

A104740


a(1) = 1; for n > 1: if n is even, a(n) = least k > 0 such that sum(i=1,n/2,a(2*i1))/sum(j=1,n,a(j))>=1/4, or 1 if there is no such k; if n is odd, a(n) = largest k > 0 such that sum(i=1,(n+1)/2,a(2*i1))/sum(j=1,n,a(j))<=1/3, or 1 if there is no such k.


0



1, 3, 1, 3, 1, 3, 1, 3, 2, 6, 3, 9, 4, 12, 6, 18, 9, 27, 14, 42, 21, 63, 31, 93, 47, 141, 70, 210, 105, 315, 158, 474, 237, 711, 355, 1065, 533, 1599, 799, 2397, 1199, 3597, 1798, 5394, 2697, 8091, 4046, 12138, 6069, 18207, 9103, 27309, 13655, 40965, 20482, 61446
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OFFSET

1,2


COMMENTS

If this sequence is interpreted as describing a 01sequence: a(1) 1's followed by a(2) 0's followed by a(3) 1's ... (alternatingly), then the portion of 1's in that 01sequence oscillates between 1/4 and 1/3, except for a slight disturbance at the beginning. Quite analogously, sequences can be constructed that describe 01sequences where the portion of 1's oscillates between arbitrary bounds 0 < r < s < 1. However, depending on the choice of the bounds, the initial disturbance may extend rather far.


LINKS



EXAMPLE

Consider n = 10; for k = 5 we have (1+1+1+1+2)/(1+3+1+3+1+3+1+3+2+k) = 6/23 < 1/4, but for k = 6 we have
(1+1+1+1+2)/(1+3+1+3+1+3+1+3+2+k) = 6/24 >= 1/4, hence a(10) = 6. Consider n = 11; for k = 3 we have
(1+1+1+1+2+k)/(1+3+1+3+1+3+1+3+2+6+k) = 9/27 <= 1/3, but for k = 4 we have (1+1+1+1+2+k)/(1+3+1+3+1+3+1+3+2+6+k) = 10/28
> 1/3, hence a(11) = 3.


PROG

(PARI) {print1(a=1, ", "); p=1; s=1; for(n=1, 28, k=1; while(((p)/(s+k))>=(1/4), k++); print1(a=max(1, k1), ", "); s=s+a; k=1; while(((p+k)/(s+k))<=(1/3), k++); print1(a=max(1, k1), ", "); s=s+a; p=p+a)}


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



