%I #6 Oct 18 2016 05:42:14
%S -1,-1,2,4,29,28,6,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,16,4,5,
%T 10,7,2,2,2,6,4,50,5,23,5,6,15,12,16,35,6,5,12,24,6,4,5,27,4,3,11,12,
%U 4,21,13,33,8,-1,3,5,22,29,9,3,12,17,14,18,2,2,2,2,2,50,3,21,4,17,6,10,9,19,32,25,11,25,16,24,4,11,10,8,3,13
%N Position of the first 2 in the decimal expansion of the cube root of n, or -1 if this digit never appears.
%H Chai Wah Wu, <a href="/A124687/b124687.txt">Table of n, a(n) for n = 0..10000</a>
%o (PARI) digitposcbrt(n,m) = \ cbrt expansion { local(x,y,r,dot); for(x=0,n, r = (x^(1/3)); if(iscube(x), y=find(Str(floor(r)),m), y=find(Str(r),m); dot=find(Str(r),"."); if(dot < y, y--); ); if(y, print1(y","),print1(-1",") ) ) } find(str,match) = \Return the position of the first occurrence of string \match in string str { local(lnm,lns,x,c,i); str=Str(str); \This allows leaving quotes off input match=Str(match); c=0; i=0; lns=length(str); lnm=length(match); if(lnm>1,i=1); x=1; while(x<=lns-lnm+1, if(mid(str,x,lnm)== match,break,x++); ); if(x>lns,return(0),return(x)) } mid(str,s,n) = \ Get a substring of length n from string str starting at position s in str. { local(v,ln,x,tmp); v =""; tmp = Vec(str); ln=length(tmp); for(x=s,s+n-1, v=concat(v,tmp[x]); ); return(v) } iscube(n) = { local(r); r = n^(1/3); if(floor(r+.5)^3== n,1,0) }
%K base,easy,sign
%O 0,3
%A _Cino Hilliard_, Dec 25 2006