|
%I #38 Feb 09 2023 10:24:01
%S 1,2,3,4,11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44,111,112,113,
%T 114,121,122,123,124,131,132,133,134,141,142,143,144,211,212,213,214,
%U 221,222,223,224,231,232,233,234,241,242,243,244,311,312,313,314,321
%N Alternate number system in base 4.
%H Hieronymus Fischer, <a href="/A084544/b084544.txt">Table of n, a(n) for n = 1..10000</a>
%H EMIS, <a href="http://www.emis.de/journals/SWJPAM/vol1-95.html">Mirror site for Southwest Journal of Pure and Applied Mathematics</a>
%H R. R. Forslund, <a href="http://www.emis.de/journals/SWJPAM/Vol1_1995/rrf01.ps">A logical alternative to the existing positional number system</a>, Southwest Journal of Pure and Applied Mathematics, Vol. 1 1995 pp. 27-29.
%H R. R. Forslund, <a href="http://my.tbaytel.net/~forslund/index.html">Positive Integer Pages</a> [Broken link]
%H James E. Foster, <a href="http://www.jstor.org/stable/3029479">A Number System without a Zero-Symbol</a>, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
%H <a href="/index/Ar#10-automatic">Index entries for 10-automatic sequences</a>.
%F From _Hieronymus Fischer_, Jun 06 and Jun 08 2012: (Start)
%F The formulas are designed to calculate base-10 numbers only using the digits 1..4.
%F a(n) = Sum_{j=0..m-1} (1 + b(j) mod 4)*10^j,
%F where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
%F Special values:
%F a(k*(4^n-1)/3) = k*(10^n-1)/9, k = 1,2,3,4.
%F a((7*4^n-4)/3) = (13*10^n-4)/9 = 10^n + 4*(10^n-1)/9.
%F a((4^n-1)/3 - 1) = 4*(10^(n-1)-1)/9, n > 1.
%F Inequalities:
%F a(n) <= (10^log_4(3*n+1)-1)/9, equality holds for n=(4^k-1)/3, k>0.
%F a(n) > (4/10)*(10^log_4(3*n+1)-1)/9, n > 0.
%F Lower and upper limits:
%F lim inf a(n)/10^log_4(3*n) = 2/45, for n --> infinity.
%F lim sup a(n)/10^log_4(3*n) = 1/9, for n --> infinity.
%F 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.
%F 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).
%F 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).
%F (End)
%F a(n) = A045926(n) / 2. - _Reinhard Zumkeller_, Jan 01 2013
%e From _Hieronymus Fischer_, Jun 06 2012: (Start)
%e a(100) = 1144.
%e a(10^3) = 33214.
%e a(10^4) = 2123434.
%e a(10^5) = 114122134.
%e a(10^6) = 3243414334.
%e a(10^7) = 211421121334.
%e a(10^8) = 11331131343334.
%e a(10^9) = 323212224213334. (End)
%o (Python)
%o def A084544(n):
%o m = (3*n+1).bit_length()-1>>1
%o 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
%Y Cf. A007931, A007932, A052382, A084545, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.
%K nonn,base
%O 1,2
%A Robert R. Forslund (forslund(AT)tbaytel.net), Jun 27 2003
%E Offset set to 1 according to A007931, A007932 by _Hieronymus Fischer_, Jun 06 2012
|
