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A007932
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Numbers that contain only 1's, 2's and 3's.
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8
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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
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OFFSET
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1,2
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COMMENTS
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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. [From Chris Gaconnet (gaconnet(AT)gmail.com), May 27 2009]
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REFERENCES
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K. Atanassov, On the 97-th, 98-th and the 99-th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 89-93.
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.
A. Salomaa, Formal Languages, Academic Press, 1973. pages 90-91. [From Chris Gaconnet (gaconnet(AT)gmail.com), May 27 2009]
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LINKS
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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
F. Smarandache, Only Problems, Not Solutions!.
Wikipedia, Bijective numeration
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FORMULA
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Contribution from Hieronymus Fischer, May 30 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)
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EXAMPLE
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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.
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MATHEMATICA
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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]
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CROSSREFS
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Cf. A007931, A052382, A084544, A084545, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.
Sequence in context: A083758 A127494 A130803 * A035122 A085305 A189818
Adjacent sequences: A007929 A007930 A007931 * A007933 A007934 A007935
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KEYWORD
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nonn,base
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AUTHOR
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R. Muller
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EXTENSIONS
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Edited and extended by Robert G. Wilson v, Dec 14 2002
Examples a(100)..a(10^9) and crossrefs added by Hieronymus Fischer, Jun 06 2012
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STATUS
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approved
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