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

0,4

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.

Links

Table of n, a(n) for n=0..64.

Example

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.

Prog

(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

Crossrefs

Cf. A247184.

Sequence in context: A233776 A195526 A153196 * A237685 A220768 A077220

Adjacent sequences: A247182 A247183 A247184 * A247186 A247187 A247188

Keyword

nonn

Author

Derek Orr, Nov 22 2014

Status

approved