[ October 14, 2021 ]
Playing around with recreational examples of various types
like combinations using squares, factorials and other functions
Vinod Muleva (email)
Ex. (4!)^{2} + (5!)^{2} = 101^{2} + 69^{2} + 3^{2} + 2^{2} + 1^{2}
5! * 6! * 7! = 757^{3} + 118^{3} + 24^{3} + 10^{3} + 3^{3} + 3 * 2^{3}
6! – 5! + 4! – 3! + 2! – 1! = 619 (Prime!)
6! + 5! – 4! + 3! – 2! – 1! = 821 (Prime!)
10! – 9! – 8! – 7! – 6! – 5! – 4! – 3! – 2! – 1! = 1794^{2} – 35^{2} – 5^{2} – 1^{2}
Vinod gave also some random examples of palindromes
that are the sum of powers.
12321^{2} + 121^{2} – 117^{2} + 11^{2} + 6^{2} + 1^{2} = 151808151
10001^{2} + 11311^{2} + 10^{4} + 10^{3} = 227969722
In an effort to structure these terms I propose the following challenge.
Find for all the squares s1 and the 'next' higher square s2 (s2 > s1) so
that their sum is palindromic.
Let me start with some initial terms of s1 + s2 = Pal
1^{2} + 2^{2} = 5
2^{2} + 20^{2} = 404
3^{2} + 18^{2} = 333
4^{2} + 14^{2} = 212
5^{2} + ? = ?
6^{2} + ? = ?
7^{2} + ? = ?
...
Maybe some squares will have an accompanying square s2 far far greater than s1 !!
I wonder if it stays easy to find these s2 squares.
Hope you like my challenge. If so can you do this also for third powers ?
In the mean time enjoy also these equations from Vinod
turning the palindromes of the form 10^n–1 into a sum of powers
9 = 3^{2}
99 = 9^{2} + 3^{2} + 3^{2}
999 = 31^{2} + 5^{2} +3^{2} + 2^{2}
9999 = 99^{2} + 13^{2} + 5^{2} + 2^{2}
99999 = 316^{2} + 11^{2} + 3^{2} + 3^{2} + 2^{2}
999999 = 947^{2} + 316^{2} + 57^{2} + 9^{2} + 2^{2}
9999999 = 3162^{2} + 41^{2} + 8^{2} + 3^{2} + 1^{2}
99999999 = 9999^{2} + 141^{2} + 8^{2} + 7^{2} + 2^{2}
999999999 = 31622^{2} + 221^{2} + 16^{2} + 3^{2} + 3^{2}
9999999999 = 99999^{2} + 447^{2} + 13^{2} + 4^{2} + 2^{2}
99999999999 = 299999^{2} + 99999^{2} + 894^{2} + 26^{2} + 9^{2} + 2^{2}
999999999999 = 999999^{2} + 1414^{2} + 23^{2} + 8^{2} + 3^{2}

I guess these are the minimum number of terms (squares)
with which to express the palindromes ?
If someone can do it with even less terms (powers besides
squares allowed), please let me know.
[ December 12, 2021 ]
The ink was not yet dry or Alexandru Petrescu
improved these Vinod equations — with only three terms.
Well done!
999 = 3^{6} + 3^{5} + 3^{3}
9999 = 21^{3} + 3^{6} + 3^{2}
99999 = 54^{2} + 43^{3} + 26^{3}
999999 = 91^{3} + 90^{2} + 62^{3}
9999999 = 194^{3} + 139^{3} + 114^{2}
99999999 = 463^{3} + 364^{2} + 28^{4}
999999999 = 1242^{2} + 1206^{2} + 999^{3}
