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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. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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.
LINKS
N. J. A. Sloane, Exciting Number Sequences (video of talk), Mar 05 2021.
N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences: An illustrated guide with many unsolved problems, Guest Lecture given in Doron Zeilberger's Experimental Mathematics Math640 Class, Rutgers University, Spring Semester, Apr 28 2022: Slides; Slides (an alternative source).
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 21.
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}.
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
See A341721 for an analog where there are n voters, not n^2.
See A341319 for a variant.
See also A290323.
Sequence in context: A050035 A306901 A355477 * A368641 A046974 A177986
KEYWORD
nonn
AUTHOR
Sean Chorney, Feb 14 2021
EXTENSIONS
Entry revised by N. J. A. Sloane, Feb 26 2021.
STATUS
approved

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Last modified July 7 02:40 EDT 2024. Contains 374062 sequences. (Running on oeis4.)