A250068 Maximum width of any region in the symmetric representation of sigma(n).
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2
Offset: 1
Keywords
Examples
a(6) = 2 since the sequence of widths at each unit step in the symmetric representation of sigma(6) = 12 is 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1. For visual examples see A237270, A237593 and sequences referenced in these.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
(* function a2[ ] is defined in A249223 *) a250068[n_]:=Max[a2[n]] a250068[{m_,n_}]:=Map[a250068,Range[m,n]] a250068[{1,100}](* data *)
-
PARI
t237048(n,k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0); kmax(n) = (sqrt(1+8*n)-1)/2; t249223(n,k) = sum(j=1, k, (-1)^(j+1)*t237048(n,j)); a(n) = my(wm = t249223(n, 1)); for (k=2, kmax(n), wm = max(wm, t249223(n, k))); wm; \\ Michel Marcus, Sep 20 2015
Formula
a(n) = max_{k=1..floor((sqrt(8*n+1) - 1)/2)} (Sum_{j=1..k}(-1)^(j+1)*A237048(n, j)), for n >= 1.
Comments