re-indexing A001650 - no!

[N. J. A. Sloane]

There was a discussion here a few days ago that proposed changing
the offset of this sequence from 1 to 0:

%I A001650
%S A001650 1,3,3,3,5,5,5,5,5,7,7,7,7,7,7,7,9,9,9,9,9,9,9,9,9,11,11,11,11,11,11,11,
%T A001650 11,11,11,11,13,13,13,13,13,13,13,13,13,13,13,13,13,15,15,15,15,15,15,
%U A001650 15,15,15,15,15,15,15,15,15,17,17,17,17,17,17,17,17,17,17,17,17,17
%N A001650 n appears n times (n odd).
%F A001650 a(n) = 1 + 2*floor(sqrt(n-1)), n > 0. - Antonio Esposito (antonio.b.esposito(AT)italtel\
.it), Jan 21 2002
%F A001650 G.f.: theta_3(x)*x/(1-x). a(n+1)=a(n)+A000122(n). - Michael Somos, Apr 29 2003.
%o A001650 (PARI) a(n)=if(n<1,0,1+2*sqrtint(n-1))
%Y A001650 Cf. A001670. Partial sums of A000122.
%Y A001650 Adjacent sequences: A001647 A001648 A001649 this_sequence A001651 A001652 A001653
%Y A001650 Sequence in context: A076566 A083574 A108025 this_sequence A101290 A080605 A077886
%K A001650 nonn,easy
%O A001650 1,2
%A A001650 njas

I don't think this works, for 0 is even.

NJAS