OFFSET
1,2
COMMENTS
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
REFERENCES
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]
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 1..10000
K. Atanassov, On Some of Smarandache's Problems
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
FORMULA
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)
EXAMPLE
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
MATHEMATICA
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 *)
PROG
(PARI) 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
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
R. Muller
EXTENSIONS
Edited and extended by Robert G. Wilson v, Dec 14 2002
Crossrefs added by Hieronymus Fischer, Jun 06 2012
STATUS
approved