A230360 - OEIS

%I #45 Jun 11 2019 22:33:10

%S 2,1,0,3,6,3,0,5,12,11,0,5,12,0,0,5,0,0,0,48,14,0,61,188,83,0,81,232,

%T 268,0,0,650,0,0,622,299,0,0,0,501,0,0,2655,602,0,6429,8990,7856,0,

%U 26187,17898,3744,0,40300,16395,0,0,0,0,0,0,124876,173552,0,0

%N a(n) is the number of base-4 n-digit numbers requiring only binary digits in bases 3 and 4.

%C 0 is included to mesh with the earlier A146025 ({0, 1, 82000}: the values in decimal with top base 4 here replaced by 5, conjectured complete).  It appears unlikely, empirically, that this sequence has a last positive term, and a heuristic approximation of terms is likely not difficult.

%C The numbers here are the counts of members of A000695 also occurring in A005836 and being n digits in length in base-4.

%H Stuart A. Burrell, Han Yu, <a href="https://arxiv.org/abs/1905.00832">Digit expansions of numbers in different bases</a>, arXiv:1905.00832 [math.NT], 2019.

%e The first 8 values are 0, 1, 4, 81, 84, 85, 256 and 273--0, 1, 11, 10000, 10010, 10011, 10000111 and 10001010 in base 3, and 0, 1, 10, 1101, 1110, 1111, 10000 and 10101 in base 4; and, from the base-4 listing, a(1)=2, a(2)=1, a(3)=0, a(4)=3, and a(5) is at least 2.

%t MapAt[# + 1 &, Array[Count[FromDigits[#, 4] & /@ IntegerDigits[Range[2^(# - 1), 2^# - 1], 2], _?(DigitCount[#, 3][[2]] == 0 &)] &, 20], 1] (* _Michael De Vlieger_, Jun 11 2019 *)

%o (PARI)

%o {

%o \\ This program finds the number of d-digit base B>b\\

%o \\ numbers not requiring digits beyond those of base b\\

%o \\ for bases b+1 through B. It runs a check in reverse\\

%o \\ down to base b+1, maintaining additions not yet done\\

%o \\ in vector S, where the digits in each base are kept\\

%o \\ in matrix N.  The value itself is kept as n, at each\\

%o \\ new base checked for n, the value in S is transfered\\

%o \\ to variable t; with the check being done of whether\\

%o \\ the criterion is satisfied for n in the base under\\

%o \\ consideration.  A flag f is used to see if n passed\\

%o \\ for all bases or there was a break.  If pass, then\\

%o \\ count variable c is incremented (as is n) for the\\

%o \\ next run through bases.  At each addition, a check\\

%o \\ of whether the base is B and the number of digits\\

%o \\ changes is done, and if so a new term is output.\\

%o \\ pos and POS are variables for the digit-positions\\

%o \\ under consideration in additions essentially mimic-\\

%o \\ king hand addition.  Flag g identifies whether or\\

%o \\ not a large addition is warranted by virtue of an\\

%o \\ addition resulting in a digit larger than b-1, the\\

%o \\ leftmost of these being the point from which this\\

%o \\ addition is made using variable s calculated four\\

%o \\ lines above the bottom one of the program.  This \\

%o \\ program is readily modified  to store a smaller #\\

%o \\ of digits (D), change the b and B values, and print\\

%o \\ specific n values as desired.\\

%o b=2;B=4;d=1;c=1;D=10000;

%o N=matrix(B-b,D);n=1;S=vector(B-b,x,1);

%o while(1,

%o   f=1;forstep(i=B,b+1,-1,

%o     t=S[i-b];if(t,

%o       S[i-b]=0;pos=0;ca=0;

%o       while(t,

%o         pos++;N[i-b,pos]+=t%i+ca;

%o         if(N[i-b,pos]>=i,ca=1;N[i-b,pos]-=i,ca=0);

%o         t\=i);

%o       if(ca,pos++;N[i-b,pos]++;if(i==B,if(pos==d+1,

%o             print1(c",");d++;c=0)));

%o       POS=pos;g=1;

%o       while(POS,

%o         if(N[i-b,POS]>=b,g=0;break(),POS--));

%o       if(g==0,

%o         f=0;POS++;while(N[i-b,POS]==b-1,POS++);

%o         N[i-b,POS]++;for(j=1,POS-1,N[i-b,j]=0);

%o         s=i^(POS-1)-n%(i^(POS-1));

%o         for(j=1,B-b,if(j!=i-b,S[j]+=s));

%o         if(i==B,if(POS==d+1,print1(c",");d++;c=0));

%o         n+=s;break())));

%o   if(f,c++;n++;S=vector(B-b,x,1)))

%o }

%Y Cf. A146025, A005836, A000695

%K nonn,base,uned

%O 1,1

%A _James G. Merickel_, Oct 16 2013