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A230399
Smallest k such that when k is divided by all numbers <= k the remainder n occurs most often.
4
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
KEYWORD
nonn,look
AUTHOR
Vladimir Letsko, Oct 18 2013
STATUS
approved