This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
For n = 9, a(9) = 4 votes are required to win. There can be either 3 districts 3+3+3 with 2 supporters in 2 of them, 6 districts 1+1+1+2+2+2 with 3 supporters in the single-voter districts and 1 in a 2-voter district, or 7 districts 1+1+1+1+1+2+2 with supporters in 4 of the single-voter districts. For n = 17, a(17) = 6 votes are required to win. This can only be achieved with 5 districts 3+3+3+4+4 with 2 supporters in each of the 3 smaller districts.
For a(3), divisors of 3^2 are 1, 3, 9: d=1: (floor(1/2)+1)*(floor(3^2/(2*1))+1) = 1*5 = 5 d=3: (floor(3/2)+1)*(floor(3^2/(2*3))+1) = 2*2 = 4 d=9: (floor(9/2)+1)*(floor(3^2/(2*9))+1) = 5*1 = 5 Party A only needs 4 cells out of 9 to win a majority of districts. For a(6), divisors of 6^2 are 1, 2, 3, 4, 6, 9, 12, 18, 36: By symmetry we can ignore d = 9, 12, 18 and 36; d=1: (floor(1/2)+1)*(floor(6^2/(2*1))+1) = 1*19 = 19 d=2: (floor(2/2)+1)*(floor(6^2/(2*2))+1) = 2*10 = 20 d=3: (floor(3/2)+1)*(floor(6^2/(2*3))+1) = 2*7 = 14 d=4: (floor(4/2)+1)*(floor(6^2/(2*4))+1) = 3*5 = 15 d=6: (floor(6/2)+1)*(floor(6^2/(2*6))+1) = 4*4 = 16 Party A only needs 14 cells out of 36 to win a majority of districts.
a:= n->min(map(d->(iquo(d, 2)+1)*(iquo(n^2, 2*d)+1), numtheory[divisors](n^2))): seq(a(n), n=1..60); # Alois P. Heinz, Feb 09 2021
a[n_] := Table[(Floor[d/2]+1)*(Floor[n^2/(2d)]+1), {d, Divisors[n^2]}] // Min;
Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Apr 04 2023 *)
a(n)={vecmin([(floor(d/2) + 1)*(floor(n^2/(2*d)) + 1) | d<-divisors(n^2)])} \\ Andrew Howroyd, Feb 09 2021
from sympy import divisors def A341319(n): return min((d//2+1)*(e//2+1) for d,e in ((v,n**2//v) for v in divisors(n**2) if v <= n)) # Chai Wah Wu, Mar 05 2021
For a(2), divisors of 2^2 are 1, 2, 4: d=1: (floor(1/2)+1)*(floor(2^2/(2*1))+1) = 1*3 = 3 d=3: (floor(2/2)+1)*(floor(2^2/(2*2))+1) = 2*2 = 4 d=9: (floor(4/2)+1)*(floor(2^2/(2*4))+1) = 3*1 = 3 OR since n is even, ((2/2)+1)^2-1=3 Party A only needs 3 cells out of 4 to win a majority of districts. For a(6), divisors of 6^2 are 1, 2, 3, 4, 6, 9, 12, 18, 36: By symmetry we can ignore d = 9, 12, 18 and 36; d=1: (floor(1/2)+1)*(floor(6^2/(2*1))+1) = 1*19 = 19 d=2: (floor(2/2)+1)*(floor(6^2/(2*2))+1) = 2*10 = 20 d=3: (floor(3/2)+1)*(floor(6^2/(2*3))+1) = 2*7 = 14 d=4: (floor(4/2)+1)*(floor(6^2/(2*4))+1) = 3*5 = 15 d=6: (floor(6/2)+1)*(floor(6^2/(2*6))+1) = 4*4 = 16 OR since n is even, ((6/2)+1)^2-1=15 Party A only needs 14 cells out of 36 to win a majority of districts.
Table[Min[Table[(Floor[d/2]+1)*(Floor[n^2/(2*d)]+1),{d,Divisors[n^2]}],If[EvenQ[n],(n/2+1)^2-1,Infinity]],{n,60}](* Stefano Spezia, Feb 15 2021 *)
from sympy import divisors def A341578(n): c = min((d//2+1)*(n**2//(2*d)+1) for d in divisors(n**2,generator=True) if d<=n) return c if n % 2 else min(c,(n//2+1)**2-1) # Chai Wah Wu, Mar 05 2021
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