A164935 -id:A164935 - cp's OEIS frontend

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

Anthony Sand, Jul 24 2014

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.

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

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

Cf. A164935.

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 *)

/* 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

A339752 Base-ten n containing a digit making its n-th appearance in the list of consecutive positive integers.

Original entry on oeis.org

1, 199981, 199982, 199983, 199984, 199985, 199986, 199987, 199988, 199989, 199990, 199991, 200001, 1599981, 1599982, 1599983, 1599984, 1599985, 1599986, 1599987, 1599988, 1599989, 1599990, 1599991, 2600001, 13199998, 13199999, 28263827, 28263828, 35000001

Offset: 1

Hans Havermann, Dec 15 2020

Similar to A216400 but an improvement in definition that does not exclude consideration of digit zero and additionally allows for a wordless description of the sequence (see "Counting digits" in links).

Conjectured to be "full": digit 0 (the final term) might have subsequents.

			28263827 contains a 2 (the third one) making its 28263827th occurrence, so 28263827 is in the sequence.
100559404366 contains a 0 (the second one) making its 100559404366th occurrence, so 100559404366 is in the sequence.

Hans Havermann, Table of n, a(n) for n = 1..662
Hans Havermann, Positive-integers digit-count coincidences
Hans Havermann, Proportionality

Cf. A000027 (consecutive positive integers), A216400 (Mathematica's Google aptitude variant), A164935 (starts with ten terms of this sequence), A245491 (contains the final term of this sequence), A339753 (English number-words variant).

cp's OEIS Frontend

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

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