PS on Leroy's sequence

[N. J. A. Sloane]

A few other comments on A058747 and A069974:

1. an explicit formula for A069974 at the start of
each block:

A069974( alpha(k) + 1 ) = A077854(k) = M(k),

where alpha(k) = A007910(k).


2. the fact that 0's regularly occur in A069974 show
that all the slots in A058747 get filled

(this follows from the fact that the parity of A077854 is 
1 1 0 0 1 1 0 0 1 1 0 0 ... )


3.  the record values in Leroy's original
sequence A058747, which is the sequence A060153,

%I A060153
%S A060153 1,2,13,26,205,410,3277,6554
%N A060153 Record values in A058747.
%Y A060153 Cf. A058747, A062500.
%K A060153 nonn,more,new
%O A060153 1,2
%A A060153 njas, Feb 23 2004

can be read off from A077854:

%I A077854
%S A077854 1,3,6,12,25,51,102,204,409,819,1638,3276,6553,13107,26214,52428,104857,
%T A077854 209715,419430,838860,1677721,3355443,6710886,13421772,26843545,53687091,
%U A077854 107374182,214748364,429496729,858993459,1717986918,3435973836,6871947673
%N A077854 Expansion of 1/((1-x)*(1-2*x)*(1+x^2)).
%F A077854 a(n) = 3*a(n-1)-3*a(n-2)+3*a(n-3)-2*a(n-4), with initial values 1 3 6 12.
%K A077854 nonn
%O A077854 0,2
%A A077854 njas, Nov 17 2002

take two terms and add 1: 12,25 -> 13, 36, skip two terms,
take two terms and add 1: ,204,409 -> 205,410, skip two terms,
repeat: 3276,6553 -> 3277,6554, ...
(Again, this is based on the location of the zeros in 69974,
which in turn depends on the parity of A077854, which as said above is
1 1 0 0 1 1 0 0 ...)

NJAS