question about A060900

Thu Sep 13 09:39:28 CEST 2007

sent me a copy of a message he sent to David Wilson.
 According to the Sloane Database, you entered this sequence, that I believe is a conjecture
 with x >= 0, y <= x.
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Date: Thu, 13 Sep 2007 15:29:30 +0100 (BST)
From: JEREMY GARDINER <jeremy.gardiner at btinternet.com>
Subject: Puzzle sequence
To: njas at research.att.com, seqfan at ext.jussieu.fr
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What number comes next in the sequence: 61, 691, 163, 487, 4201, ?

  Found at:  http://nces.ed.gov/nceskids/index.asp

  Scroll down for the answer ...

  Jeremy Gardiner

  9631. The sequence consists of the prime numbers which, when their digits are reversed, are perfect squares. 

  <correctness of the answer has not been checked>

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<div>What number comes next in the sequence: 61, 691, 163, 487, 4201, ?</div>  <div> </div>  <div>Found at:  <A href="http://nces.ed.gov/nceskids/index.asp">http://nces.ed.gov/nceskids/index.asp</A></div>  <div> </div>  <div>Scroll down for the answer ...</div>  <div> </div>  <div>Jeremy Gardiner</div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div> </div>  <div>9631. The sequence consists of the prime numbers which, when their digits are reversed, are perfect squares. <BR></div>  <div><correctness of the answer has not been checked></div>  <div> </div>  <div> </div> 
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I finally sat down and figured the equations that will, when applied to all
of the factors n_i of x, give the total number of odd-length figurate
representations of x. I should note that this does not account for the case
where the difference between successive terms of a representation is zero.
In other words, the equations count representations like this:

54 57 59
53 55 57
52 53 54
51 51 51 . . .
50 49 48
49 47 45
48 45 42

but not like this:

51
51
51
51
51
51
51

The equations for each of the figurate representations of x with length n
are:

x = n(n+1)/2 + n(r) = n(2n)/2 + n(r - ((n-1)/2)) = n(3n - 1)/2 + n(r -
2((n-2)/2)) = n(4n - 2)/2 + n(r - 3((n-2)/2)) = . . .

each of these reduces to the form (n^2 + 2nr + n)/2 = x. Solving this for r
gives us:

1.) r = (2x - n - n^2) / 2n

In these equations r acts as a limiting value, thus:

2.) floor(r / ((n-1)/2)) = c

where c represents the number of representations of length n.

For example the number 385 has three prime factors -- 5, 7, and 11. By
application of eqn 1 we find the following r values:

r_1 = (770 - 5 - 5^2) / 10 = 74
r_2 = (770 - 7 - 7^2) / 14 = 51
r_3 = (770 - 11 - 11^2) / 22 = 29

Then, by application of 2 we find c values:

c_1 = 74 / ((5-1)/2) = 74 / 2 = 37
c_2 = 51 / ((7-1)/2) = 51 / 3 = 17
c_3 = 29 / ((11-1)/2) = 29 / 5 = floor(5.8) = 5

Therefore, 385 has 37 + 17 + 5 = 59 representations as an odd-length sum of
a figurate number. Note that all of the compound factors of 385 (i.e. 35,
55, and 77) produce negative values for r. When this is not the case,
compound factors may be used. The next task is to figure out an equation or
method for finding all of the even-length sums of figurate numbers equal to
x, then combine these approaches to compute the sequence of all figurate
representations of a number x.

On a side note, I recently enrolled in a multivariable calculus class with
my wife (who needs it as a prereq for grad school). It's been almost ten
years since I've had a math class; I am very grateful to this list and the
OEIS for rekindling my interest in math and providing a relatively
non-threatening environment where I could learn to enjoy it. Thanks!

(Sigh.<;-)

The answer as worded inadequately constrains the sequence ordering.

Instead of
it should say
(content redacted to prevent puzzle spoilage).

And of course it should be branded with a scarlet "base" keyword.