Multimagic series for squares
See also the multimagic series for cubes

As we have seen for the smallest bimagic and the smallest trimagic squares, it may be interesting, in order to try to construct a p-multimagic square of order-n, to find all the p-multimagic series of order n, that is to say the series of n different integers, from 1 to n², having the correct magic, bimagic,... until p-multimagic sums.

The order 4 is the smallest order allowing to get bimagic series. Here are the 2 bimagic series, which are also surprising trimagic:

So that means that:

  1. 15 + 9 + 8 + 2 = 14 + 12 + 5 + 3 = 34 = S1
  2. 152 + 92 + 82 + 22 = 142 + 122 + 52 + 32 = 374 = S2
  3. 153 + 93 + 83 + 23 = 143 + 123 + 53 + 33 = 4624 = S3

For the order 5, there are 8 bimagic series for which the list is given in the Smallest bimagic square page.

Here is a summary of the number of multimagic series, some of these lists being downloadable as Excel files, from 28Kb to 800Kb each. For the order 12, the numbers of the trimagic, tetramagic and pentamagic series are kindly communicated by Walter Trump, Germany. In 2004, the numbers of Walter Trump are confirmed by Fredrik Jansson, Finland. And Fredrik goes further: he is the first to have computed the huge number of bimagic series of order 12, and the number of multimagic series of order 13 (excepted bimagic series of order 13).

New steps on bimagic series, between July and October 2005, from Germany:

In April 2008, Michael Quist, USA, computed the numbers of tetramagic and pentamagic series of order 16. And in May 2008, he computed the number of trimagic series of order 15 (and also confirmed the number of trimagic series of order 13 previously computed by Fredrik Jansson). Then in August 2008, he computed the number of bimagic series of orders 18, 19 and 20 (and also confirmed the number of bimagic series of smaller orders previously computed).

So, the order 12 is the smallest order allowing tetra and pentamagic series. In the 106 tetramagic series, 4 of them are also pentamagic. These 4 series are symmetrical left/right, that is to say that i-th number + (13-i)th number = 12² + 1 = 145.

What is the smallest order allowing hexamagic series?

The bimagic, trimagic, tetramagic and pentamagic series are referenced respectively under the numbers A052457, A052458, A090037 and A106646 in the Neil Sloane's Encyclopedia of Integer Sequences, AT&T Research.

See also the "Multimagic Series" topic in the Eric Weisstein's World of Mathematics, Wolfram Research.

In April-May 2005, Walter Trump had computed an estimation of the number of various magic series, including bimagic series from order 13 to order 20, and using Monte Carlo methods. More details at http://www.trump.de/magic-squares/magic-series/index.html. The true numbers of bimagic series, computed after his estimation (see above), are very close: its estimation method is proved excellent!

In January 2006, Robert Gerbicz, Hungary, proved that there is no tetramagic series of order 15. Here is his proof.

The magic sums are:

S4==15 mod 16. Because (2x+1)^4==1 mod 16 and (2x)^4==0 mod 16, all 15 numbers have to be odd.

Let a(k)=2b(k)+1 (where every b(k) is an integer) and T1=sum(b(k), k=1..15), T2=sum(b(k)^2, k=1..15), and so on...

By binomial theorem we can get that:

It is easy to solve this linear system of equations:

There is a contradiction. Because x==x^4 mod 2, we should have T1==T4 mod 2: but T1 is even and T4 is odd!

So there is no tetramagic (and pentamagic, hexamagic...) series for squares of order 15.

In July 2008, Jaroslaw Wroblewski, Poland, proved that there is no hexamagic series of order 17. Here is his proof.

From S4=1(mod 17) we conclude that the series must contain 17 odd terms or one odd term and 16 even terms.

***** In case of 17 odd terms, each of the form 4k+1 (where k may be negative), after denoting by Ki the sum of i-th powers of k's, we get

which leads to

giving a contradiction on parity of K1. There are no tetramagic series in this case.

***** In case of 1 odd term of the form 4k+1 (where k may be negative) and 16 even terms of the form 2p, after denoting by Pi the sum of i-th powers of p's, we get from equations on S2 and S4 respectively:

As P2 and P4 have the same parity, k must be even, say k = 2m. We get from equation on S6:

which allows us to write m=4r. We get from equation on S2 and S6:

Since P2=P6(mod 4), we can take r=4q. At this point the odd term of the magic series is equal to 128q+1, q = -2,-1,0,1,2.

From the equation on S4, which takes form

we conclude that P4 is divisible by 32 and in view of P4 = P6 (mod 8), P6 is divisible by 8.

Equation on S6 gives:

This forces q to be odd, i.e. q must be plus or minus 1.

In case q=1 we get:

In case q=-1 we get:

In either case P4 is divisible by 16, which means that all p's are odd or all are even. Hovever P6 is not divisible by 64, which rules out possibility of all even p's. Let p=4x+1 (x being integer of any sign).

From equations on S2 and S4 we get for q=1:

and for q=-1:

In both cases there is mod 4 contradiction.

And again in July 2008, Jaroslaw Wroblewski proved that there is no hexamagic series of order 19. Here is his proof.

S4(mod 16)=7, implies that each series must have 7 odd and 12 even terms, of the form 4k+1 and 2p respectively. Then equations on S2, S4, S6 give:

Last equation gives even K1 and then P2 and P4 get contradictory parity, which proves there are no hexamagic series of order 19.

Return to the home page http://www.multimagie.com