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A245491
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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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2
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9, 87, 1068, 16022, 284704, 5834024, 135430302, 3511116537, 100559404366, 3152738985032, 107400330425888, 3950024143546665, 155996847068247395, 6584073072068125453, 295764262988176583800, 14088968131538370019982, 709394716006812244474473
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OFFSET
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2,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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9 < zero(9,base=2) = 10.
87 < zero(87,3) = 88.
1068 < zero(1068,4) = 1069.
100559404366 < zero(100559404366,10) = 100559404367.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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