World!OfNumbers |
HOME plate WON | |
||
---|---|---|---|

Extraordinary squares and powers(non-palindromic allowed) | |||

triangle square penta hexa hepta octa nona |

A palindromic chain reaction starting from 1001

A palindromic chain reaction starting from 109

109 is a basenumber for the following palindromic triangular **5995**.

The sum of the squares of the digits of **5995** is :

This ** palindrome ** is also the __smallest__ square containing exactly four digits 4's ! (Sloane's A036511)

SPDS palindromes

by Felice Russo (email) [ *July 5, 1999* ]

By solving a problem in Vol. 29 of *"Journal of Recreational Mathematics"*

Felice Russo stumbled on two interesting palindromic numbers: **121** and **212**.

If we take the square of these two palindromes we obtain: **14641** and **44944**.

Both these palindromic squares are in SPDS (**Square-Partial-Digital-Subsequences**).

A number is said to be an element of the set SPDS if it is a square

which can be partitioned into two or more numbers that are also squares.

For example **14641** can be partitioned into the following squares: 1, 4, 64, 4

More interesting the palindrome **44944** that can be partitioned in five squares

that are also single digit palindromes: 4, 4, 9, 4, 4.

By a computer search these are the only palindromes in SPDS up to 3*10^8.

How many SPDS palindromes exist beyond 3*10^8 ?

A response from Alain Bex (email) [ *December 28, 1999* ]

"Infinitely many. (if 0 is a square)

every number of the form :

2 ... n zero's ... 1 ... n zero's ... 2(which is a palindrome)

has a square :

4 ... n zero's ... 4 ... n zero's ... 9 ... n zero's ... 4 ... n zero's ... 4

(which is also a palindrome).

I searched all squares from 0 to the square of 270*10^6 and did not found any others."

Fascinating square of 1003 digits

One constructs this as follows. Start with the initial square '**9**'.

Calculate the next smallest non trivial square starting with '**9**'.

This gives the square '**961**' or 31^2. (900 is excluded as the next square

must be non trivial meaning not ending in zero's)

Repeat this procedure now not with square '**9**' but with this new square '**961**'.

After a tenfold iteration we arrive at this 1003-digit square !

The root of this monstersquare is the next number of9616201620200016202000200000001889198463265350032175809769802525931907463586839 53372943202402012699567212284633760835227930246202000200000002000000000000000286 74570609183461343006847306064000000000000000000000000000000000000000000000000000 00000000000165762890459108592015259933448776693776451231520215691182485548960535 13407845648040095751097710143387381095940620098289505535125614470450939160397698 76896979086347754805195901529462026613507501648758045675165710149033232580911054 95275391508140259645504424098617924718273963483282386760738531854344564367702075 18569743231858613025709254857930447199794751022405596837264849453479919042878461 21230650258983014894678321267198660655507316299804535757670938028350277863843458 14818258133214645546046187760712527279785484496736454520815281195847787074116812 26823301300771758521612706290200804538893679740002426955408231457680947810963158 34124835975647700891238330554238156360187301492206636065480004091531578283770622 92595687694293607658268932513211592034772025

3101000100000001000000000000000143372853045917306715034236530320000000000000000 00000000000000000000000000000000000000000000001000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000010603496991036628926304444713741834555298799667027754796246140889648 28492192175159864060941684234739538923381863482793853164978801504095322725503902 27748540628752666464970620036253149337706479443351572242228953285666037295270306 10939087202964086953045

[ *October 24, 2003* ]

Jeff Heleen (email) assembled some Ubasic code

and came up with the following program and **the eleventh iteration**.

" The only odd thing I noticed was on the step where the number ended in '...6455044'10 input "Input #:";N:if N=0 then end 20 C=1 30 for X=1 to 999 40 A=N*10^X:M=isqrt(A) 50 B=N*10^X+10^X-1:K=isqrt(B) 60 if K-M>0 then print C;M+1;:else 90 70 color 4:print (M+1)^2:color 7:print 80 N=(M+1)^2:C+=1:cancel for:goto 30 90 nextThe very large (2006 digit) square is: 96162016202000162020002000000018891984632653500321758097698025259319074635868395/ 33729432024020126995672122846337608352279302462020002000000020000000000000002867/ 45706091834613430068473060640000000000000000000000000000000000000000000000000000/ 00000000001657628904591085920152599334487766937764512315202156911824855489605351/ 34078456480400957510977101433873810959406200982895055351256144704509391603976987/ 68969790863477548051959015294620266135075016487580456751657101490332325809110549/ 52753915081402596455044240986179247182739634832823867607385318543445643677020751/ 85697432318586130257092548579304471997947510224055968372648494534799190428784612/ 12306502589830148946783212671986606555073162998045357576709380283502778638434581/ 48182581332146455460461877607125272797854844967364545208152811958477870741168122/ 68233013007717585216127062902008045388936797400024269554082314576809478109631583/ 41248359756477008912383305542381563601873014922066360654800040915315782837706229/ 25956876942936076582689325132115920347720258914994974518600054221669090816529552/ 61437894236214173621326322905481649506236046822896582106849756893167828306680432/ 84252106595764741153008678797865772530987719762684436486962268943451632788553907/ 31656642177431506759136251659632668095875384355654848275886209152908202640645654/ 85309633115453107968421137508354794003355919210333172509687721211355572517254015/ 84966273404725097750293678477710394740725438561482888229734954251750517670696829/ 31197667454547117994586808831353669444718073268237891569601608146879839980577855/ 68498634230572956495503173144474822989944662967196243585206000154855148644388166/ 99332772926750077032264526806408733028864379845899337956585494518830134664037042/ 60272665189844312896602443928808140300712823333043585779414780090526790082673840/ 39874613154649303834964454713467632342245246537690796683760984228012389608036027/ 58114886616724551236161055317248176721364665155501095064349159986496584253872772/ 51376287320282709627901684380119477746357992201754685746359044847610551198174793/ 674225and its titanic root (with 1003 digits) is:98062233404099134876441622495550000000000000000000000000000000132221684961658493/ 56044245355497113216355576978978796197322817480193272005001656175142188205512607/ 86641784654818466088559784060192956988106411704239660892984201575933925927257050/ 93722895752000000000000000000000000000000000000000000000000000000000000000000000/ 00000000000000000000000000000000000000000000000000000000000000000000000000000000/ 00000000000000000000000000000000000000000000000000000000000000000000000000000000/ 00000000000000000000002162395262695446493305898867218755368290302794627516422912/ 97852058496812878294382635141749116922272185336610080590262350721905475050291580/ 25019094606974160494878769448599322323358462223322770851845953571023019388923016/ 69998725035652930764159885003020458848439834918776188362960878122551312120192964/ 14090175134199456183573744310455126367521842569362771952871094631058349558872104/ 94580220340673831425337024896747506901142043562828036442324931521033685432413885/ 2469485859733815887082314200288040945972185

on your page, my program ended that step with '...645504'.

However, when looking at the following step at the juncture where this step joins it,

the end is given correctly with '...6455044'. And your final step matched my next to

last step. So if someone can verify that I have a correct 11th step...

I couldn't go any further as UBASIC gave me an overflow error. Jeff. "

Three digits only squares

An overview of all squares with threedigit combinations is compiled at the following page :

Squares containing at most three distinct digits

Equal Sums of Like Powers

A remarkable collection of Equal Sums of Like Powers is presented by Chen Shuwen (email) from the People's Republic of China.

His website deals with Integer Solutions of the Diophantine System !

Primes whose reversal is a square

(See Sloane's On-Line Encyclopedia of Integer Sequences A007488)

Two of them have a palindromic root :

Pandigital squares

Note that the basenumber 99066 can be rotated for 180° and remains unchanged !

Squares with identical halves

Note that 36363636364 + 63636363637 = 100000000001 ... speaking of palindromes !

Justin Chan's (email) contribution [ July 11, 2008 ] |
Is a 2n-digit number with identical halves a square number? Note that such number is c(10^n+1) where 10^(n-1) <= c < 10^n. Let 10^n+1 = m*t^2, where t^2 is the largest square dividing 10^n+1. Then c >= m. In other words, there exists a 2n-digit square with identical halves if and only if 10^n+1 is non-squarefree. Example: n = 11 (the least n such that 10^n+1 is not squarefree) 10^11+1 = 100000000001 = (11^2)*23*4093*8779 = 826446281*11^2 Therefore 36363636364^2 = 13223140496_13223140496. Note that mbt + m(t-b)t = m*t^2 = 10^n+1. Hence the 10^n+1_complement of a solution's base is the base of another solution (if t-b is greater than t/sqrt(10)). e.g. 63636363637^2 = 40495867769_40495867769 (t-b = 11-4 = 7) The next n with a solution is n = 21 (t = 7) with 4 solutions (3 <= b <= 6). Other n with solutions: n = 33 (t = 11), n = 39 (t = 13), n = 55 (t = 11), n = 63 (t = 7) In general, there are families of solutions: n = 22k+11 (11|t), n = 42k+21 (7|t), n = 78k+39 (13|t), and so on. Are there any 2n-digit cubes or higher powers with identical halves? Such number requires that 10^n+1 has at most |

Concatenations of two successive integers

Squares showing up as substring of their decimal image

From Sloane's OEIS database

More digital silliness or rotating squares

by Bob Hearn (from the SEQFAN mailing list) [ *January 10, 2008* ]

This is a cute pair (but still only a pair):

rotating the result by one place.

Justin Chan's (email) contribution [ July 11, 2008 ] |
The first property: 1222222222^2 = 1493827159950617284 also occurs for 12 and 21, 122 and 221, and so on, since: 22...21^2 = (22...20)(22...22)+1 = 10(2...22)(22...22)+1 = 10(12...22^2-10...00^2)+1 Since 12...22^2 has same number of digits as 10...00^2 and begins with 1, |

Pieter Post's (email) contribution [ September 16, 2017 ] |
There is another infinite subsequence (Source OEIS sequence A134514) :
The cyclic pair (201023, 320102)*k (for k = 2 and 3) and its squares (40410256529, 102465290404)*k^2.
For example n = 6 gives lower bound k = 1581139 with lower cyclic pair:
And upperbound k = 3333333 with upper cyclic pair: |

And here's a solution in __base 23__:

Exclusionary Squares and Cubes

E.g.:

203879^{2} = 41566646641

639172^{2} = 408540845584

Here is an interesting webpage from the collection of Clif Pickover

Clif Pickover's Extreme Challenges in Mathematics and Morals

Exclusionary Squares and Cubes

Here are some corresponding __OEIS__ entries

With no repeating digits in the base numbers

A112321 Least n-digit number such that its square is exclusionary, or 0 if no such number exists. - Lekraj Beedassy

A112322 Exclusionary square associated to corresponding smallest n-digit number (A112321), or 0 if no such number exists. - Lekraj Beedassy and Klaus Brockhaus

A113951 Largest number whose n-th power is exclusionary (or 0 if no such number exists). - Lekraj Beedassy

A113952 Largest exclusionary n-th power (or 0 if no such number exists). - Lekraj Beedassy

A112993 Exclusionary cubes: cubes of the terms in A112994. - Lekraj Beedassy

A112994 Numbers whose cubes are exclusionary: numbers n such that n and n^3 have no digits in common. - Lekraj Beedassy

With repeating digits in the base numbers

A029783 Exclusionary squares: numbers n such that digits of n are not present in n^2. - Patrick De Geest

A029784 Squares such that digits of sqrt(n) are not present in n. - Patrick De Geest

A029785 Digits of n are not present in n^3. - Patrick De Geest

A029786 Cubes such that digits of cube root of n are not present in n. - Patrick De Geest

[ TOP OF PAGE]

Patrick De Geest - Belgium - Short Bio - Some Pictures

E-mail address : pdg@worldofnumbers.com