magic series - Magic series History

Magic Series

History of magic series formulae

Read how the magic series formulae were found.

Consider a more general problem

The problem of Magic Series is equivalent to a certain partition problem:
How many partitons of S=m(mⁿ+1)/2 into m distinct positive integers each less than or equal to mⁿ are there?
This is the same as the number of partitions of s=mⁿ(m-1)/2 into up to m positive integers (not necessarily distinct) where each is less than or equal to m^n-1(m-1).

In order to get a greater amount of data, I consider a random positive integer x as limit instead of mⁿ.
Then we search for:
Number N(x,m) of partitions of S=(x+1)·m/2 into m distinct positive integers each less than or equal to x.
It's interesting that S is the integer with the maximal number of (x,m)-partitions.
(If x is even and m is odd then S=(x+1)·m/2-1/2 has as many partitions as S=(x+1)·m/2+1/2.)

The most easy cases

The case m=1 is trivial:
N(x,1) = 1

The case m=2 is easy:
If x is even: N(x,2) = (S-1)/2 = x/2
If x is odd : N(x,2) = (S-2)/2 = (x-1)/2 = x/2 -1/2

The case m=3 is more difficult:
If x is even: N(x,3) = x²/8 -x/4
If x ≡ 1 mod 4: N(x,3) = x²/8 -x/4 + 1/8
If x ≡ 3 mod 4: N(x,3) = x²/8 -x/4 + 5/8

Main assumptions

1. N(x,m) is equal to a polynomial of order (m-1) in x.
Unfortunately the polynomial is dependent on the value of x.

2. Equal polynomials occur periodically with increasing x.
The length L=L(m) of the period depends on m. We already know L(2)=2 and L(3)=4.

Suitable equation systems

For small m we can calculate a lot of data N(x,m) by partition algorithms.
The best one I know is Gerbicz's algorithm.
In order to find the coefficients of an order-(m-1) polynomial p(x) we need m equations.
I was successful with the equations p(x)=N(x,m) ; p(x+L(m))=N(x+L(m)) ; p(x+2L(m))=N(x+2L(m)); ... p(x+(m-1)L(m))=N(x+(m-1)L(m))
Where L(m) was found by trial and error.

Length of the periods

L(1) = 1
L(2) = 2
L(3) = 4 = 2·(2)
L(4) = 6 = 4·(3/2)
L(5) = 24 = 6·(4)
L(6) = 60 = 24·(5/2)
L(7) = 120 = 60·(6/3)
L(8) = 420 = 120·(7/2)
L(9) = 1680 = 420·(8/2)
L(10) = 2520 = 1680·(9/6)
L(11) = 5040 = 2520·(10/5)
L(12) = 27720 = 5040·(11/2)

The set of polynomials

For any order m we get a set of L(m) not necessarily distinct polynomials.
See text-files for various orders m where the polynomials are divided in odd and even x:
The coefficients (without substitution u) are listed in the oder c_m-1, c_m-2,... c₀:
m = 02 03 04 05 06 07 08 09

The basic polynomials

The polynomials p(L(m)) = p(2·L(m)) = p(3·L(m)) = ... have the shortest denominators of the coefficients.
I call them basic polynomials. They are shown here without substitution u:

p1(x)=1
p2(x)=1/2*x
p3(x)=1/8*x^2-1/4*x
p4(x)=1/36*x^3-1/8*x^2+1/4*x
p5(x)=23/4608*x^4-23/576*x^3+1/9*x^2-1/8*x
p6(x)=11/14400*x^5-11/1152*x^4+97/2160*x^3-13/144*x^2+7/60*x
p7(x)=841/8294400*x^6-841/460800*x^5+10727/829440*x^4-1579/34560*x^3+2429/28800*x^2-19/240*x
p8(x)=151/12700800*x^7-151/518400*x^6+3523/1209600*x^5-317/20736*x^4+221/4800*x^3-197/2400*x^2+29/280*x
p9(x)=259723/208089907200*x^8-259723/6502809600*x^7+5966137/11147673600*x^6-603523/154828800*x^5+3336977/199065600*x^4-59837/1382400*x^3+1078471/16934400*x^2-51/1120*x
p10(x)=15619/131681894400*x^9-15619/3251404800*x^8+1825279/21946982400*x^7-4001/4976640*x^6+5325293/1119744000*x^5-1538693/87091200*x^4+9224989/228614400*x^3-349/6615*x^2+2/35*x
p11(x)=34706647/3371056496640000*x^10-34706647/67421129932800*x^9+628203301/56184274944000*x^8-388740427/2809213747200*x^7+2698198091/2508226560000*x^6-1369369523/250822656000*x^5+2664722173/146313216000*x^4-560549/14288400*x^3+2654003/50803200*x^2-409/10080*x
p12(x)=655177/796675461120000*x^11-655177/13168189440000*x^10+16927/12773376000*x^9-35894891/1755758592000*x^8+162739487/804722688000*x^7-29327/21870000*x^6+36563450893/6035420160000*x^5-113626159/6096384000*x^4+139564127/3592512000*x^3-57493/1058400*x^2+67/1260*x

According to Dirk Kinnaes p13(x) can be written in the following way:

p13(x)=P(x/L), where L = 27720 and
P(k)=111653889847265554500729596226977332663543450828800000000 k^12
       - 24167508624949254220937142040471284126308106240000000 k^11
           + 2328187879504943249538089057906974828658688000000 k^10
               - 131639191818571768708422620420500537344000000 k^9
                    + 4850210223790767137895100111324385280000 k^8
                        - 122235937732065745487116042721280000 k^7
                             + 2151579691119069577040067993600 k^6
                                  - 26523081378202904684703360 k^5
                                       + 226080799815574918368 k^4
                                            - 1293553803575880 k^3
                                                  + 4702336578 k^2
                                                       - 10131 k

A general partition conjecture

Say f(i) are the terms of Cloître's sequence.
Let S, m, x be positive integers.
Say N is the number of partitions of S into m distinct positive integers each less than or equal to x.
Conjecture: N ≤ f(m)·u^m-1/(m-1)!² with u = (x-(m-1)/2)/2

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