cp's OEIS Frontend

The sums-complement of a sequence s(1), s(2), ... of positive integers is introduced here as the set of numbers c(1), c(2), ... such that no c(n) is a sum s(j)+s(j+1)+...+s(k) for any j and k satisfying 1 <= j <= k. If this set is not empty, the term "sums-complement" also applies to the (possibly finite) sequence of numbers c(n) arranged in increasing order. In particular, the difference sequence D(r) of a Beatty sequence B(r) of an irrational number r > 2 has an infinite sums-complement, abbreviated as SC(r) in the following table:

r B(r) D(r) SC(r)

----------------------------------------------------

sqrt(5) A022839 A081427 A276871

sqrt(6) A022840 A276856 A276872

sqrt(7) A022841 A276857 A276873

sqrt(8) A022842 A276858 A276874

e A022843 A276859 A276875

2*e A276853 A276860 A276876

Pi A022844 A063438 A276877

2*Pi A028130 A276861 A276878

1+sqrt(2) A003151 A276862 A276879

1+sqrt(3) A054088 A007538 A276880

1+sqrt(5) A276854 A276863 A276881

2+sqrt(2) A001952 A276864 A276882

2+sqrt(3) A003512 A276865 A276883

2+sqrt(5) A004976 A276866 A276884

1+tau A001950 2 + A003849 A276885

2+tau A003231 A276867 A276886

3+tau A276855 A276868 A276887

2+sqrt(1/2) A182769 A276869 A276888

sqrt(2)+sqrt(3) A110117 A276870 A276889

From Jeffrey Shallit, Aug 15 2023: (Start)

Simpler description: this sequence represents those positive integers that CANNOT be expressed as a difference of two elements of A022839.

There is a 20-state Fibonacci automaton for the terms of this sequence (see a276871.pdf). It takes as input the Zeckendorf representation of n and accepts iff n is a member of A276871. (End)

Conjecture: a(n)/n -> 2 + sqrt(2), and 0 < 2 + sqrt(2) - a(n)/n < -1 + sqrt(2) for n >= 1.

From Michel Dekking, Mar 20 2022: (Start)

Proof of (a part of) Kimberling's conjecture. This is based on the formula for the recursion in the FORMULA section of A293077, where we use that the final-letter-removed version of the String-Replace map has the same fixed point A289035.

Let N1(n) be the number of 1's in the n-th iterate theta(n) of the final-letter-removed mapping defined in A289035. It was observed in A293077 that N1(n) = L(n-1)/2, where L(n) = A293077(n) is the length of the n-th iterate theta(n) of the final-letter-removed mapping. As usual, proving a(n)/n -> 2 + sqrt(2) is the same as proving that the frequency of 1's in A293077 is equal to 1/(2 + sqrt(2)) = 1 - sqrt(2)/2 =: mu. Ignoring a proof of the existence of the limit, this means that we have to show that

N1(n) /L(n) -> 1 - sqrt(2)/2 as n->oo.

By the observation above this is the same as proving that

L(n-1)/2L(n) -> 1 - sqrt(2)/2 as n->oo.

We know from A293077 that

L(n) = 2 L(n-1) - L(n-2) + 2 floor(L(n-2)/4).

For the determination of the limit we replace the term 2 floor(L(n-2)/4) by L(n-2)/2, as we may.

So we have L(n) ~ 2 L(n-1) - L(n-2)/2, which leads to

L(n)/2L(n+1) ~ 2 L(n-1)/2L(n+1) - L(n-2)/4L(n+1).

Replacing L(n-1)/L(n+1) by L(n-1)/2L(n)*2L(n)/L(n+1), and similarly for the last term, we obtain an equation for mu as n->oo:

mu = 4mu^2 - 2mu^3.

The solutions to this equation are 0, 1+sqrt(2)/2, and 1-sqrt(2)/2. Since 0

Note: The full conjecture of Kimberling would follow from the conjecture A289035(n) = A171588(n-3) for n>3. (End)

Conjecture: a(n+1) - a(n) = A276864(n) for all n. - Michel Dekking, Mar 20 2022

From Michel Dekking, Mar 22 2022: (Start)

The arguments given in the proof above carry over to the positions of 1 in the sequences A288997, A289001, A289011, A289025, A289035, A289071, A289239, and A289242. It follows that for all these sequences x(n)/n -> 2 + sqrt(2), where x(n) is the position of the n-th 1 in these sequences.

Note that this does not apply to the sequence A289165 where the String-Replace map is not equal to a two-block map because the 2-block 11 occurs at position 85 in A289165. (End)