OFFSET
0,7
COMMENTS
a(n) = number of real zeros of the polynomial P(n,x) = sum_{k=0..n-1} d(k) x^k, where d(k) are the digits of the decimal expansion of floor(Pi*10^n), n=0,1,2,...
From Robert Israel, Dec 19 2018: (Start)
If d(n) = 0 then P(n,x)=P(n-1,x) so a(n)=a(n-1).
If d(n) <> 0 and P(n,x) has nonzero discriminant, then a(n) == n (mod 2).
Conjecture: P(n,x) has nonzero discriminant for all n >= 1.
Record values: a(0)=0, a(1)=1, a(6)=2, a(135)=3, a(374)=4. (End)
LINKS
Robert Israel, Table of n, a(n) for n = 0..545
EXAMPLE
a(0) = 0 because 3 => P(0,x)=3 is a constant and has 0 real root;
a(1) = 1 because 31 => P(1,x) = 1+3x has 1 real root;
a(6) = 2 because 3141592 => P(6,x) = 2 + 9x + 5x^2 + x^3 + 4x^4 + x^5 + 3x^6 has 2 real roots.
MAPLE
L:= convert(floor(10^100*Pi), base, 10):
f:= proc(n) local P, x, i;
P:=add(L[-i]*x^(i-1), i=1..n+1);
sturm(sturmseq(P, x), x, -infinity, infinity)
end proc:
map(f, [$0..100]); # Robert Israel, Dec 19 2018
PROG
(PARI) A175799(n)={ default(realprecision)>n || default(realprecision, n+1); sum(k=1, #n=factor(1.*Pol(eval(Vec(Str(Pi*10^n\1)))))~, (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, Dec 04 2010
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Dec 04 2010
EXTENSIONS
Corrected and extended by Robert Israel, Dec 19 2018
STATUS
approved