A014261 Numbers that contain odd digits only.

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, 33, 35, 37, 39, 51, 53, 55, 57, 59, 71, 73, 75, 77, 79, 91, 93, 95, 97, 99, 111, 113, 115, 117, 119, 131, 133, 135, 137, 139, 151, 153, 155, 157, 159, 171, 173, 175, 177, 179, 191, 193, 195, 197, 199, 311, 313, 315, 317, 319

Offset

1,2

Comments

Or, numbers whose product of digits is odd.

Complement of A007928; A196563(a(n)) = 0. - Reinhard Zumkeller, Oct 04 2011

If n is represented as a zerofree base-5 number (see A084545) 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) = 1, 3, 5, 7, 9 for k = 1..5. - Hieronymus Fischer, Jun 06 2012

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Index entries for 10-automatic sequences.

Formula

A121759(a(n)) = a(n); A000035(A007959(a(n))) = 1. - Reinhard Zumkeller, Nov 30 2007

From Reinhard Zumkeller, Aug 30 2009: (Start)

a(n+1) - a(n) = A164898(n). - Reinhard Zumkeller, Aug 30 2009

a(n+1) = h(a(n)) with h(x) = 1 + (if x mod 10 < 9 then x + x mod 2 else 10*h(floor(x/10)));

a(n) = f(n, 1) where f(n, x) = if n = 1 then x else f(n-1, h(x)). (End)

From Hieronymus Fischer, Jun 06 2012: (Start)

a(n) = Sum_{j = 0..m-1} ((2*b_j(n)+1) mod 10)*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) = 1*(10^n-1)/3.

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

a(4*(5^n-1)/4) = 7*(10^n-1)/9.

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

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

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

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

a(n) <= 27/(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) if and only if all digits of A084545(n) are 1, else a(n) > A084545(n).

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

a(n) = A225985(A226091(n)). - Reinhard Zumkeller, May 26 2013

Sum_{n>=1} 1/a(n) = A194181. - Bernard Schott, Jan 13 2022

Example

a(10^3) = 13779.
a(10^4) = 397779.
a(10^5) = 11177779.
a(10^6) = 335777779.

Mathematica

Select[Range[400], OddQ[Times@@IntegerDigits[#]] &] (* Alonso del Arte, Feb 21 2014 *)

Prog

(Magma) [ n : n in [1..129] | IsOdd(&*Intseq(n, 10)) ];
(Haskell)
a014261 n = a014261_list !! (n-1)
a014261_list = filter (all (`elem` "13579") . show) [1, 3..]
-- Reinhard Zumkeller, Jul 05 2011
(PARI) is(n)=Set(digits(n)%2)==[1] \\ Charles R Greathouse IV, Jul 06 2017
(PARI) a(n)={my(k=1); while(n>5^k, n-=5^k; k++); fromdigits([2*d+1 | d<-digits(5^k+n-1, 5)]) - 3*10^k} \\ Andrew Howroyd, Jan 17 2020
(Python)
from itertools import islice, count
def A014261(): return filter(lambda n: set(str(n)) <= {'1', '3', '5', '7', '9'}, count(1, 2))
A014261_list = list(islice(A014261(), 20)) # Chai Wah Wu, Nov 22 2021
(Python)
from itertools import count, islice, product
def agen(): yield from (int("".join(p)) for d in count(1) for p in product("13579", repeat=d))
print(list(islice(agen(), 60))) # Michael S. Branicky, Jan 13 2022

Crossrefs

Similar to but different from A066640.

Subsequence of A059708. Cf. A005408, A010879, A014263, A046034, A084545, A089581, A084984, A001743, A001744, A202267, A202268, A196564.

Cf. A192107, A194181.

Cf. A030096 (primes).

Subsequence of A225985.

Sequence in context: A225105 A143451 A352547 * A066640 A137507 A061808

Adjacent sequences: A014258 A014259 A014260 * A014262 A014263 A014264

Keyword

nonn,base,easy,look

Author

N. J. A. Sloane

Extensions

More terms from Robert G. Wilson v, Oct 18 2002

Examples and crossrefs added by Hieronymus Fischer, Jun 06 2012

Status

approved