A230399 Smallest k such that when k is divided by all numbers <= k the remainder n occurs most often.

1, 7, 26, 23, 34, 53, 118, 167, 188, 69, 178, 179, 372, 349, 374, 375, 376, 377, 498, 499, 356, 501, 502, 503, 1284, 1285, 746, 747, 748, 749, 1038, 1039, 1112, 753, 754, 755, 2136, 2137, 2138, 2139, 2140, 2141, 2562, 2443, 1484, 2445, 1486, 1487, 2568, 2569, 2570, 2571, 2572, 2573, 2934, 2575

Offset

0,2

Comments

Number of row in A072528 where it happens first that the n-th column contains the maximum value in the row. - Ralf Stephan, Oct 21 2013

Any nonnegative integer n is the most common remainder during dividing m by all numbers not exceeding m for infinitely many different positive integers m.

Ties for most common are not allowed. Thus a(1) is not 5, although 0 and 1 both occur twice among 5 mod 1 to 5 mod 5. - Robert Israel, May 10 2020

Vladimir Letsko (see Letsko Link) asks if a(2013) exists (paraphrasing his question) without asking its value. Indeed, for m = 526173 we have T(m, 2013) = 68 where 68 is the largest value of T(526173, k) for k >= 0 and T as defined in A072528. - David A. Corneth, May 12 2020

Links

David A. Corneth, Table of n, a(n) for n = 0..10080 (first 468 terms from Robert Israel)

Vladimir Letsko, Mathematical Marathon, Problem 181 (in Russian).

Example

a(1) = 7 because
(1) 7 mod 2 = 7 mod 3 = 7 mod 6 = 1 and the other remainders occur fewer times;
(2) 7 is the least number k for which r = k mod b yields the remainder r=1 for more bases b < k than any other remainder.

Maple

maxrem:=proc(n) local r, n1, i, mx, M, R; n1:=`if`(n mod 2 = 0, n/2-1, (n-1)/2);
R:=Array(0..n1, fill=1):
if n mod 2 = 0 then R[0]:=2 fi:
for i to n1 do r:=n mod i: R[r]:=R[r]+1 od:
mx:=R[0]:M:={0}:
for i to n1 do
if R[i]> mx then mx:=R[i]:M:={i} elif mx=R[i] then M:=M union {i} fi od:
M; end;
Rs:={0}:S:=[[0, 1]]:for n to 6000 do r:=maxrem(n):if nops(r)=1 then r:=op(r):
if not member(r, Rs) then Rs:=Rs union {r}:S:=[op(S), [r, n]] fi fi od:
S:=sort(S);
T:=[]:for i to nops(S) do if S[i, 1]=i-1 then T:=[op(T), S[i, 2]] else break fi od:T;

Mathematica

a[n_] := Module[{k, rems}, For[k = 1, True, k++, rems = SortBy[Tally[Mod[k, Range[k]]], Last]; If[rems[[-1, 1]] == n && rems[[-1, 2]] != rems[[-2, 2]], Print[n, " ", k]; Return[k]]]];
a[0] = 1;
a /@ Range[0, 100] (* Jean-François Alcover, Jun 18 2020 *)

Prog

(PARI) record(n)=v=vector(n+1); for(d=1, n, t=(n%d)+1; v[t]=v[t]+1); m=0; p=0; for(i=1, n, if(v[i]>m, m=v[i]; p=i)); p
for(n=1, 100, for(j=1, 10^6, if(record(j)==n, print1(j, ", "); break))) \\ Ralf Stephan, Oct 21 2013

Crossrefs

Cf. A230374, A002182, A072528.

Sequence in context: A284786 A262109 A276614 * A246910 A098127 A283427

Adjacent sequences: A230396 A230397 A230398 * A230400 A230401 A230402

Keyword

nonn,look

Author

Vladimir Letsko, Oct 18 2013

Status

approved