%I #14 Aug 01 2024 14:12:56
%S 0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,0,0,0,0,0,0,0,0,0,2,0,0
%N Number of real zeros of the polynomial whose coefficients are the decimal expansion of n.
%C From the example a(121)=2, P(121,x)=(x+1)^2, it is seen that the roots are counted with multiplicity. For n=0, the polynomial would be P=0 with infinitely many real roots. - _M. F. Hasler_, Nov 24 2010
%C a(n) is the number of real zeros of the polynomial P(n,x) = Sum_{k=0..p} d(k)
%C x^k where d(k) are the digits of the decimal expansion of n = d(p) d(p-1)...d(0), n =0,1,2,...
%e a(121) = 2 because 1+2x+x^2 = 0 has 2 real roots.
%e a(1597200)=6 because x^6 + 5x^5 + 9x^4+7x^3+2x^2 = x^2 (x+2)(x+1)^3 has 6 real roots.
%p with(numtheory):T:=array(1..10000000): x2:=0: printf(`%d, `, x2):for n from
%p 1 to 300 do: for i from 1 to 8 do: T[i]:=0:od: l:=length(n) : n0:=n:for m from
%p 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :u: T[m]:=u:od:x:=fsolve(T[1]+
%p T[2]*z + T[3]*z^2+ T[4]*z^3+ T[5]*z^4 + T[6]*z^5 + T[7]*z^6 + T[8]*z^7, z, real):x1:=[x]: x2:=nops(x1): printf(`%d, `, x2):od:
%o (PARI) A173667(n)=sum(k=1,#n=factor(1.*Pol(eval(Vec(Str(n)))))~,(poldegree(n[1,k])==1)*n[2,k] ) /* factorization over the reals => linear factor for each root. poldegree()==1 could be replaced by poldisc()>=0 */ \\ _M. F. Hasler_, Nov 25 2010
%K nonn,base
%O 1,100
%A _Michel Lagneau_, Nov 24 2010