[ *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 = 80846392**1**

28434 * 28435 – 1 = 80852078**9**

28435 * 28436 – 1 = 80857765**9**

28436 * 28437 – 1 = 80863453**1**

__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:

last digit of n | last digit of n*(n+1)–1 |

0 | 9 |

1 | 1 |

2 | 5 |

3 | 1 |

4 | 9 |

5 | 9 |

6 | 1 |

7 | 5 |

8 | 1 |

9 | 9 |

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