PSEUDO PALINDROMES

Let b be a numeric base (such as 10).  Let S be a set of positive
integers such that
 
    I.  2(number of elements of S)+1 < b (for base b = 10, this
        implies that S has at most 4 elements).

    II. Every member of S is either the sum or difference of two
        distinct members of S (which implies that the S must contain
        at least 3 elements).
 
Then
 
    N(b, S) = b^max(S) ( (SUM, s in S, b^s) - 1 + (SUM, s in S, b^-s) )
 
is the pseudopalindrome base of a palindromic square.
 
Condition I guarantees that squaring N(b, S) will produce no carries.
Condition II guarantees that when squaring N(b, S), all columns will
be positive.
 
For example, let b = 10.  Then set S = { 3, 5, 8 } meets conditions
I and II above.  We compute
 
    N(10, S)
        = 10^8 * (10^8 + 10^5 + 10^3 - 1 + 10^-3 + 10^-5 + 10^-8)
        = 10010100n00101001
        = 10010099900101001
 
which is the pseudopalindrome base of the palindromic square
 
        100202100010002070200010001202001
 
(in the pseudopalindromic representation 10010100n00101001, note
that the 1's occur at the 3rd, 5th, and 8th positions to left and
right of the central "n" digit of value -1).
 
Similarly, the set S = { 1, 2, 4, 5 } meets conditions I and II above,
so that
 
        11010911011^2 = 121240161292161042121
 
is a palindrome.
 
However, S = { 1, 2, 3, 4, 5 } does not meet condition I, since
2(5)+1 = 11 >= b = 10.  Thus
 
      11111n11111^2 
    = 11110911111^2
    = 123452345716543254321
 
is not palindromic (the central digit carries).

S = { 1, 2, 5, 6 } meets condition I, but not Condition II, since
2 is not the sum or difference of distinct members of S.  Hence
 
        110011n110011
        1100109110011^2 = 1210240053929194500420121
 
is not palindromic (some columns in the pseudopalindromic
multiplication add up negative).



Dave Wilson (email) [ February 20, 1998 ].