|
%I #17 Jul 12 2018 04:28:27
%S 1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,
%T 0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,
%U 1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,0
%N Increments in the number of decimal digits of 4^n.
%C This sequence is not periodic because log(4)/log(10) is an irrational number. - _T. D. Noe_, Jan 25 2007
%H Robert Israel, <a href="/A125144/b125144.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n)=Number_of_digits{4^(n+1)}-Number_of digits{4^(n)} with n>=0 and where "Number_of digits" is a hypothetical function giving the number of digits of the argument.
%e a(1)=1 because 4^(1+1)=16 (two digits) 4^1=4 (one digit) and the difference is 1.
%e a(2)=0 because 4^(2+1)=64 (two digits) 4^(2)=16 (two digits) and the difference is 0.
%p P:=proc(n) local i,j,k,w,old; k:=4; for i from 1 by 1 to n do j:=k^i; w:=0; while j>0 do w:=w+1; j:=trunc(j/10); od; if i>1 then print(w-old); old:=w; else old:=w; fi; od; end: P(1000);
%p # alternative:
%p H:= [seq(ilog10(4^i),i=1..1001)]:
%p H[2..-1]-H[1..-2]; # _Robert Israel_, Jul 12 2018
%o (PARI) a(n) = #digits(4^(n+1)) - #digits(4^n); \\ _Michel Marcus_, Jul 12 2018
%Y Cf. A125117, A125122.
%Y First differences of A210434.
%K easy,nonn,base
%O 1,1
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jan 11 2007
%E Offset corrected by _Robert Israel_, Jul 11 2018
|
