A115779 Consider the Levenshtein distance between k considered as a decimal string and k considered as a binary string. Then a(n) is the greatest number m such that the Levenshtein distance is n or 0 if no such number exists.

1, 0, 11, 15, 111, 121, 1011, 1111, 2011, 11111, 16111, 111111, 131011, 1011111, 1111111, 2011111, 11111111, 16111111, 111111111, 131111111, 1011111111, 1111111111, 2111111111, 11111111111

Offset: 0

Author

Robert G. Wilson v, Jan 26 2006

Keywords

• more
• nonn
• base

Comments

Difference between A115779&A115778: 1, 0, 9, 11, 103, 99, 979, 1047, 1789, 10855, 15599, 109067, 128789, 1006889, 1102919, 1988889, 11078343, ...,.

Crossrefs

Cf. A000027, A007088, A115777.

Programs

Mathematica

levenshtein[s_List, t_List] := Module[{d, n = Length@s, m = Length@t}, Which[s === t, 0, n == 0, m, m == 0, n, s != t, d = Table[0, {m + 1}, {n + 1}]; d[[1, Range[n + 1]]] = Range[0, n]; d[[Range[m + 1], 1]] = Range[0, m]; Do[d[[j + 1, i + 1]] = Min[d[[j, i + 1]] + 1, d[[j + 1, i]] + 1, d[[j, i]] + If[s[[i]] === t[[j]], 0, 1]], {j, m}, {i, n}]; d[[ -1, -1]]]];
t = Table[0, {25}]; f[n_] := levenshtein[ IntegerDigits[n], IntegerDigits[n, 2]]; Do[ t[[f@n+1]] = n, {n, 10^6}]; t

Formula

a(1)=0 since no number satisfies the definition and generally a(n)>= 2^(n+1).

Extensions

a(18)-a(23) from Lars Blomberg, Jul 16 2015