WON plate145 | World!OfNumbers [ 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)=3102112013 This means that our palindrome (with the embedded2112) is the result of summing the squares of the2002 consecutive numbers from 102 up to 2103. 1022 + 1032 + ... + 21022 + 21032 = 3102112013 Shifting the construction 15 places to the righteven 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 constantwhile 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 ! [12+...+52][62+...+162][172+...+2132]•••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 A000145 Prime Curios! Prime Puzzle Wikipedia 145 Le nombre 145
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