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%I #15 Sep 11 2024 00:57:59
%S 23,29,35,41,47,53,59,65,71,77,79,83,89,95,99,101,107,111,113,119,125,
%T 131,137,139,143,149,155,159,161,167,173,179,185,191,197,199,203,209,
%U 215,219,221,223,227,233,239,245,251,257,259,263,269
%N Numbers m for which the sum of all values of k satisfying the equation: m mod k = floor((m - k)/k) mod k (1 <= k <= m) exceeds 2*m.
%C The first even element of this sequence is a(817) = 3464.
%e Let T(i,j) be the triangle read by rows: T(i,j) = 1 if i mod j = floor((i - j)/j) mod j, T(i,j) = 0 otherwise, for 1 <= j <= i. The triangle begins:
%e i\j| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
%e -----------------------------------------
%e 1| 1
%e 2| 1 1
%e 3| 1 0 1
%e 4| 1 0 0 1
%e 5| 1 1 0 0 1
%e 6| 1 1 0 0 0 1
%e 7| 1 0 1 0 0 0 1
%e 8| 1 0 0 0 0 0 0 1
%e 9| 1 1 0 1 0 0 0 0 1
%e 10| 1 1 0 0 0 0 0 0 0 1
%e 11| 1 0 1 0 1 0 0 0 0 0 1
%e 12| 1 0 1 0 0 0 0 0 0 0 0 1
%e 13| 1 1 0 0 0 1 0 0 0 0 0 0 1
%e 14| 1 1 0 1 0 0 0 0 0 0 0 0 0 1
%e 15| 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1
%e ...
%e The j-th column has period j^2. Consecutive elements of this period are j X j identity matrix entries, read by rows.
%e 11 is not in this sequence because only k's <= 11 satisfying the equation 11 mod k = floor((11 - k)/k) mod k are: 1, 3, 5, 11, hence 1+3+5+11 = 20 and 20 < 2*11.
%e 23 is in this sequence because only k's <= 23 satisfying the equation 23 mod k = floor((23 - k)/k) mod k are: 1, 5, 7, 11, 23, hence 1+5+7+11+23 = 47 and 47 > 2*23.
%o (Maxima)
%o (f(i,j):=mod(i-floor((i-j)/j),j),
%o (n:0, for m:2 thru 500 do
%o (s:0, for k:1 thru floor(m/2) do
%o (if f(m,k)=0 then
%o (s:s+k)), if s>m then
%o (n:n+1, print(n , "" , m)))));
%Y Cf. A005101, A051731, A375007.
%K nonn
%O 1,1
%A _Lechoslaw Ratajczak_, Aug 20 2024