*9 + floor(4321/1000)*81) = 4321 - (432*1 + 43*9 + 4*81) = 4321 - (432 + 387 + 324) = 3178. (End)
The sum of the reciprocals of these numbers from a(1)=1 to infinity, called the Kempner series, is convergent towards a limit: 23.103447... whose decimal expansion is in A082839. - Bernard Schott, Feb 23 2019
Integer n > 0 is encoded using bijective base-9 numeration, see Wikipedia link below. - Alois P. Heinz, Feb 16 2020

			For k >= 0, a(10^k) = (1, 11, 121, 1331, 14641, 162151, 1783661, 19731371, ...) = A325203(k). - _Hieronymus Fischer_, May 30 2012 and Jun 06 2012; edited by _M. F. Hasler_, Jan 13 2020

Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, p. 258.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
K. Mahler, On the generating function of the integers with a missing digit, J. Indian Math. Soc. 15A (1951), 34-40.
Eric Weisstein's World of Mathematics, Kempner Series.
Eric Weisstein's World of Mathematics, Zerofree
Wikipedia, Bijective numeration
Index entries for 10-automatic sequences.

Cf. A004719, A052040, different from A067251.
Cf. A055640, A046034, A007931, A007932, A084544, A084545.
Column k=9 of A214676.
Cf. A011540 (complement), A043489, A054054, A168046.
Cf. A052383 (without 1), A052404 (without 2), A052405 (without 3), A052406 (without 4), A052413 (without 5), A052414 (without 6), A052419 (without 7), A052421 (without 8), A007095 (without 9).
Zeroless numbers in some other bases <= 10: A000042 (base 2), A032924 (base 3), A023705 (base 4), A248910 (base 6), A255805 (base 8), A255808 (base 9).
Cf. A082839 (sum of reciprocals).
Cf. A038618 (subset of primes)
Subsequences: A007602, A038186, A052041, A052043, A052045, A059405, A066484, A099542.

a052382 n = a052382_list !! (n-1)
a052382_list = iterate f 1 where
f x = 1 + if r < 9 then x else 10 * f x' where (x', r) = divMod x 10
-- Reinhard Zumkeller, Mar 08 2015, Apr 07 2011

[ n: n in [1..114] | not 0 in Intseq(n) ]; // Bruno Berselli, May 28 2011

a:= proc(n) local d, l, m; m:= n; l:= NULL;
      while m>0 do d:= irem(m, 9, 'm');
        if d=0 then d:=9; m:= m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 11 2015
is_zeroless := n -> not is(0 in convert(n, base, 10)):
select(is_zeroless, [seq(1..113)]);  # Peter Luschny, Jun 20 2025

A052382 = Select[Range[100], DigitCount[#, 10, 0] == 0 &] (* Alonso del Arte, Mar 10 2011 *)

select( {is_A052382(n)=n&&vecmin(digits(n))}, [0..111]) \\ actually: is_A052382 = (bool) A054054. - M. F. Hasler, Jan 23 2013, edited Jan 13 2020

a(n) = for (w=0, oo, if (n >= 9^w, n -= 9^w, return ((10^w-1)/9 + fromdigits(digits(n, 9))))) \\ Rémy Sigrist, Jul 26 2017

apply( {A052382(n,L=logint(n,9))=fromdigits(digits(n-9^L>>3,9))+10^L\9}, [1..100])
next_A052382(n, d=digits(n+=1))={for(i=1, #d, d[i]|| return(n-n%(d=10^(#d-i+1))+d\9)); n} \\ least a(k) > n. Used in A038618.
( {A052382_vec(n,M=1)=M--;vector(n, i, M=next_A052382(M))} )(99) \\ n terms >= M
\\ See OEIS Wiki page (cf. LINKS) for more programs. - M. F. Hasler, Jan 11 2020

A052382 = [n for n in range(1,10**5) if not str(n).count('0')]
# Chai Wah Wu, Aug 26 2014

from sympy import integer_log
def A052382(n):
    m = integer_log(k:=(n<<3)+1,9)[0]
    return sum((1+(k-9**m)//(9**j<<3)%9)*10**j for j in range(m)) # Chai Wah Wu, Jun 27 2025

A052382
"Answers the n-th term of A052382, where n is the receiver."
^self zerofree: 10
A052382_inverse
"Answers that index n which satisfy A052382(n) = m, where m is the receiver.”
^self zerofree_inverse: 10
zerofree: base
"Answers the n-th zerofree number in base base, where n is the receiver. Valid for base > 2.
Usage: n zerofree: b [b = 10 for this sequence]
Answer: a(n)"
| n m s c bi ci d |
n := self.
c := base - 1.
m := (base - 2) * n + 1 integerFloorLog: c.
d := n - (((c raisedToInteger: m) - 1)//(base - 2)).
bi := 1.
ci := 1.
s := 0.
1 to: m
do:
[:i |
s := (d // ci \\ c + 1) * bi + s.
bi := base * bi.
ci := c * ci].
^s
zerofree_inverse: base
"Answers the index n such that the n-th zerofree number in base base is = m, where m is the receiver. Valid for base > 2.
Usage: m zerofree_inverse: b [b = 10 for this sequence]
Answer: n"
| m p q s |
m := self.
s := 0.
p := base.
q := 1.
[p < m] whileTrue:
[s := m // p * q + s.
p := base * p.
q := (base - 1) * q].
^m - s
"by Hieronymus Fischer, May 28 2014"

seq 0 1000 | grep -v 0; # Joerg Arndt, May 29 2011

a(n+1) = f(a(n)) with f(x) = 1 + if x mod 10 < 9 then x else 10*f([x/10]). - Reinhard Zumkeller, Nov 15 2009
From Hieronymus Fischer, Apr 30, May 30, Jun 08 2012, Feb 17 2019: (Start)
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 9)*10^j, where m = floor(log_9(8*n + 1)), b(j) = floor((8*n + 1 - 9^m)/(8*9^j)).
Also: a(n) = Sum_{j=0..m-1} (1 + A010878(b(j)))*10^j.
a(9*n + k) = 10*a(n) + k, k=1..9.
Special values:
a(k*(9^n - 1)/8) = k*(10^n - 1)/9, k=1..9.
a((17*9^n - 9)/8) = 2*10^n - 1.
a((9^n - 1)/8 - 1) = 10^(n-1) - 1, n > 1.
Inequalities:
a(n) <= (1/9)*((8*n+1)^(1/log_10(9)) - 1), equality holds for n=(9^k-1)/8, k>0.
a(n) > (1/10)*((8*n+1)^(1/log_10(9)) - 1), n > 0.
Lower and upper limits:
lim inf a(n)/10^log_9(8*n) = 1/10, for n -> infinity.
lim inf a(n)/n^(1/log_10(9)) = 8^(1/log_10(9))/10, for n -> infinity.
lim sup a(n)/10^log_9(8*n) = 1/9, for n -> infinity.
lim sup a(n)/n^(1/log_10(9)) = 8^(1/log_10(9))/9, for n -> infinity.
G.f.: g(x) = (x^(1/8)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(9/8)*(1 - 10z(j)^9 + 9z(j)^10)/((1-z(j))(1-z(j)^9)), where z(j) = x^9^j.
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1 - 10(x^9^j)^9 + 9(x^9^j)^10)*x^9^j*f_j(x)/(1-x^9^j), where f_j(x) = 10^j*x^((9^j-1)/8)/(1-(x^9^j)^9). Here, the f_j obey the recurrence f_0(x) = 1/(1-x^9), f_(j+1)(x) = 10x*f_j(x^9).
Also: g(x) = (1/(1-x))*((Sum{k=0..8} h_(9,k)(x)) - 9*h_(9,9)(x)), where h_(9,k)(x) = Sum_{j>=0} 10^j*x^((9^(j+1)-1)/8)*x^(k*9^j)/(1-x^9^(j+1)).
Generic formulas for analogous sequences with numbers expressed in base p and only using the digits 1, 2, 3, ... d, where 1 < d < p:
a(n) = Sum_{j=0..m-1} (1 + b(j) mod d)*p^j, where m = floor(log_d((d-1)*n+1)), b(j) = floor(((d-1)*n+1-d^m)/((d-1)*d^j)).
Special values:
a(k*(d^n-1)/(d-1)) = k*(10^n-1)/9, k=1..d.
a(d*((2d-1)*d^(n-1)-1)/(d-1)) = ((d+9)*10^n-d)/9 = 10^n + d*(10^n-1)/9.
a((d^n-1)/(d-1)-1) = d*(10^(n-1)-1)/9, n > 1.
Inequalities:
a(n) <= (10^log_d((d-1)*n+1)-1)/9, equality holds for n = (d^k-1)/(d-1), k > 0.
a(n) > (d/10)*(10^log_d((d-1)*n+1)-1)/9, n > 0.
Lower and upper limits:
lim inf a(n)/10^log_d((d-1)*n) = d/90, for n -> infinity.
lim sup a(n)/10^log_d((d-1)*n) = 1/9, for n -> infinity.
G.f.: g(x) = (1/(1-x)) Sum_{j>=0} (1 - (d+1)(x^d^j)^d + d(x^d^j)^(d+1))*x^d^j*f_j(x)/(1-x^d^j), where f_j(x) = p^j*x^((d^j-1)/(d-1))/(1-(x^d^j)^d). Here, the f_j obey the recursion f_0(x) = 1/(1-x^d), f_(j+1)(x) = px*f_j(x^d).
(End)
A052382 = { n | A054054(n) > 0 }. - M. F. Hasler, Jan 23 2013
From Hieronymus Fischer, Feb 20 2019: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 0.696899720...
Sum_{n>=1} 1/a(n)^2 = 1.6269683705819...
Sum_{n>=1} 1/a(n) = 23.1034479... = A082839. This so-called Kempner series converges very slowly. For the calculation of the sum, it is helpful to use the following fraction of partial sums, which converges rapidly:
lim_{n->infinity} (Sum_{k=p(n)..p(n+1)-1} 1/a(k)) / (Sum_{k=p(n-1)..p(n)-1} 1/a(k)) = 9/10, where p(n) = (9^n-1)/8, n > 1.
(End)

Typos in formula section corrected by Hieronymus Fischer, May 30 2012

Name clarified by Peter Luschny, Jun 20 2025

A046034 Numbers whose digits are primes.

Original entry on oeis.org

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

Eric W. Weisstein

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

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

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.

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).

a046034 n = a046034_list !! (n-1)
a046034_list = filter (all (`elem` "2357") . show ) [0..]
-- Reinhard Zumkeller, Jul 19 2011

[n: n in [2..532] | Set(Intseq(n)) subset [2, 3, 5, 7]];  // Bruno Berselli, Jul 19 2011

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

is_A046034(n)=Set(isprime(digits(n)))==[1] \\ M. F. Hasler, Oct 12 2013

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

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

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

A007931 Numbers that contain only 1's and 2's. Nonempty binary strings of length n in lexicographic order.

Original entry on oeis.org

1, 2, 11, 12, 21, 22, 111, 112, 121, 122, 211, 212, 221, 222, 1111, 1112, 1121, 1122, 1211, 1212, 1221, 1222, 2111, 2112, 2121, 2122, 2211, 2212, 2221, 2222, 11111, 11112, 11121, 11122, 11211, 11212, 11221, 11222, 12111, 12112, 12121, 12122

Offset: 1

R. Muller

Numbers written in the dyadic system [Smullyan, Stillwell]. - N. J. A. Sloane, Feb 13 2019
Logic-binary sequence: prefix it by the empty word to have all binary words on the alphabet {1,2}.
The least binary word of length k is a(2^k - 1).
See Mathematica program for logic-binary sequence using (0,1) in place of (1,2); the sequence starts with 0,1,00,01,10. - Clark Kimberling, Feb 09 2012
A007953(a(n)) = A014701(n+1); A007954(a(n)) = A048896(n). - Reinhard Zumkeller, Oct 26 2012
a(n) is n written in base 2 where zeros are not allowed but twos are. The two distinct digits used are 1, 2 instead of 0, 1. To obtain this sequence from the "canonical" base 2 sequence with zeros allowed, just replace any 0 with a 2 and then subtract one from the group of digits situated on the left: (10-->2; 100-->12; 110-->22; 1000-->112; 1010-->122). - Robin Garcia, Jan 31 2014
For numbers made of only two different digits, see also A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340(digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases). Numbers with exactly two distinct (but unspecified) digits in base 10 are listed in A031955, for other bases in A031948-A031954. - M. F. Hasler, Apr 04 2015
The variant with digits {0, 1} instead of {1, 2} is obtained by deleting all initial digits in sequence A007088 (numbers written in base 2). - M. F. Hasler, Nov 03 2020

			Positive numbers may not start with 0 in the OEIS, otherwise this sequence would have been written as: 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 00000, 00001, 00010, 00011, 00100, 00101, 00110, 00111, 01000, 01001, 01010, 01011, ...
From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(10)   = 122.
a(100)  = 211212.
a(10^3) = 222212112.
a(10^4) = 1122211121112.
a(10^5) = 2111122121211112.
a(10^6) = 2221211112112111112.
a(10^7) = 11221112112122121111112.
a(10^8) = 12222212122221111211111112.
a(10^9) = 22122211221212211212111111112. (End)

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 2. - From N. J. A. Sloane, Jul 26 2012
K. Atanassov, On the 97th, 98th and the 99th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 89-93.
R. M. Smullyan, Theory of Formal Systems, Princeton, 1961.
John Stillwell, Reverse Mathematics, Princeton, 2018. See p. 90.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000 (terms up to 2^10-2 from T. D. Noe, corrected by Sean A. Irvine, April 18 2019)
K. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995.
R. R. Forslund, Positive Integer Pages
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
F. Smarandache, Only Problems, Not Solutions!.
Index entries for 10-automatic sequences.

Cf. A007932 (digits 1-3), A059893, A045670, A052382 (digits 1-9), A059939, A059941, A059943, A032924, A084544, A084545, A046034 (prime digits 2,3,5,7), A089581, A084984 (no prime digits); A001742, A001743, A001744: loops; A202267 (digits 0, 1 and primes), A202268 (digits 1,4,6,8,9), A014261 (odd digits), A014263 (even digits).
Cf. A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340 (digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases).
Cf. A020450 (primes).

a007931 n = f (n + 1) where
   f x = if x < 2 then 0 else (10 * f x') + m + 1
     where (x', m) = divMod x 2
-- Reinhard Zumkeller, Oct 26 2012

[n: n in [1..100000] | Set(Intseq(n)) subset {1,2}]; // Vincenzo Librandi, Aug 19 2016

# Maple program to produce the sequence:
a:= proc(n) local m, r, d; m, r:= n, 0;
      while m>0 do d:= irem(m, 2, 'm');
        if d=0 then d:=2; m:= m-1 fi;
        r:= d, r
      od; parse(cat(r))/10
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 26 2016
# Maple program to invert this sequence: given a(n), it returns n. - N. J. A. Sloane, Jul 09 2012
invert7931:=proc(u)
local t1,t2,i;
t1:=convert(u,base,10);
[seq(t1[i]-1,i=1..nops(t1))];
[op(%),1];
t2:=convert(%,base,2,10);
add(t2[i]*10^(i-1),i=1..nops(t2))-1;
end;

f[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[f, 42] (* Robert G. Wilson v Sep 14 2006 *)
(* Next, A007931 using (0,1) instead of (1,2) *)
d[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[FromCharacterCode[ToCharacterCode[ToString[d[#]]] - 1] &, 100] (* Peter J. C. Moses, at request of Clark Kimberling, Feb 09 2012 *)
Flatten[Table[FromDigits/@Tuples[{1,2},n],{n,5}]] (* Harvey P. Dale, Sep 13 2014 *)

apply( {A007931(n)=fromdigits([d+1|d<-binary(n+1)[^1]])}, [1..44]) \\ M. F. Hasler, Nov 03 2020, replacing older code from Mar 26 2015

/* inverse function */ apply( {A007931_inv(N)=fromdigits([d-1|d<-digits(N)],2)+2<M. F. Hasler, Nov 09 2020

def a(n): return int(bin(n+1)[3:].replace('1', '2').replace('0', '1'))
print([a(n) for n in range(1, 45)]) # Michael S. Branicky, May 13 2021

def A007931(n): return int(s:=bin(n+1)[3:])+(10**(len(s))-1)//9 # Chai Wah Wu, Jun 13 2025

To get a(n), write n+1 in base 2, remove initial 1, add 1 to all remaining digits: e.g., eleven (11) in base 2 is 1011; remove initial 1 and add 1 to remaining digits: a(10)=122. - Clark Kimberling, Mar 11 2003
Conversely, given a(n), to get n: subtract 1 from all digits, prefix with an initial 1, convert this binary number to base 10, subtract 1. E.g., a(6)=22 -> 11 -> 111 -> 7 -> 6. - N. J. A. Sloane, Jul 09 2012
a(n) = A053645(n+1)+A002275(A000523(n)) = a(n-2^b(n))+10^b(n) where b(n) = A059939(n) = floor(log_2(n+1)-1). - Henry Bottomley, Feb 14 2001
From Hieronymus Fischer, Jun 06 2012 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1 and 2.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 2)*10^j, where m = floor(log_2(n+1)), b(j) = floor((n+1-2^m)/(2^j)).
Special values:
a(k*(2^n-1)) = k*(10^n-1)/9, k= 1,2.
a(3*2^n-2) = (11*10^n-2)/9 = 10^n+2*(10^n-1)/9.
a(2^n-2) = 2*(10^(n-1)-1)/9, n>1.
Inequalities:
a(n) <= (10^log_2(n+1)-1)/9, equality holds for n=2^k-1, k>0.
a(n) > (2/10)*(10^log_2(n+1)-1)/9.
Lower and upper limits:
lim inf a(n)/10^log_2(n) = 1/45, for n --> infinity.
lim sup a(n)/10^log_2(n) = 1/9, for n --> infinity.
G.f.: g(x) = (1/(x(1-x)))*sum_{j=0..infinity} 10^j* x^(2*2^j)*(1 + 2 x^2^j)/(1 + x^2^j).
Also: g(x) = (1/(1-x))*(h_(2,0)(x) + h_(2,1)(x) - 2*h_(2,2)(x)), where h_(2,k)(x) = sum_{j>=0} 10^j*x^(2^(j+1)-1)*x^(k*2^j)/(1-x^2^(j+1)).
Also: g(x) = (1/(1-x)) sum_{j>=0} (1 - 3(x^2^j)^2 + 2(x^2^j)^3)*x^2^j*f_j(x)/(1-x^2^j), where f_j(x) = 10^j*x^(2^j-1)/(1-(x^2^j)^2). The f_j obey the recurrence f_0(x) = 1/(1-x^2), f_(j+1)(x) = 10x*f_j(x^2). (End)

Some crossrefs added by Hieronymus Fischer, Jun 06 2012

Edited by M. F. Hasler, Mar 26 2015

A014263 Numbers that contain even digits only.

Original entry on oeis.org

0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88, 200, 202, 204, 206, 208, 220, 222, 224, 226, 228, 240, 242, 244, 246, 248, 260, 262, 264, 266, 268, 280, 282, 284, 286, 288, 400, 402, 404, 406, 408, 420, 422, 424

Offset: 1

N. J. A. Sloane

The set of real numbers between 0 and 1 that contain no odd digits in their decimal expansion has Hausdorff dimension log 5 / log 10.
Integers written in base 5 and then doubled (in base 10). - Franklin T. Adams-Watters, Mar 15 2006
The carryless mod 10 "even" numbers (cf. A004529) sorted and duplicates removed. - N. J. A. Sloane, Aug 03 2010.
Complement of A007957; A196564(a(n)) = 0; A103181(a(n)) = 0. - Reinhard Zumkeller, Oct 04 2011
If n-1 is represented as a base-5 number (see A007091) 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)=0,2,4,6,8 for k=0..4. - Hieronymus Fischer, Jun 03 2012

			a(1000) = 24888.
a(10^4) = 60888.
a(10^5) = 22288888.
a(10^6) = 446888888.

K. J. Falconer, The Geometry of Fractal Sets, Cambridge, 1985; p. 19.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for 10-automatic sequences.
Index entries for sequences related to carryless arithmetic

Subsequence of A059708.

Cf. A061810, A061811, A007091, A014261, A046034, A052382, A084544, A089581, A084984, A017042, A001743, A202267, A202268, A194182, A196563.

a014263 n = a014263_list !! (n-1)
a014263_list = filter (all (`elem` "02468") . show) [0,2..]
-- Reinhard Zumkeller, Jul 05 2011

[n: n in [0..424] | Set(Intseq(n)) subset [0..8 by 2]];  // Bruno Berselli, Jul 19 2011

a:= proc(m) local L,i;
  L:= convert(m-1,base,5);
  2*add(L[i]*10^(i-1),i=1..nops(L))
end proc:
seq(a(i),i=1..100); # Robert Israel, Apr 07 2016

Select[Range[450], And@@EvenQ[IntegerDigits[#]]&] (* Harvey P. Dale, Jan 30 2011 *)
FromDigits/@Tuples[{0,2,4,6,8},3] (* Harvey P. Dale, Jul 07 2025 *)

a(n) = 2*fromdigits(digits(n-1, 5), 10); \\ Michel Marcus, Nov 04 2022

is(n)=#setminus(Set(digits(n)), [0,2,4,6,8])==0 \\ Charles R Greathouse IV, Mar 03 2025

from sympy.ntheory.digits import digits
def a(n): return int(''.join(str(2*d) for d in digits(n, 5)[1:]))
print([a(n) for n in range(58)]) # Michael S. Branicky, Jan 13 2022

from itertools import count, islice, product
def agen(): # generator of terms
    yield 0
    for d in count(1):
        for first in "2468":
            for rest in product("02468", repeat=d-1):
                yield int(first + "".join(rest))
print(list(islice(agen(), 58))) # Michael S. Branicky, Jan 13 2022

A045888(a(n)) = 0. - Reinhard Zumkeller, Aug 25 2009
a(n) = A179082(n) for n <= 25. - Reinhard Zumkeller, Jun 28 2010
From Hieronymus Fischer, Jun 06 2012: (Start)
a(n) = ((2*b_m(n)) mod 8 + 2)*10^m + Sum_{j=0..m-1} ((2*b_j(n)) mod 10)*10^j, where n>1, b_j(n) = floor((n-1-5^m)/5^j), m = floor(log_5(n-1)).
a(1*5^n+1) = 2*10^n.
a(2*5^n+1) = 4*10^n.
a(3*5^n+1) = 6*10^n.
a(4*5^n+1) = 8*10^n.
a(n) = 2*10^log_5(n-1) for n=5^k+1,
a(n) < 2*10^log_5(n-1), else.
a(n) > (8/9)*10^log_5(n-1) n>1.
a(n) = 2*A007091(n-1), iff the digits of A007091(n-1) are 0 or 1.
G.f.: g(x) = (x/(1-x))*Sum_{j>=0} 10^j*x^5^j *(1-x^5^j)* (2+4x^5^j+ 6(x^2)^5^j+ 8(x^3)^5^j)/(1-x^5^(j+1)).
Also: g(x) = 2*(x/(1-x))*Sum_{j>=0} 10^j*x^5^j * (1-4x^(3*5^j)+3x^(4*5^j))/((1-x^5^j)(1-x^5^(j+1))).
Also: g(x) = 2*(x/(1-x))*(h_(5,1)(x) + h_(5,2)(x) + h_(5,3)(x) + h_(5,4)(x) - 4*h_(5,5)(x)), where h_(5,k)(x) = Sum_{j>=0} 10^j*(x^5^j)^k/(1-(x^5^j)^5). (End)
a(5*n+i-4) = 10*a(n) + 2*i for n >= 1, i=0..4. - Robert Israel, Apr 07 2016
Sum_{n>=2} 1/a(n) = A194182. - Bernard Schott, Jan 13 2022

Examples and crossrefs added by Hieronymus Fischer, Jun 06 2012

A001744 Numbers n such that every digit contains a loop (version 2).

Original entry on oeis.org

0, 4, 6, 8, 9, 40, 44, 46, 48, 49, 60, 64, 66, 68, 69, 80, 84, 86, 88, 89, 90, 94, 96, 98, 99, 400, 404, 406, 408, 409, 440, 444, 446, 448, 449, 460, 464, 466, 468, 469, 480, 484, 486, 488, 489, 490, 494, 496, 498, 499, 600, 604, 606, 608, 609, 640, 644, 646

Offset: 1

N. J. A. Sloane

See A001743 for the other version.

If n-1 is represented as a base-5 number (see A007091) 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)=0,4,6,8,9 for k=0..4. - Hieronymus Fischer, May 30 2012

			a(1000) = 46999.
a(10^4) = 809999.
a(10^5) = 44499999.
a(10^6) = 668999999.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
Index entries for 10-automatic sequences.

Cf. A061371, A029581, A007091, A046034, A084544, A084984, A017042, A001743, A014261, A014263, A202267, A202268.

FromDigits/@Tuples[{0,4,6,8,9},3] (* Harvey P. Dale, Aug 16 2018 *)

is(n) = #setintersect(vecsort(digits(n), , 8), [1, 2, 3, 5, 7])==0 \\ Felix Fröhlich, Sep 09 2019

From Hieronymus Fischer, May 30 2012: (Start)
a(n) = ((2*b_m(n)) mod 8 + 4 + floor(b_m(n)/4) - floor((b_m(n)+1)/4))*10^m + sum_{j=0..m-1} ((2*b_j(n))) mod 10 + 2*floor((b_j(n)+4)/5) - floor((b_j(n)+1)/5) -floor(b_j(n)/5)))*10^j, where n>1, b_j(n)) = floor((n-1-5^m)/5^j), m = floor(log_5(n-1)).
a(1*5^n+1) = 4*10^n.
a(2*5^n+1) = 6*10^n.
a(3*5^n+1) = 8*10^n.
a(4*5^n+1) = 9*10^n.
a(n) = 4*10^log_5(n-1) for n=5^k+1,
a(n) < 4*10^log_5(n-1), otherwise.
a(n) > 10^log_5(n-1) n>1.
a(n) = 4*A007091(n-1), iff the digits of A007091(n-1) are 0 or 1.
G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^5^j*(1-x^5^j)*(4 + 6x^5^j + 8(x^2)^5^j + 9(x^3)^5^j)/(1-x^5^(j+1)).
Also: g(x) = (x/(1-x))*(4*h_(5,1)(x) + 2*h_(5,2)(x) + 2*h_(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)^k/(1-(x^5^j)^5). (End)

Ambiguous comment deleted by Zak Seidov, May 25 2010

Examples added by Hieronymus Fischer, May 30 2012

A007932 Numbers that contain only 1's, 2's and 3's.

Original entry on oeis.org

1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333, 1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221

Offset: 1

R. Muller

This sequence is the alternate number system in base 3. - Robert R. Forslund (forslund(AT)tbaytel.net), Jun 27 2003
a(n) is the "bijective base-k numeration" or "k-adic notation" for k=3. - Chris Gaconnet (gaconnet(AT)gmail.com), May 27 2009
a(n) = n written in base 3 where zeros are not allowed but threes are. The three distinct digits used are 1, 2 and 3 instead of 0, 1 and 2. To obtain this sequence from the "canonical" base 3 sequence with zeros allowed, just replace any 0 with a 3 and then subtract one from the group of digits situated on the left: (20-->13; 100-->23; 110-->33; 1000-->223; 1010-->233). This can be done in any integer positive base b, replacing zeros with positive b's and subtracting one from the group of digits situated on the left. And zero is the only digit that can be replaced, since there is always a more significant digit greater than 0, on the left, from which to subtract one. - Robin Garcia, Jan 07 2014

			a(100)  = 3131.
a(10^3) = 323231.
a(10^4) = 111123331.
a(10^5) = 11231311131.
a(10^6) = 1212133131231.
a(10^7) = 123133223331331.
a(10^8) = 13221311111312131.
a(10^9) = 2113123122313232231.
- _Hieronymus Fischer_, Jun 06 2012

K. Atanassov, On the 97th, 98th and the 99th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 89-93.
A. Salomaa, Formal Languages, Academic Press, 1973. pages 90-91. [From Chris Gaconnet (gaconnet(AT)gmail.com), May 27 2009]

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
K. Atanassov, On Some of Smarandache's Problems
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
Robert R. Forslund, A Logical Alternative to the Existing Positional Number System, Southwest Journal of Pure and Applied Mathematics. Vol. 1 1995 pp. 27-29.
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
F. Smarandache, Only Problems, Not Solutions!.
Wikipedia, Bijective numeration
Index entries for 10-automatic sequences.

Cf. A007931, A052382, A084544, A084545, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.

NextNbr[n_] := Block[{d = IntegerDigits[n + 1], l}, l = Length[d]; While[l != 1, If[ d[[l]] > 3, d[[l - 1]]++; d[[l]] = 1]; l-- ]; If[ d[[1]] > 3, d[[1]] = 11]; FromDigits[d]]; NestList[ NextNbr, 1, 51]
Table[FromDigits/@Tuples[{1,2,3},n],{n,4}]//Flatten (* Harvey P. Dale, Mar 29 2018 *)

a(n) = my (w=3); while (n>w, n -= w; w *= 3); my (d=digits(w+n-1, 3)); d[1] = 0; fromdigits(d) + (10^(#d-1)-1)/9 \\ Rémy Sigrist, Aug 28 2018

From Hieronymus Fischer, May 30 2012 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1, 2, 3.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 3)*10^j,
where m = floor(log_3(2*n+1)), b(j) = floor((2*n+1-3^m)/(2*3^j)).
Special values:
a(k*(3^n-1)/2) = k*(10^n-1)/9, k=1,2,3.
a((5*3^n-3)/2) = (4*10^n-1)/3 = 10^n + (10^n-1)/3.
a((3^n-1)/2 - 1) = (10^(n-1)-1)/3, n>1.
Inequalities:
a(n) <= (10^log_3(2*n+1)-1)/9, equality holds for n=(3^k-1)/2, k>0.
a(n) > (3/10)*(10^log_3(2*n+1)-1)/9, n>0.
Lower and upper limits:
lim inf a(n)/10^log_3(2*n) = 1/30, for n --> infinity.
lim sup a(n)/10^log_3(2*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/2)*(1-x))^(-1) Sum_{j=>0} 10^j*(x^3^j)^(3/2) * (1-x^3^j)*(1 + 2x^3^j + 3x^(2*3^j))/(1 - x^3^(j+1)).
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1 - 4(x^3^j)^3 + 3(x^3^j)^4)*x^3^j*f_j(x)/(1-x^3^j), where f_j(x) = 10^j*x^((3^j-1)/2)/(1-(x^3^j)^3). The f_j obey the recurrence f_0(x) = 1/(1-x^3), f_(j+1)(x) = 10x*f_j(x^3).
Also: g(x) = (1/(1-x))*(h_(3,0)(x) + h_(3,1)(x) + h_(3,2)(x) - 3*h_(3,3)(x)), where h_(3,k)(x) = Sum_{j>=0} 10^j*x^((3^(j+1)-1)/2) * (x^3^j)^k/(1-(x^3^j)^3).
(End)

Edited and extended by Robert G. Wilson v, Dec 14 2002

Crossrefs added by Hieronymus Fischer, Jun 06 2012

A214676 A(n,k) is n represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 1, 11, 1, 2, 111, 1, 2, 11, 1111, 1, 2, 3, 12, 11111, 1, 2, 3, 11, 21, 111111, 1, 2, 3, 4, 12, 22, 1111111, 1, 2, 3, 4, 11, 13, 111, 11111111, 1, 2, 3, 4, 5, 12, 21, 112, 111111111, 1, 2, 3, 4, 5, 11, 13, 22, 121, 1111111111

Offset: 1

Alois P. Heinz, Jul 25 2012

The digit set for bijective base-k numeration is {1, 2, ..., k}.

			Square array A(n,k) begins:
:         1,   1,  1,  1,  1,  1,  1,  1, ...
:        11,   2,  2,  2,  2,  2,  2,  2, ...
:       111,  11,  3,  3,  3,  3,  3,  3, ...
:      1111,  12, 11,  4,  4,  4,  4,  4, ...
:     11111,  21, 12, 11,  5,  5,  5,  5, ...
:    111111,  22, 13, 12, 11,  6,  6,  6, ...
:   1111111, 111, 21, 13, 12, 11,  7,  7, ...
:  11111111, 112, 22, 14, 13, 12, 11,  8, ...

Alois P. Heinz, Antidiagonals n = 1..18, flattened
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995, 27-29.
Eric Weisstein's World of Mathematics, Zerofree
Wikipedia, Bijective numeration

Columns k=1-9 give: A000042, A007931, A007932, A084544, A084545, A057436, A214677, A214678, A052382.

A(n+1,n) gives A010850.

A:= proc(n, b) local d, l, m; m:= n; l:= NULL;
      while m>0 do  d:= irem(m, b, 'm');
        if d=0 then d:=b; m:=m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);

A[n_, b_] := Module[{d, l, m}, m = n; l = Nothing; While[m > 0, {m, d} = QuotientRemainder[m, b]; If[d == 0, d = b; m--]; l = {d, l}]; FromDigits @ Flatten @ l];
Table[A[n, d-n+1], {d, 1, 12}, {n, 1, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A084545 Alternate number system in base 5.

Original entry on oeis.org

1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55, 111, 112, 113, 114, 115, 121, 122, 123, 124, 125, 131, 132, 133, 134, 135, 141, 142, 143, 144, 145, 151, 152, 153, 154, 155, 211, 212, 213, 214, 215, 221, 222

Offset: 1

Robert R. Forslund (forslund(AT)tbaytel.net), Jun 27 2003

			From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(100)  = 345.
a(10^3) = 12445.
a(10^4) = 254445.
a(10^5) = 11144445.
a(10^6) = 223444445.
a(10^7) = 4524444445.
a(10^8) = 145544444445.
a(10^9) = 3521444444445. (End)

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1 1995 pp. 27-29.
R. R. Forslund, Positive Integer Pages [Wayback Machine link]
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
Index entries for 10-automatic sequences.

Cf. A007931, A007932, A052382, A084544, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.

a(n) = my (w=5); while (n>w, n -= w; w *= 5); my (d=digits(w+n-1, 5)); d[1] = 0; fromdigits(d) + (10^(#d-1)-1)/9 \\ Rémy Sigrist, Dec 04 2019

From Hieronymus Fischer, Jun 06 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1..5.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 5)*10^j, where m = floor(log_5(4*n+1)), b(j) = floor((4*n+1-5^m)/(4*5^j)).
a(k*(5^n-1)/4) = k*(10^n-1)/9, for k = 1,2,3,4,5.
a((9*5^n-5)/4) = (14*10^n-5)/9 = 10^n + 5*(10^n-1)/9.
a((5^n-1)/4 - 1) = 5*(10^(n-1)-1)/9, n>1.
a(n) <= (10^log_5(4*n+1)-1)/9, equality holds for n=(5^k-1)/4, k>0.
a(n) > (5/10)*(10^log_5(4*n+1)-1)/9, n>0.
lim inf a(n)/10^log_5(4*n) = 1/18, for n --> infinity.
lim sup a(n)/10^log_5(4*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/4)*(1-x))^(-1) sum_{j>=0} 10^j*z(j)^(5/4)*(1 - 6z(j)^5 + 5z(j)^6)/((1-z(j))(1-z(j)^5)), where z(j) = x^5^j.
Also: g(x) = (1/(1-x)) sum_{j>=0} (1-6(x^5^j)^5+5(x^5^j)^6)*x^5^j*f_j(x)/(1-x^5^j), where f_j(x) = 10^j*x^((5^j-1)/4)/(1-(x^5^j)^5). The f_j obey the recurrence f_0(x) = 1/(1-x^5), f_(j+1)(x) = 10x*f_j(x^5).
Also: g(x) = 1/(1-x))*(h_(5,0)(x) + h_(5,1)(x) + h_(5,2)(x) + h_(4,1)(x) + h_(5,4)(x) - 5*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)

Offset set to 1 according to A007931, A007932 and more terms added by Hieronymus Fischer, Jun 06 2012

A029581 Numbers in which all digits are composite.

Original entry on oeis.org

4, 6, 8, 9, 44, 46, 48, 49, 64, 66, 68, 69, 84, 86, 88, 89, 94, 96, 98, 99, 444, 446, 448, 449, 464, 466, 468, 469, 484, 486, 488, 489, 494, 496, 498, 499, 644, 646, 648, 649, 664, 666, 668, 669, 684, 686, 688, 689, 694, 696, 698, 699, 844, 846, 848

Offset: 1

N. J. A. Sloane

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)=4,6,8,9 for k=1..4. - Hieronymus Fischer, May 30 2012

			From _Hieronymus Fischer_, May 30 2012: (Start)
a(1000) = 88649.
a(10^4) = 6468989
a(10^5) = 449466489. (End)

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.

Cf. A002808, A001744, A046034, A084544, A084984, A017042, A001743, A014261, A014263, A202267, A202268.

[n: n in [1..1000] | Set(Intseq(n)) subset [4, 6, 8, 9]]; // Vincenzo Librandi, Dec 17 2018

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

From Hieronymus Fischer, May 30 and Jun 25 2012: (Start)
a(n) = Sum_{j=0..m-1} (2*b(j) mod 8 + 4 + 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)).
Also: a(n) = Sum_{j=0..m-1} (A010877(2*b(j)) + 4 + A002265(b(j)) - A002265(b(j)+1))*10^j.
Special values:
a(1*(4^n-1)/3) = 4*(10^n-1)/9.
a(2*(4^n-1)/3) = 2*(10^n-1)/3.
a(3*(4^n-1)/3) = 8*(10^n-1)/9.
a(4*(4^n-1)/3) = 10^n-1.
a(n) < 4*(10^log_4(3*n+1)-1)/9, equality holds for n=(4^k-1)/3, k > 0.
a(n) < 4*A084544(n), equality holds iff all digits of A084544(n) are 1.
a(n) > 2*A084544(n).
Lower and upper limits:
lim inf a(n)/10^log_4(n) = 1/10*10^log_4(3) = 0.62127870, for n --> inf.
lim sup a(n)/10^log_4(n) = 4/9*10^log_4(3) = 2.756123868970, for n --> inf.
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)*(1-z(j))*(4 + 6z(j) + 8*z(j)^2 + 9*z(j)^3)/(1-z(j)^4), where z(j) = x^4^j.
Also: g(x) = (1/(1-x))*(4*h_(4,0)(x) + 2*h_(4,1)(x) + 2*h_(4,2)(x) + h_(4,3)(x) - 9*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) = 1.039691381254753739202528087006945643166147087095114911673083135126969046250... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Offset corrected by Arkadiusz Wesolowski, Oct 03 2011

cp's OEIS Frontend

A052382 Numbers without 0 in the decimal expansion, colloquial 'zeroless numbers'.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

Python

Python

Smalltalk

sh

Formula

Extensions

A046034 Numbers whose digits are primes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

Python

Formula

Extensions

A007931 Numbers that contain only 1's and 2's. Nonempty binary strings of length n in lexicographic order.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Python

Python

Formula

Extensions

A014263 Numbers that contain even digits only.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Python

Python

Formula

Extensions