%I #18 Dec 15 2020 01:57:20

%S 4,13,20,53,56,92,109,120,160,200,221,268,325,389,420,497,561,616,684,

%T 725,813,901,969,1064,1132,1197,1329,1421,1516,1581,1740,1849,1904,

%U 2060,2189,2288,2444,2560,2696,2849,2985,3128,3261,3404,3564,3744,3904,4044,4204,4381,4585,4725

%N a(n) is the smallest number k that can be partitioned into a set of n distinct positive integers {e(1), e(2), ..., e(n)} where Sum_{i=1..n} e(i)*(e(i)-1) = k*(k-1)/2.

%C These numbers solve the problem of what is the required minimum number of socks of n colors such that a random drawing of two socks has a 50% chance of matching. In this version the number of socks of each color is distinct, but there may be a color with only one sock.

%e For n = 3, {1, 3, 9} is the set with the smallest sum that has this property. With 1 socks of one color, 3 socks of another color, and 9 socks of a third color, there is exactly a 50% chance that a random draw of two socks will produce a matching pair. (1*0 + 3*2 + 9*8) = (13*12) / 2.

%e n = 2, sum = 4, set = {1, 3}

%e n = 3, sum = 13, set = {1, 3, 9}

%e n = 4, sum = 20, set = {1, 2, 3, 14}

%e n = 5, sum = 53, set = {1, 2, 3, 11, 36}

%e n = 6, sum = 56, set = {1, 2, 3, 5, 6, 39}

%o (PARI) \\ See 'Faster PARI Program' link in A246750 for PartsByWeight.

%o a(n)={local(FC=Map()); for(k=1, oo, if(PartsByWeight(n, k-n*(n-1)/2, k*(k-1)/2, (i,v)->(i+v-1)*(i+v-2)), return(k))); oo} \\ _Andrew Howroyd_, Nov 30 2020

%Y Cf. other variations of the problem: A246750, A332105, A339272.

%K nonn

%O 2,1

%A _Dean D. Ballard_, Nov 29 2020

%E a(16)-a(24) from _Michael S. Branicky_, Nov 29 2020

%E a(25)-a(30) from _Andrew Howroyd_, Nov 30 2020

%E a(31)-a(53) from _Michael S. Branicky_, Dec 03 2020