

A347944


Numbers that cannot be written as the sum of two (not necessarily distinct) terms in A000009.


2



0, 1, 63, 71, 75, 83, 85, 87, 89, 113, 115, 117, 120, 129, 133, 138, 139, 141, 151, 155, 156, 159, 161, 162, 163, 172, 179, 181, 182, 184, 185, 189, 190, 191, 199, 201, 205, 209, 212, 213, 215, 216, 217, 220, 221, 222, 233, 235, 236, 239, 242, 243, 245, 247, 248, 250
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OFFSET

1,3


COMMENTS

Most natural numbers are here: this sequence has natural density 1. Proof: Let S be the set of numbers can be written as the sum of two terms in A000009. For A000009(n1) < N <= A000009(n), at most n^2 numbers among [0,N] are in S (since if N = A000009(m) + A000009(k) then m,k <= n1), so #(S intersect {0,1,...,N})/#{0,1,...,N} <= n^2/(A000009(n1)+2) > 0 as N goes to infinity.
This also means that lim_{n>oo} a(n)/n = 1. Proof: Since this is a list we have a(n) >= n1, so liminf_{n>oo} a(n)/n >= 1. If limsup_{n>oo} a(n)/n > 1, there exists eps > 0 and n_1 < n_2 < ... < n_i < ... such that a(n_i)/(n_i) > 1+eps for all i, then #({0,1,...,a(n_i)}\S)/#{0,1,...,a(n_i)} = #{a(1),a(2),...,a(n_i)}/#{0,1,...,a(n_i)} = (n_i)/(a(n_i)+1) < 1/(1+eps) for all i, contradicting with the fact that this sequence has natural density 1. Hence limsup_{n>oo} a(n)/n <= 1, so lim_{n>oo} a(n)/n = 1.
Note that although 89 is in A000009, it is not the sum of two terms there.


LINKS



EXAMPLE

63 is a term since it is not the sum of two terms in A000009.
61 is not a term since 61 = 15 + 46.
73 is not a term since 73 = 27 + 46.


MATHEMATICA

Select[Range[0, 250], !ContainsAny[p, k=1; While[Max[p=PartitionsQ/@Range@k++]<#]; #Union@Most@p]&] (* Giorgos Kalogeropoulos, Sep 21 2021 *)


PROG

v(n) = my(l=leng(n), v=[]); for(i=0, l1, v=concat(v, vector(l, j, A000009(i)+A000009(j1)))); v=vecsort(v); v
list_zero(n) = setminus([0..n], Set(v(n)))


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



