cp's OEIS Frontend

A380153 Numbers m for which the sum of all values of k satisfying the equation: (m - floor((m - k)/k)) mod k = 0 (1 <= k <= m) equals 2*m.

%I A380153 #12 Feb 05 2025 22:18:42
%S A380153 39,4395,29055,57931,81115,152571,164955,410731,664747,877435,2080875,
%T A380153 2521087,2539515
%N A380153 Numbers m for which the sum of all values of k satisfying the equation: (m - floor((m - k)/k)) mod k = 0 (1 <= k <= m) equals 2*m.
%C A380153 a(14) > 3*10^6 (if it exists). Is there any even term?
%e A380153 Let T(i,j) be the triangle read by rows: T(i,j) = (i - floor((i - j)/j)) mod j for 1 <= j <= i. The triangle begins:
%e A380153  i\j | 1 2 3 4 5 6 7 8 9 10 11 ...
%e A380153 -----+------------------------
%e A380153    1 | 0
%e A380153    2 | 0 0
%e A380153    3 | 0 1 0
%e A380153    4 | 0 1 1 0
%e A380153    5 | 0 0 2 1 0
%e A380153    6 | 0 0 2 2 1 0
%e A380153    7 | 0 1 0 3 2 1 0
%e A380153    8 | 0 1 1 3 3 2 1 0
%e A380153    9 | 0 0 1 0 4 3 2 1 0
%e A380153   10 | 0 0 2 1 4 4 3 2 1  0
%e A380153   11 | 0 1 0 2 0 5 4 3 2  1  0
%e A380153   ...
%e A380153 The j-th column has period j^2, r-th element of this period has the form (r - 1 - floor((r - 1)/j)) mod j (1 <= r <= j^2). The period of j-th column consists of the sequence (0,1,2,...,j-1) and its consecutive j-1 right rotations (moving rightmost element to the left end).
%e A380153 39 is in this sequence because the only k's <= 39 satisfying the equation (39 - floor((39 - k)/k)) mod k = 0 are: 1, 3, 7, 9, 19, 39, hence: 1+3+7+9+19+39 = 78 = 2*39.
%o A380153 (Maxima)
%o A380153 (f(i, j):=mod(i-floor((i-j)/j), j),
%o A380153 (n:0, for m:2 thru 5000 do
%o A380153 (s:0, for k:1 thru floor(m/2) do
%o A380153 (if f(m, k)=0 then
%o A380153 (s:s+k)), if s=m then
%o A380153 (n:n+1, print(n , "" , m)))));
%Y A380153 Cf. A048158, A375595, A378275.
%K A380153 nonn,more
%O A380153 1,1
%A A380153 _Lechoslaw Ratajczak_, Jan 13 2025
%E A380153 a(9)-a(13) from _Jinyuan Wang_, Jan 14 2025