n such that there is one sequence of length n having equal sum and product

[Christian G.Bower]

Here is some of the original correspondence on that subject:

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Date: Mon, 20 Apr 1998 12:21:28 -0400
From: wilson at cabletron.com ("David W. Wilson")
Subject: Equal sum-product sequences
X-Sequence: 44

For n = 1 to 10000, I computed the number f(n) of nondecreasing sequences of
n positive integers whose sum and product are equal (I have already sent the
sequence to Sloane).  I found

       n    f(n)    Sequences of length n with equal sum and product

       1     inf    (k) for any k >= 1
       2       1    (2,2)
       3       1    (1,2,3)
       4       1    (1,1,2,4)
       5       3    (1,1,1,2,5) (1,1,1,3,3) (1,1,2,2,2)
       6       1    (1,1,1,1,2,6)
       7       2    (1,1,1,1,1,2,7) (1,1,1,1,1,3,4)
       8       2    (1,1,1,1,1,1,2,8) (1,1,1,1,1,2,2,3)
       9       2    (1,1,1,1,1,1,1,2,9) (1,1,1,1,1,1,1,3,5)
      10       2    (1,1,1,1,1,1,1,1,2,10) (1,1,1,1,1,1,1,1,4,4)

Clearly, for n >= 2, f(n) >= 1, as the seqence (1x(n-2),2,n) has length n and
equal sum and product 2n.

There seem to be only a few n for which f(n) = 1, that is, for which
(1x(n-2),2,n) is the unique sequence length n having equal sum and product
2n.  Up to 10000, these n are 2, 3, 4, 6, 24, 114, 174, and 444.  There is no
easy way to identify these n, is there?

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Date: Wed, 22 Apr 1998 15:10:23
From: bowerc at usa.net ("Christian G.Bower")
Subject: [seqfan] Re: Equal sum-product sequences
X-Sequence: 45

Dave Wilson computed the number of nondecreasing sequences of n
positive integers whose sum and product are equal for n=1 to 10000.

I computed it up to 100000.

> There seem to be only a few n for which f(n) = 1, that is, for which
> (1x(n-2),2,n) is the unique sequence length n having equal sum and product
> 2n.  Up to 10000, these n are 2, 3, 4, 6, 24, 114, 174, and 444.  There is
no
> easy way to identify these n, is there?

Up to 100000, I didn't find any others. I considered the posibility that
lim (n -> inf) f(n)=inf.

I catalogued the least and greatest values of f(n) within ranges.

Here are my results:

      Range        min        max 
      2 -   1000 : 1(444)   : 22(793)
   1001 -   2000 : 2(1102)  : 28(1261)
   2001 -   3000 : 2(2820)  : 35(2161)
   3001 -   4000 : 3(3954)  : 36(3421)
   4001 -   5000 : 2(4354)  : 37(4001)
   5001 -   6000 : 2(5274)  : 41(5441)
   6001 -   7000 : 2(6324)  : 42(6481)
   7001 -   8000 : 3(7220)  : 45(7561)
   8001 -   9000 : 4(8946)  : 47(8033)
   9001 -  10000 : 4(9804)  : 51(9361)

  10001 -  20000 : 3(11874) : 61(18113)
  20001 -  30000 : 2(27744) : 71(27721)
  30001 -  40000 : 5(39294) : 76(39313)
  40001 -  50000 : 4(47644) : 74(40321)
  50001 -  60000 : 5(50196) : 77(54601)
  60001 -  70000 : 5(68112) : 85(65521)
  70001 -  80000 : 7(78819) : 83(71605)
  80001 -  90000 : 7(89400) : 92(80581)
  90001 - 100000 : 5(92382) : 95(95761)

The numbers in paretheses are the values of n at which the min or
max value of f(n) was attained. In the case of a tie, the greatest
value was chosen for the min and the least value was chosen for the
max.

The results seem to support the conjecture, but it is still VERY
speculative. Obviously the mins increase very slowly if at all.

All of the n's for which f(n) is a local max are odd.
All but one of the min is even, the sole exception being 78819.

It is easy to show that for odd n>3 f(n)>=2.

Besides (1,1,...,1,2,n) there is (1,1,...,1,3,(n+1)/2).

I also thought it might be interesting to evaluate:

the least k such that f(n)=k:

That sequence begins:

2 7 5 13 25 37 41 89 61 85...

and is not strictly increasing. All the values except the first seem
to be odd.

Christian

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And no, I haven't worked on it since.