%I #16 Jan 24 2022 16:15:00
%S 0,1,3,3,2,4,1,2,1,2,1,1,1,1,1,5,1,2,1,2,1,1,2,1,5,1,2,1,1,2,1,2,6,1,
%T 2,1,2,2,5,2,5,2,1,3,2,1,2,2,4,3,3,4,1,2,5,1,1,1,1,1,3,1,3,2,1,3,2,2,
%U 3,1,2,1,3,1,2,1,4,1,2,2,4,3,3,1,1,1,1,1,2,4,1,1,1,1,1,1,3,1,1,2,2,1,2,2,2,1,2
%N Run lengths of even terms in A350877 (half if even, add next prime if odd).
%C Equals first differences of indices of odd terms (A350616) minus one.
%C After the initial 0, also equals the 2-valuation (A007814) of A350618, the terms following odd terms in A350877.
%C Record values are a(1) = 0, a(2) = 1, a(3) = 3, a(6) = 4, a(16) = 5, a(33) = 6, a(146) = 7, a(243) = 11, a(1596) = 12, a(2092) = 13, ... The first occurrences of the other values are: a(5) = 2, a(8) = 510, a(9) = 667, a(10) = 1526. No others up to n = 33000. - _M. F. Hasler_, Jan 24 2022
%H M. F. Hasler, <a href="/A350833/b350833.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A350616(n+1) - A350616(n) - 1 = A007814(A350618(n-1)) for n > 1.
%e Between the first odd term, A350877(1) = 1, and second odd term, A350877(2) = 3, there are no even terms, therefore a(1) = 0.
%e Between the second and third odd term, A350877(4) = 3, there is one even term, A350877(3) = 6, therefore a(2) = 1.
%o (PARI) A350833_upto(N)={N=A350616_upto(N);[N[k]-N[k-1]-1|k<-[2..#N]]}
%Y Cf. A007814, A350877, A350616, A350618.
%K nonn
%O 1,3
%A _M. F. Hasler_, Jan 23 2022