A383732 a(n) is the smallest k such that every digit from 0 to 9 appears at least n times among the first k digits of Pi (after the decimal point).

32, 50, 54, 65, 71, 77, 96, 99, 120, 139, 156, 166, 209, 224, 232, 235, 242, 288, 299, 301, 306, 320, 343, 351, 405, 407, 412, 429, 439, 452, 458, 463, 468, 475, 478, 486, 506, 538, 540, 544, 548, 556, 559, 560, 567, 569, 575, 577, 584, 591, 609, 621, 622, 625, 626, 631, 633, 634, 641

Offset: 1

Author

Guy Amit, May 07 2025

Keywords

• nonn
• base

Comments

The first 6 terms also appear in A037008, the position of the digit 0 in the decimal expansion of Pi.

From Pontus von Brömssen, May 13 2025: (Start)

If the digit "3" before the decimal point is included (but a(n) still represents the number of digits after the decimal point), the first difference to this sequence (i.e., the first time the extra "3" is useful) would be a(229) = 2583 instead of 2597.

For d = 0, 1, ..., 9, the smallest n >= 1 for which d is the last digit to occur n times is 1, 146359, 2100, 229, 94, 61, 118, 7, 794, 9734, respectively.

(End)

Examples

			For n = 1, a(1) = 32 since this is the first position in the decimal expansion of Pi such that every digit from 0 to 9 has appeared at least once among the first 32 digits.
For n = 2, a(2) = 50 since this is the first position in the decimal expansion of Pi such that every digit from 0 to 9 has appeared at least twice among the first 50 digits.
		

Links

• Guy Amit, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000796, A037008.

Programs

MATLAB

% Assuming x contains the digits of Pi, x = [1, 4, 1, 5, 9, ...]
a = nan(1);
counts = zeros(10, 1);
n = 1;
for i = 1:length(x)
    for j = 1:10
        if x(i) == (j-1)
            counts(j) = counts(j) + 1;
        end
    end
    if sum(counts>=n) == 10
        a(n) = i;
        n = n + 1;
    end
end

Mathematica

piDigit = Rest[RealDigits[Pi, 10, 700][[1]]];
f[pd_List]:=Module[{res,len,count,n=1},count=CreateDataStructure["FixedArray",ConstantArray[0,10]];res=CreateDataStructure["Queue"];len=Length[pd]; Do[Do[If[pd[[i]]==j-1,count["SetPart",j,1+count["Part",j]]],{j,10}];If[Total[Boole[#>=n]&/@(count["Elements"])]==10,res["Push",i];n++],{i,len}];res["Elements"]];f[piDigit] (* Shenghui Yang, May 19 2025 *) (* or *)
upto[nd_] := Block[{pi=Rest@ RealDigits[Pi,10,nd][[1]], cnt=0 Range[10], n=1}, Reap[Do[cnt[[pi[[j]] + 1]]++; If[Min[cnt-n] == 0, n++; Sow@j], {j, Length@ pi}]][[2,1]]]; upto[700] (* for older Mma, Giovanni Resta, May 22 2025 *)