A246750 a(n) is the smallest number k that can be partitioned into a set of n distinct integers {e(1), e(2), ..., e(n)} where all e(i) >= 2 and the sum of e(i)*(e(i)-1) for i = 1 to n equals k*(k-1)/2.

9, 28, 41, 65, 85, 96, 149, 176, 200, 244, 281, 332, 389, 400, 497, 565, 609, 657, 745, 833, 884, 989, 1060, 1132, 1217, 1312, 1441, 1536, 1621, 1740, 1832, 1961, 2080, 2189, 2308, 2424, 2533, 2669, 2832, 2948, 3128, 3244, 3441, 3557, 3717, 3901, 4064, 4204, 4408

Offset: 2

Author

Dean D. Ballard, Nov 20 2020

Keywords

• nonn

Comments

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.

Examples

			For n = 3, {3, 6, 19} is the set with the smallest sum that has this property. With 3 socks of one color, 6 socks of another color, and 19 socks of a third color, there is exactly a 50% chance that a random draw of two socks will produce a matching pair. (3*2 + 6*5 + 19*18) = (28*27) / 2.
n = 2, sum = 9, set = {3, 6}
n = 3, sum = 28, set = {3, 6, 19}
n = 4, sum = 41, set = {2, 3, 8, 28}
n = 5, sum = 65, set = {2, 4, 6, 8, 45}
n = 6, sum = 85, set = {2, 3, 5, 6, 10, 59}
		

Links

• Andrew Howroyd, Faster PARI Program

Crossrefs

Cf. A332105.

Programs

PARI

\\ See Links for a faster program.
a(n)={for(k=(n+1)*(n+2)/2-1, oo, my(t=k*(k-1)/2); forpart(p=k-n*(n+1)/2, if(sum(i=1, n, (p[i]+i)*(p[i]+i-1))==t, return(k)), , [n,n]))} \\ Andrew Howroyd, Nov 20 2020

Extensions

a(13)-a(50) from Andrew Howroyd, Nov 22 2020