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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.

A341578 a(n) is the minimum number of total votes needed for one party to win if there are n^2 voters divided into equal districts.

Original entry on oeis.org

1, 3, 4, 8, 9, 14, 16, 24, 25, 33, 36, 45, 49, 60, 64, 80, 81, 95, 100, 117, 121, 138, 144, 165, 169, 189, 196, 224, 225, 247, 256, 288, 289, 315, 324, 350, 361, 390, 400, 429, 441, 473, 484, 528, 529, 564, 576, 624, 625, 663, 676, 728, 729, 770, 784, 825, 841, 885, 900, 943
Offset: 1

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Author

Sean Chorney, Feb 14 2021

Keywords

Comments

Comments from Jack W Grahl and Andrew Weimholt, Feb 26 2021: (Start):
This is a two-party election. The size d of each district must divide n^2, so there are n^2/d equal districts.
The districts are winner-takes-all, and tied districts go to neither candidate. For an even number of districts, it is enough to win half the districts and tie in one further district.
So for 5 districts of 5 votes each, one party could win with 3 votes in each of 3 districts, and 0 in all other districts, for a total of a(5) = 9 votes.
For 8 districts of size 8, 5 votes in each of 4 districts and 4 votes in a fifth district are enough, for a total of a(8) = 24 votes.
d need not equal n. For n=6, it is better to gerrymander the 36 votes into 3 districts with 12 votes each, and then a(6) = 14 = 7+7+0 votes are enough to win. (End)
This is related to the gerrymandering question. What is the asymptotic behavior of a(n)? - N. J. A. Sloane, Feb 20 2021. Answer from Don Reble, Feb 26 2020: The lower bound is [(n^2+1)/4 + n/2]; the upper bound is [n^2/4 + n]. Each bound is reached infinitely often. In general the best choice for d is not unique, since d and n/d give the same answer.

Examples

			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.
		

Crossrefs

See A341721 for an analog where there are n voters, not n^2.
See A341319 for a variant.
See also A290323.

Programs

  • Mathematica
    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 *)
  • Python
    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

Formula

a(n) is the minimum value of {(floor(d/2)+1)*(floor(n^2/(2*d))+1) over all divisors d of n^2 AND (n/2+1)^2-1, if n is even}.

Extensions

Entry revised by N. J. A. Sloane, Feb 26 2021.