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A263078 a(n) = greatest k for which A155043(n+k) < A155043(n); a(n) = A263077(n)-n.

%I A263078 #13 Oct 24 2015 12:22:11
%S A263078 -1,-2,-1,-2,1,-4,5,-2,-3,-4,1,-6,5,-2,3,2,5,-6,11,-2,9,-4,11,-2,-3,
%T A263078 -4,15,-6,19,-8,29,-2,27,-4,37,12,47,-4,45,-6,55,-8,65,-2,51,-4,61,-6,
%U A263078 -1,-2,69,-4,79,-6,77,-8,83,2,81,-12,79,10,77,76,75,6,73,16,71,14,69,-12,67,22,65,20,73,18,77,16,27,26,37,-12,35,34,45,20,51,18,49,40,47,26,45,-12,43,42,41,40,39,30
%N A263078 a(n) = greatest k for which A155043(n+k) < A155043(n); a(n) = A263077(n)-n.
%H A263078 Antti Karttunen, <a href="/A263078/b263078.txt">Table of n, a(n) for n = 1..124340</a>
%F A263078 a(n) = A263077(n)-n.
%e A263078 For n=1 we have A049820(1) = 0, thus A155043(1) = 1, and 0 is the only (and thus the largest) number from which zero can be reached with less steps (namely in zero steps, A155043(0) = 0), thus a(1) = 0 - 1 = -1.
%e A263078 For n=7, we have A155043(7) = 4 [as A049820(7) = 5, A049820(5) = 3, A049820(3) = 1, A049820(1) = 0], but there exists x=12 for which we have A049820(12) = 6, A049820(6) = 2, A049820(2) = 0, and this is the largest x such that A155043(x) < A155043(7), thus a(7) = 12 - 7 = 5.
%t A263078 a[0] = 0; a[n_] := a[n] = 1 + a[n - DivisorSigma[0, n]]; Table[k = 3 n;
%t A263078 While[a@ k >= a@ n, k--]; k - n, {n, 102}] (* _Michael De Vlieger_, Oct 13 2015 *)
%o A263078 (PARI)
%o A263078 A263078 = n -> A263077(n) - n;
%o A263078 for(n=1,124340,write("b263078.txt",n," ",A263078(n)));
%o A263078 \\ Other code as in A263077
%Y A263078 Cf. A155043, A261089, A262503, A262909, A263077.
%Y A263078 Cf. A263079 (indices of the negative terms), A263080 (of the positive terms).
%K A263078 sign
%O A263078 1,2
%A A263078 _Antti Karttunen_, Oct 09 2015