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*i-1))/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*i-1))/sum(j=1,n,a(j))<=1/3, or 1 if there is no such k.

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

Offset

1,2

Comments

If this sequence is interpreted as describing a 0-1-sequence: 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 0-1-sequence oscillates between 1/4 and 1/3, except for a slight disturbance at the beginning. Quite analogously, sequences can be constructed that describe 0-1-sequences 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.

Interleaving of A073941 and A081848 from a(3) onward.

Links

Table of n, a(n) for n=1..56.

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, k-1), ", "); s=s+a; k=1; while(((p+k)/(s+k))<=(1/3), k++); print1(a=max(1, k-1), ", "); s=s+a; p=p+a)}

Crossrefs

Cf. A073941, A081848.

Sequence in context: A334464 A316780 A338730 * A111736 A330889 A035652

Adjacent sequences: A104737 A104738 A104739 * A104741 A104742 A104743

Keyword

nonn

Author

Klaus Brockhaus, Mar 21 2005

Status

approved