[ *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