[ September 6, 2002 ]
1|9|9|1 a vertical palindrome
Enoch Haga (email)

Enoch noticed something interesting. Take for example
the following sequence of two consecutive numbers n (n > 3).

8 * 9 – 1 = 71
9 * 10 – 1 = 89
10 * 11 – 1 = 109
11 * 12 – 1 = 131

Look at the vertical palindrome formed
by the last digits of the four primes 1 9 9 1

This happens invariably at n = 8, 53, 548, 1758,
5268, 13078, 14913, 15958, 28433, 37563, 39498, 50863,
57463, 72918, 74808, 76063, 82763, 91618, 101978, 103113,
117543, and maybe forever ?

Another example :

28433 * 28434 – 1 = 808463921
28434 * 28435 – 1 = 808520789
28435 * 28436 – 1 = 808577659
28436 * 28437 – 1 = 808634531

Questions
1. Is this 1991-pattern finite or infinite
for such a sequence of four consecutive n's ?
2. Is it possible to find another vertical palindrome
consisting of 5 or more digits ?

[ September 14, 2002 ]
Jean Claude Rosa (email) writes

While examining the enddigits of n and n*(n+1)–1
one gets the following table:

so the only possible vertical palindrome is 1991 since
all other palindromic constructions contain the digit 5
(151, 91519, ...) and thus n*(n+1)–1 cannot be prime.

See my webpage sumpower.htm
for the search of Palindromic Quasi_Under_Squares.
n*(n+1)–1 is also of the form m+(m+1)^2