A245491 The least x > 0 such that x < the number of zero digits in the base-n expansions of the numbers 1 through x.

9, 87, 1068, 16022, 284704, 5834024, 135430302, 3511116537, 100559404366, 3152738985032, 107400330425888, 3950024143546665, 155996847068247395, 6584073072068125453, 295764262988176583800, 14088968131538370019982, 709394716006812244474473

Offset

2,1

Comments

If the function zeros(n,b) returns the number of zeros in the numbers 1 through n in base b, then:

zeros(2,2) = zeros_in(10) = 1.

zeros(4,2) = zeros_in(10,100) = 3.

zeros(5,2) = zeros_in(10,100,101) = 4.

zeros(6,2) = zeros_in(10,100,101,110) = 5.

zeros(8,2) = zeros_in(10,100,101,110,1000) = 8.

zeros(9,2) = zeros_in(10,100,101,110,1000,1001) = 10.

Therefore 9 < zeros(9,2) and 9 is the first entry in the list.

Links

Anthony Sand, Table of n, a(n) for n = 2..100

Example

9 < zero(9,base=2) = 10.
87 < zero(87,3) = 88.
1068 < zero(1068,4) = 1069.
100559404366 < zero(100559404366,10) = 100559404367.

Mathematica

a245491[n_Integer] := Module[{x = 0, z = 0},
While[x >= z, x++; z += Count[IntegerDigits[x, n], 0]]; x]; Map[a245491, Range[2, 12]] (* Michael De Vlieger, Aug 06 2014 *)

Prog

(PARI) /* formula for calculating n such that zero(n) > n, zero(n-1) <= (n-1) */
{estimate(x, b) = m1=b; est=x\b; nn=est; while(nn>0, d=nn%b; m2 = nn\b; if(d==0, est+=(x%m1)+1; if(m2>0, m2--)); est+=m1*m2; m1*=b; nn=nn\b); return(est)}
{bmin=2; bmx=20; for(bs=bmin, bmx, ni=bs^bs; n=bs+1; ez1=0; ez2=0; until(ez1>n && ez2<=n-1, ez = estimate(n, bs); if(n>=ez, n+=ni, n-=ni; if(ni>1, ni=ni\bs)); ez1 = estimate(n, bs); ez2 = estimate(n-1, bs)); print1(n, ", ")) } \\ Anthony Sand, Aug 11 2014

Crossrefs

Cf. A164935.

Sequence in context: A152264 A035101 A351525 * A160466 A362509 A015583

Adjacent sequences: A245488 A245489 A245490 * A245492 A245493 A245494

Keyword

nonn,base

Author

Anthony Sand, Jul 24 2014

Status

approved