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A173667 Number of real zeros of the polynomial whose coefficients are the decimal expansion of n. 7
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, 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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,100

COMMENTS

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

a(n) is the number of real zeros of the polynomial P(n,x) = Sum_{k=0..p} d(k)

  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,...

LINKS

Table of n, a(n) for n=1..112.

EXAMPLE

a(121) = 2 because 1+2x+x^2 = 0 has 2 real roots.

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.

MAPLE

with(numtheory):T:=array(1..10000000): x2:=0: printf(`%d, `, x2):for n from

  1 to 300 do: for i from 1 to 8 do: T[i]:=0:od: l:=length(n) : n0:=n:for m from

  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]+

  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:

PROG

(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

CROSSREFS

Sequence in context: A251570 A292241 A332015 * A241541 A086008 A083910

Adjacent sequences:  A173664 A173665 A173666 * A173668 A173669 A173670

KEYWORD

nonn,base

AUTHOR

Michel Lagneau, Nov 24 2010

STATUS

approved

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Last modified June 19 03:22 EDT 2021. Contains 345125 sequences. (Running on oeis4.)