

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(n1)], chosen without replacement.


2



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
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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



