A365415 Number of decimal digits of e after its decimal point needed to contain all digit substrings of length n.
20, 371, 8091, 102127, 1061612, 12108840, 198150340, 1929504533
Offset: 1
Examples
a(1) = 20, since 20 is the smallest number of digits in decimal expansion of e in with every digit 0..9 (or, to be more precise, every digit string of length 1) appears: 2.71828182845904523536. a(2) = 371, since the first appearance of the digit string "12" is at positions 370-371 of the decimal expansion of e and the remaining digit strings of length 2 appear at least once before that position. a(3) = 8091, since the first appearance of the digit string "548" is at positions 8089-8091 of the decimal expansion of e and the remaining digit strings of length 3 appear at least once before that position. a(4) = 102127, since the first appearance of the digit string "1769" is at positions 102124-102127 of the decimal expansion of e and the remaining digit strings of length 4 appear at least once before that position.
Programs
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Mathematica
dok = 300000; an = {}; For[li = 1, li <= 3, li++, p = ToString[N[E, dok]]; cyf = {}; par = 0; For[i = 3, i <= dok, i++, If[par == 0, a = StringTake[p, {i, i + li - 1}]; If[MemberQ[cyf, a] == False, cyf = Append[cyf, a]; If[Length[cyf] == 10^li, an = Append[an, i + li - 3]; par = 1]], Break[]] ]]; Print[an]
Formula
a(n) = A152182(n) + n - 2.
Extensions
a(6)-a(8) from Michael S. Branicky, Oct 04 2023
Comments