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

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

1,2

Comments

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

References

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.

Links

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.

Formula

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)

Example

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)

Maple

# 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;

Mathematica

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 *)

Prog

(Haskell)
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
(PARI) 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
(PARI) /* inverse function */ apply( {A007931_inv(N)=fromdigits([d-1|d<-digits(N)], 2)+2<<logint(N, 10)-1}, [1, 2, 11, 12, 21, 22, 111]) \\ M. F. Hasler, Nov 09 2020
(Magma) [n: n in [1..100000] | Set(Intseq(n)) subset {1, 2}]; // Vincenzo Librandi, Aug 19 2016
(Python)
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

Crossrefs

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).

Sequence in context: A038115 A089604 A038114 * A136407 A214215 A136999

Adjacent sequences: A007928 A007929 A007930 * A007932 A007933 A007934

Keyword

nonn,base,nice,easy

Author

R. Muller

Extensions

Some crossrefs added by Hieronymus Fischer, Jun 06 2012

Edited by M. F. Hasler, Mar 26 2015

Status

approved