%I
%S 1,2,1,3,10,2,8,26,52,85,39,19,87,170,79,186,64,68,214,367,
%T 198,385,182,396,628,876,1145,865,566,882,1216,1560,1916,
%U 2289,1895,1478,1915,1462,1928,2414,2911,2401
%N a(1) = 1; a(n+1) = a(n) + (sum of digits of a(1) up to a(n)), with "+" when a(n) is odd, or "" if even.
%C The graph appears to have a shape similar to that of Mertens function A002321, with increasingly large "mountains" and "valleys":
%C Successive record values of opposite sign are a(2) = 2, a(3) = 1, a(5) = 10, a(10) = 85, a(16) = 186, a(222) = 75573, a(391) = 26186, a(658) = 341791, a(987) = 134304, a(1831) = 1820815, a(2476) = 393048, a(2692) = 2089141, a(3321) = 1816290, a(6114) = 8650189, ...
%H M. F. Hasler, <a href="/A332058/b332058.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Angelini, <a href="http://list.seqfan.eu/pipermail/seqfan/2020February/020508.html">Re: Add or subtract my cumulative sum of digits</a>, SeqFan list, Feb 24 2020.
%e a(1) = 1 is odd, so we add the partial sum (so far equal to a(1)) to get the next term, a(2) = 2.
%e Now a(2) = 2 is even, so we subtract the sum of the digits of a(1) and a(2), 1 + 2 = 3 to get a(3) = 1.
%e Since a(3) = 1 is odd, we add the sum of the digits of a(1), a(2) and a(3), 1 + 2 + 1 = 4 to get a(4) = 3.
%e And so on.
%t Nest[Append[#, #[[1]] + (2 Boole[OddQ@ #[[1]] ]  1)*Total[Flatten@ IntegerDigits[#]] ] &, {1}, 41] (* _Michael De Vlieger_, Feb 25 2020 *)
%o (PARI) A332058_vec(N,a=1,s=a)={vector(N,n, a=(1)^a*s+=sumdigits(a))}
%Y See A332056 for the variant considering sum of a(n) instead of digits.
%K sign,base
%O 1,2
%A _Eric Angelini_ and _M. F. Hasler_, Feb 24 2020
