A247185 a(0) = 0. a(n) is the number of repeating sums in the collection of all sums of two elements in [a(0), ... a(n-1)], chosen without replacement.

0, 0, 0, 2, 4, 6, 9, 11, 16, 19, 22, 26, 32, 38, 43, 50, 56, 67, 75, 81, 89, 97, 109, 119, 130, 140, 154, 166, 178, 194, 205, 220, 233, 250, 264, 283, 296, 312, 327, 345, 359, 378, 397, 415, 432, 456, 481, 504, 523, 547, 569, 591, 617, 641, 664, 689, 718, 744, 769, 797, 824, 847, 878, 910, 945

Offset: 0

Author

Derek Orr, Nov 22 2014

Keywords

• nonn

Comments

Without replacement means that a(i)+a(i) is not a valid sum to include. However, if a(i) = a(j), a(i)+a(j) is still a valid sum to include because they have different indices.

a(i)+a(j) and a(j)+a(i) are regarded as the same sum for all indices i and j.

a(n) <= A000217(n)-n.

Examples

			a(1) gives the number of repeating sums in the collection of all possible sums of two elements in [0]. There are no sums between two elements here, so a(1) = 0.
a(2) gives the number of repeating sums in the collection of all possible sums of two elements in [0,0]. There is only one sum, 0, thus there are no repeats. So a(2) = 0.
a(3) gives the number of repeating sums in the collection of all possible sums of two elements in [0,0,0]. The possible sums are 0+0, 0+0, or 0+0, thus there are two repeats. So a(3) = 2.
a(4) gives the number of repeating sums in the collection of all possible sums of two elements in [0,0,0,2]. The possible sums are 0+0, 0+0, 0+2, 0+0, 0+2, and 0+2. There are 4 repeating sums (2 extra zeros and 2 extra twos). So a(4) = 4.
		

Crossrefs

Cf. A247184.

Programs

PARI

v=[0];n=1;while(n<75,w=[];for(i=1,#v,for(j=i+1,#v,w=concat(w,v[i]+v[j])));v=concat(v,#w-#vecsort(w,,8));n++);v