[ *December 29, 2002* ]

Palindromic sums of powers of

(at least two) consecutive numbers

The palindromic year is almost at his end.

I hope you all enjoyed it very much!

The next palindromic year will be **2112**.

Here is one way how 2002 announces **2112** !

(in short format notation)

2(102,2103,#2002)=**310**__2112__013

This means that our palindrome (with the embedded

2112) is the result of summing the squares of the

2002 consecutive numbers from 102 up to 2103.

102^{2} + 103^{2} + ... + 2102^{2} + 2103^{2} = **3102112013**

Shifting the construction 15 places to the right

even produces another palindrome !

2(117,2118,#2002)=**3168778613**

And guess what... the difference between these two palindromes

stripped off from their unlucky 13 appendices results in

**31687786_** – **31021120_** = 666666 a Double Beast.

The Number of the Beast, my loyal companion over the years,

produces other palindromes as well as f.i. in this following example

2(59,724,#666)=126696621

Lots and lots of interesting combinations can be detected.

Powers 2 (the squares) offer almost endless results

when searching for them. Higher powers become very rare (why?).

Many playful things can be discovered. We are only standing

at the very brink of the potential that will be revealed.

There are quite a lot of search routes or paths.

I hope you want to take at least one of them.

Generally stated we have the equation

m(a,z,#n)=**p** with (m > 1)

(ps. this WONplate will not consider powers 1 (unity)

and at least two consecutives must be added. Perfect powers

have already been extensively studied elsewhere.)

Look for larger power m.

My current record is only a mere 5...

5(5,6,#2)=**10901**

Look for larger starting numbers a.

Current record from Jean Claude Rosa is

2(12615243893562,12615243893563,#2)=

**318288756988131889657882813**

Look for the longest sum of consecutives n.

Current record is

2(30,2429258,#2429229)=**4778591852581958774**

Look for more friends for each value a.

E.g. integer 4 has at least 4 z-friends.

2(4,6,#3)=**77**

2(4,12,#9)=**636**

2(4,14,#11)=**1001**

2(4,17,#14)=**1771**

is there a fifth one, a sixth one, etc. ?

Look for palindromic beauty.

Some nice repdigit examples are these

2(71,78,#8)=**44444**

2(51,113,#63)=**444444**

Look for duplicate palindromes.

I.e. palindromes expressible in two or more ways.

2(9,118,#110) = **554455** = 2(331,335,#5)

or keeping the consecutives constant

while changing their powers as in

3(1,2,#2) = **9**

5(1,2,#2) = **33**

or

3(2,4,#3) = **99**

4(2,4,#3) = **353**

Look for longer consecutive strings.

And I am not saying you should begin with 1 !

[1^{2}+...+5^{2}][6^{2}+...+16^{2}][17^{2}+...+213^{2}]•••

or

[2(1,5,#5)=**55**][2(6,16,#11)=**595**][2(17,213,#197)=**3242423**]•••

Consult the database for these kind of palindromes at

Palindromic Sums of Powers of Consecutives