A365314 Number of unordered pairs of distinct positive integers <= n that can be linearly combined using nonnegative coefficients to obtain n.
0, 0, 1, 3, 6, 8, 14, 14, 23, 24, 33, 28, 52, 36, 55, 58, 73, 53, 95, 62, 110, 94, 105, 81, 165, 105, 133, 132, 176, 112, 225, 123, 210, 174, 192, 186, 306, 157, 223, 218, 328, 180, 354, 192, 324, 315, 288, 216, 474, 260, 383, 311, 404, 254, 491, 338, 511, 360
Offset: 0
Keywords
Examples
We have 19 = 4*3 + 1*7, so the pair (3,7) is counted under a(19). The a(2) = 1 through a(7) = 14 pairs: (1,2) (1,2) (1,2) (1,2) (1,2) (1,2) (1,3) (1,3) (1,3) (1,3) (1,3) (2,3) (1,4) (1,4) (1,4) (1,4) (2,3) (1,5) (1,5) (1,5) (2,4) (2,3) (1,6) (1,6) (3,4) (2,5) (2,3) (1,7) (3,5) (2,4) (2,3) (4,5) (2,5) (2,5) (2,6) (2,7) (3,4) (3,4) (3,5) (3,7) (3,6) (4,7) (4,6) (5,7) (5,6) (6,7)
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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Mathematica
combs[n_,y_]:=With[{s=Table[{k,i},{k,y},{i,0,Floor[n/k]}]},Select[Tuples[s],Total[Times@@@#]==n&]]; Table[Length[Select[Subsets[Range[n],{2}], combs[n,#]!={}&]],{n,0,30}]
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Python
from itertools import count from sympy import divisors def A365314(n): a = set() for i in range(1,n+1): if not n%i: a.update(tuple(sorted((i,j))) for j in range(1,n+1) if j!=i) else: for j in count(0,i): if j > n: break k = n-j for d in divisors(k): if d>=i: break a.add((d,i)) return len(a) # Chai Wah Wu, Sep 12 2023
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