A365315 Number of unordered pairs of distinct positive integers <= n that can be linearly combined using positive coefficients to obtain n.
0, 0, 0, 1, 2, 4, 5, 8, 10, 12, 15, 18, 20, 24, 28, 28, 35, 37, 42, 44, 49, 49, 60, 59, 66, 65, 79, 74, 85, 84, 93, 93, 107, 100, 120, 104, 126, 121, 142, 129, 145, 140, 160, 150, 173, 154, 189, 170, 196, 176, 208, 193, 223, 202, 238, 203, 241, 227, 267, 235
Offset: 0
Keywords
Examples
We have 19 = 4*3 + 1*7, so the pair (3,7) is counted under a(19). For the pair p = (2,3), we have 4 = 2*2 + 0*3, so p is counted under A365314(4), but it is not possible to write 4 as a positive linear combination of 2 and 3, so p is not counted under a(4). The a(3) = 1 through a(10) = 15 pairs: (1,2) (1,2) (1,2) (1,2) (1,2) (1,2) (1,2) (1,2) (1,3) (1,3) (1,3) (1,3) (1,3) (1,3) (1,3) (1,4) (1,4) (1,4) (1,4) (1,4) (1,4) (2,3) (1,5) (1,5) (1,5) (1,5) (1,5) (2,4) (1,6) (1,6) (1,6) (1,6) (2,3) (1,7) (1,7) (1,7) (2,5) (2,3) (1,8) (1,8) (3,4) (2,4) (2,3) (1,9) (2,6) (2,5) (2,3) (3,5) (2,7) (2,4) (3,6) (2,6) (4,5) (2,8) (3,4) (3,7) (4,6)
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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Mathematica
combp[n_,y_]:=With[{s=Table[{k,i},{k,y},{i,1,Floor[n/k]}]},Select[Tuples[s],Total[Times@@@#]==n&]]; Table[Length[Select[Subsets[Range[n],{2}],combp[n,#]!={}&]],{n,0,30}]
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Python
from itertools import count from sympy import divisors def A365315(n): a = set() for i in range(1,n+1): for j in count(i,i): if j >= n: break for d in divisors(n-j): if d>=i: break a.add((d,i)) return len(a) # Chai Wah Wu, Sep 13 2023
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