A084544 Alternate number system in base 4.
1, 2, 3, 4, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44, 111, 112, 113, 114, 121, 122, 123, 124, 131, 132, 133, 134, 141, 142, 143, 144, 211, 212, 213, 214, 221, 222, 223, 224, 231, 232, 233, 234, 241, 242, 243, 244, 311, 312, 313, 314, 321
Offset: 1
Examples
From _Hieronymus Fischer_, Jun 06 2012: (Start) a(100) = 1144. a(10^3) = 33214. a(10^4) = 2123434. a(10^5) = 114122134. a(10^6) = 3243414334. a(10^7) = 211421121334. a(10^8) = 11331131343334. a(10^9) = 323212224213334. (End)
Links
- Hieronymus Fischer, Table of n, a(n) for n = 1..10000
- 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 pp. 27-29.
- R. R. Forslund, Positive Integer Pages [Broken link]
- James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
- Index entries for 10-automatic sequences.
Crossrefs
Programs
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Python
def A084544(n): m = (3*n+1).bit_length()-1>>1 return int(''.join((str(((3*n+1-(1<<(m<<1)))//(3<<((m-1-j)<<1))&3)+1) for j in range(m)))) # Chai Wah Wu, Feb 08 2023
Formula
From Hieronymus Fischer, Jun 06 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1..4.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 4)*10^j,
where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
Special values:
a(k*(4^n-1)/3) = k*(10^n-1)/9, k = 1,2,3,4.
a((7*4^n-4)/3) = (13*10^n-4)/9 = 10^n + 4*(10^n-1)/9.
a((4^n-1)/3 - 1) = 4*(10^(n-1)-1)/9, n > 1.
Inequalities:
a(n) <= (10^log_4(3*n+1)-1)/9, equality holds for n=(4^k-1)/3, k>0.
a(n) > (4/10)*(10^log_4(3*n+1)-1)/9, n > 0.
Lower and upper limits:
lim inf a(n)/10^log_4(3*n) = 2/45, for n --> infinity.
lim sup a(n)/10^log_4(3*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1 - 5z(j)^4 + 4z(j)^5)/((1-z(j))(1-z(j)^4)), where z(j) = x^4^j.
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1-5(x^4^j)^4 + 4(x^4^j)^5)*x^4^j*f_j(x)/(1-x^4^j), where f_j(x) = 10^j*x^((4^j-1)/3)/(1-(x^4^j)^4). The f_j obey the recurrence f_0(x) = 1/(1-x^4), f_(j+1)(x) = 10x*f_j(x^4).
Also: g(x) = (1/(1-x))* (h_(4,0)(x) + h_(4,1)(x) + h_(4,2)(x) + h_(4,3)(x) - 4*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3) * (x^4^j)^k/(1-(x^4^j)^4).
(End)
a(n) = A045926(n) / 2. - Reinhard Zumkeller, Jan 01 2013