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%I #3 Oct 01 2013 21:35:19
%S -1,2,47,1,3,12,2,10,3,2,159,45,348,12,129,144,31,56,196,113,57,326,
%T 19,5,33,163,75,19,69,162,8,67,176,30,88,62,64,22,49,33,170,2,97,24,
%U 183,420,119,198,241,36,114,49,6,42,17,339,92,26,16,14,68,117,350,124,8,64,69,16,20,147,76,60,17,68,188,37,105,35,356,79
%N Position of the first n in the decimal expansion of Pi^n, or -1 if the digits of n never appear.
%C Conjecture: -1 only occurs in the first entry of the sequence. 2000 digits of precision was used in the calculation.
%F Pi = 3.1415926535...
%o (PARI) digitpospi(n) = \ Pi^n expansion first occurrence of n { local(x,y,r,dot); for(x=0,n, r = Pi^x; if(r==floor(r), y=find(Str(floor(r)),x), y=find(Str(r),x); 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 0,2
%A _Cino Hilliard_, Dec 25 2006
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