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
Examples
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)
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.
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).
Programs
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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
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Magma
[n: n in [1..100000] | Set(Intseq(n)) subset {1,2}]; // Vincenzo Librandi, Aug 19 2016
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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;
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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 *)
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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
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PARI
/* inverse function */ apply( {A007931_inv(N)=fromdigits([d-1|d<-digits(N)],2)+2<
M. F. Hasler, Nov 09 2020 -
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
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Python
def A007931(n): return int(s:=bin(n+1)[3:])+(10**(len(s))-1)//9 # Chai Wah Wu, Jun 13 2025
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)
Extensions
Some crossrefs added by Hieronymus Fischer, Jun 06 2012
Edited by M. F. Hasler, Mar 26 2015
Comments