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Numbers whose digits occur with same frequency ~ Classification P2
Class_P3 Class_P4 Class_P5

Concept originated dd.[ December 27, 1999 ]

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Find more smallest Squares, Cubes and higher Powers whose digits occur a same number of times ?
E.g. 10011² = 100220121 has exactly three zero's, three ones and three two's !
Note that the number of different digits in the decimal expansion may differ from their required frequency.
The next tables already give most solutions for frequencies 1 to 6 for the powers 2 to 5.
Though the numbers are definitely 'base-related' they surely display a notable beauty !

Some of the first questions I came up with are :
Are there always solutions possible for any frequency ?
What are the possible shortcuts one can make while programming this puzzle ?

Freq- uency	Square Root A052069	Square A052070 General Classification for case P2	Class
1	0	0	P2(1.1)A
2	88	7744	P2(2.2)A
3	10.011	100220121	P2(3.3)A
4	31.646.191	1001481404808481	P2(4.4)A
5	16.431.563 3.180.566.996	269996262622969 10116006416044464016 (is next one)	P2(3.5)A P2(4.5)A
6	667.567.716	445646655445456656 by Jeff Heleen - [ January 17, 2000 ]	P2(3.6)A
7	10.715.008.859	114811414848448481881 by Jeff Heleen - [ January 20, 2000 ]	P2(3.7)A
8	652.246.443.112	425425422552255452244544 by Jon E. Schoenfield - [ August 18, 2007 ]	P2(3.8)A
9	15.647.628.653.832	244848282488224248488284224 by Jon E. Schoenfield - [ August 18, 2007 ]	P2(3.9)A
10	?	?	?
11	?	?	?

Freq- uency	Cube Root A052071	Cube A052072 General Classification for case P3	Class
1	0	0	P3(1.1)A
2	11	1331	P3(2.2)A
3	888	700227072	P3(3.3)A
4	2.830.479	22676697737363992239	P3(5.4)A
5	120.023.142	1728999927211172788179288 by Jeff Heleen - [ January 21, 2000 ]	P3(5.5)A
6	6.351.783.105	256263633328535368685258882625 by Jeff Heleen - [ January 25, 2000 ]	P3(5.6)A
7	?	?	?
8	?	?	?
9	?	?	?
10	?	?	?
11	?	?	?

Freq- uency	4th Root A052093	4th Power (by Jeff Heleen) A052094 General Classification for case P4	Class
1	0	0	P4(1.1)A
2	207	1836036801	P4(5.2)A
3	130.398	289123718973983667216	P4(7.3)A
4	5.694.207	1051315345334684056886604801	P4(7.4)A
5	426.424.828	33065106952901022329359695121613056	P4(7.5)A
6	562.379.596.107	100027225332362313560971766691937509957725591601	P4(8.6)A
7	?	?	?
8	?	?	?
9	?	?	?
10	?	?	?
11	?	?	?

Freq- uency	5th Root A054212	5th Power (by Jeff Heleen) A054213 General Classification for case P5	Class
1	0	0	P5(1.1)A
2	2.955	225313610074846875	P5(9.2)A
3	49.995	312343781246875156246875	P5(8.3)A
4	10.365.589	119665765800843104737370354851986949	P5(9.4)A
5	75.418.384	2439979134100773706931016420916722663424	P5(8.5)A
6	2.592.877.410	117195225794292252449115584887987847895470100000	P5(8.6)A
7	?	?	?
8	?	?	?
9	?	?	?
10	?	?	?
11	?	?	?

Take a look at the following table and decide if you want to partake to complete it.
The task looks formidable but with joined efforts I think we can make a success.

	Frequency with which the digits occur in Pn
'd' differ. digits	1	2	3	4	5	6	7	...
1	1.1	1.2	1.3	1.4	1.5	1.6	1.7	...
2	2.1	2.2	2.3	2.4	2.5	2.6	2.7	...
3	3.1	3.2	3.3	3.4	3.5	3.6	3.7	...
4	4.1	4.2	4.3	4.4	4.5	4.6	4.7	...
5	5.1	5.2	5.3	5.4	5.5	5.6	5.7	...
6	6.1	6.2	6.3	6.4	6.5	6.6	6.7	...
7	7.1	7.2	7.3	7.4	7.5	7.6	7.7	...
8	8.1	8.2	8.3	8.4	8.5	8.6	8.7	...
9	9.1	9.2	9.3	9.4	9.5	9.6	9.7	...
10	10.1	10.2	10.3	10.4	10.5	10.6	10.7	...

What it means is best explained with an example.
Take for instance cell 3.2 of the square table P₂.
This means we want to find the smallest, the largest and the total of all squares
whose decimal expansion contains exactly 3 different digits (from the 10 available)
and these 3 digits each occur 2 times (the order is irrelevant - but leading zero's are not allowed).
Clearly the first solution for P₂(3.2) is 478 which gives square 228484.
To indicate it's the smallest solution I'll add a superscript 'A' to the notation,
a 'Z' for the largest or greatest one, and a 'T' for the total number of solutions.
The final notation for the above example becomes thus
P₂(3.2)^A = 478² = 228484

	Frequency with which the digits occur in P2
'd' differ. digits	1	2	3	4	5	6	7	...
1	4	0	0	0	0	0	0	...
2	6	1	0	0	0	0	0	...
3	13	8	12	0	1	4	2	...
4	36	38	71	102	210	?	?	...
5	66	165	992	5527	?	?	?	...
6	96	1020	20700	?	?	?	?	...
7	123	5360	?	?	?	?	?	...
8	97	24553	?	?	?	?	?	...
9	83	98442	?	?	?	?	?	...
10	87	468372	?	?	?	?	?	...

JH = Jeff Heleen
JES = Jon E. Schoenfield

The smallest solutions

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P₂(1.1)^A = 0² = 0
P₂(1. >1)^A = nihil

P₂(2.1)^A = 4² = 16
P₂(2.2)^A = 88² = 7744 = P₂(2.2)^Z
P₂(2. >2)^A = nonexistant ?

P₂(3.1)^A = 13² = 169
P₂(3.2)^A = 478² = 228484
P₂(3.3)^A = 10011² = 100220121
P₂(3.4)^A = nihil
P₂(3.5)^A = 16431563² = 269996262622969 = P₂(3.5)^Z
P₂(3.6)^A = 667567716² = 445646655445456656
P₂(3.7)^A = 10715008859² = 114811414848448481881
P₂(3.8)^A = 652246443112² = 425425422552255452244544 (JES)
P₂(3.9)^A = 15647628653832² = 244848282488224248488284224 (JES)

P₂(4.1)^A = 32² = 1024
P₂(4.2)^A = 3362² = 11303044
P₂(4.3)^A = 333303² = 111090889809
P₂(4.4)^A = 31646191² = 1001481404808481
P₂(4.5)^A = 3180566996² = 10116006416044464016
P₂(4.6)^A = 317314986546² = 100688800686688161010116
P₂(4.7)^A = 31694023812290² = 1004511145414005545155044100

P₂(5.1)^A = 113² = 12769
P₂(5.2)^A = 32151² = 1033686801
P₂(5.3)^A = 10006422² = 100128481242084
P₂(5.4)^A = 3162432161² = 10000977172927129921
P₂(5.5)^A = 1000024959196² = 1000049919014961464966416

P₂(6.1)^A = 322² = 103684
P₂(6.2)^A = 316528² = 100189974784
P₂(6.3)^A = 316327677² = 100063199236216329
P₂(6.4)^A = 316229895221² = 100001346631484638638841
P₂(6.5)^A = 316227983736319² = 100000137697937633971319669761

P₂(7.1)^A = 1017² = 1034289
P₂(7.2)^A = 3165959² = 10023296389681
P₂(7.3)^A = 10000586856² = 100011737464399964736

P₂(8.1)^A = 3206² = 10278436
P₂(8.2)^A = 31645258² = 1001422353886564
P₂(8.3)^A = 316245723114² = 100011357387896753856996

P₂(9.1)^A = 10124² = 102495376
P₂(9.2)^A = 316424617² = 100124538243596689
P₂(9.3)^A = 10000562214135² = 100011244598784733593798225

P₂(10.1)^A = 32043² = 1026753849
P₂(10.2)^A = 3164252736² = 10012495377283485696
P₂(10.3)^A = 316245509988426² = 100011222587839648934653957476

The largest solutions

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P₂(1.1)^Z = 3² = 9
P₂(1. >1)^Z = nihil

P₂(2.1)^Z = 9² = 81
P₂(2.2)^Z = 88² = 7744 = P₂(2.2)^A
P₂(2. >2)^Z = nihil ?

P₂(3.1)^Z = 31² = 961
P₂(3.2)^Z = 893² = 797449
P₂(3.3)^Z = 31086² = 966339396
P₂(3.4)^Z = nihil
P₂(3.5)^Z = 16431563² = 269996262622969 = P₂(3.5)^A
P₂(3.6)^Z = 895439100² = 801811181808810000
P₂(3.7)^Z = 15008515000² = 225255522505225000000

P₂(4.1)^Z = 99² = 9801
P₂(4.2)^Z = 9933² = 98664489
P₂(4.3)^Z = 999702² = 999404088804
P₂(4.4)^Z = 99993333² = 9998666644448889
P₂(4.5)^Z = 9999933333² = 99998666664444488889
P₂(4.6)^Z = 999999333333² = 999998666666444444888889
P₂(4.7)^Z = 99999993333333² = 9999998666666644444448888889

P₂(5.1)^Z = 311² = 96721
P₂(5.2)^Z = 99330² = 9866448900
P₂(5.3)^Z = 31620450² = 999852858202500
P₂(5.4)^Z = 9999941572² = 99998831443413831184
P₂(5.5)^Z = 3162213167485² = 9999592116615516661225225 (JH)
P₂(5.6)^Z = 999999364485939² = 999998728972281878121728711721 (JH)

P₂(6.1)^Z = 968² = 937024
P₂(6.2)^Z = 997535² = 995076076225
P₂(6.3)^Z = 999937749² = 999875501875187001
P₂(6.4)^Z = 999994157200² = 999988314434138311840000
P₂(6.5)^Z = 999999403278226² = 999998806556808076875565707076 (JH)

P₂(7.1)^Z = 3142² = 9872164
P₂(7.2)^Z = 9994302² = 99886072467204
P₂(7.3)^Z = 31620987204² = 999886831755531737616

P₂(8.1)^Z = 9916² = 98327056
P₂(8.2)^Z = 99937789² = 9987561670208521
P₂(8.3)^Z = 999944361681² = 999888726457622541145761

P₂(9.1)^Z = 30384² = 923187456
P₂(9.2)^Z = 999416681² = 998833702261055761
P₂(9.3)^Z = 31621017808182² = 999888767225363175346145124

P₂(10.1)^Z = 99066² = 9814072356
P₂(10.2)^Z = 9994363488² = 99887301530267526144
P₂(10.3)^Z = 999944387118711² = 999888777330214565264406301521

The total number of solutions

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P₂(1.1)^T = 4

0² = 0

1² = 1

2² = 4

3² = 9

P₂(1. >1)^T = 0

From Charles W. Trigg's book “Mathematical Quickies”, Dover Publications 1985, Q27 page 10 and 88.

I quote the reasoning that there exist no squares composed of only identical digits.

1. A square can end only in 0, 1, 4, 5, 6 or 9.

2. The tens' digit of a square number is even unless the units' digit is 6,

in which event it is odd.

Thus the digits cannot all be the same unless they are all 4's. (A square

consisting only of zeros is obviously meaningless.) But ...444 = 4(...111)

and since the tens' digit inside the parentheses is odd, no square integer

in the decimal system can consist of digits all alike.

P₂(2.1)^T = 6

4² = 16

5² = 25

6² = 36

7² = 49

8² = 64

9² = 81

P₂(2.2)^T = 1    a unique solution !

88² = 7744

P₂(2. >2)^T = 0    all nonexistant ?

The following source states that squares using only two distinct digits, not ending in 0,

are probably finite (no other term < 10^41) - A016069 and A018884.

The largest known such square is 81619² = 6661661161.

P₂(3.1)^T = 13

13² = 169  |  18² = 324  |  25² = 625  |  31² = 961

14² = 196  |  19² = 361  |  27² = 729

16² = 256  |  23² = 529  |  28² = 784

17² = 289  |  24² = 576  |  29² = 841

P₂(3.2)^T = 8

478² = 228484  |  836² = 698896

577² = 332929  |  880² = 774400

583² = 339889  |  881² = 776161

715² = 511225  |  893² = 797449

P₂(3.3)^T = 12

10011² = 100220121  |  11010² = 121220100  |  22002² = 484088004

10110² = 102212100  |  14499² = 210221001  |  22020² = 484880400

10401² = 108180801  |  20022² = 400880484  |  28998² = 840884004

11001² = 121022001  |  20220² = 408848400  |  31086² = 966339396

P₂(3.4)^T = 0
P₂(3.5)^T = 1    a unique Prime Curios! solution !

16431563² = 269996262622969

P₂(3.6)^T = 4

667567716² = 445646655445456656

696037692² = 484468468684686864

817851954² = 668881818661618116

895439100² = 801811181808810000

P₂(3.7)^T = 2

10715008859² = 114811414848448481881

15008515000² = 225255522505225000000

P₂(4.1)^T = 36
P₂(4.2)^T = 38
P₂(4.3)^T = 71
P₂(4.4)^T = 102
P₂(4.5)^T = 210 (JH)

P₂(5.1)^T = 66
P₂(5.2)^T = 165
P₂(5.3)^T = 992
P₂(5.4)^T = 5527 (JH)
P₂(5.5)^T = ?

P₂(6.1)^T = 96
P₂(6.2)^T = 1020
P₂(6.3)^T = 20700 (JH)

P₂(7.1)^T = 123
P₂(7.2)^T = 5360
P₂(7.3)^T = ?

P₂(8.1)^T = 97
P₂(8.2)^T = 24553
P₂(8.3)^T = ?

P₂(9.1)^T = 83
P₂(9.2)^T = 98442 (JH)

P₂(10.1)^T = 87
P₂(10.2)^T = 468372 (JH)

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While working on the data for this page I came up with the following
infinite pattern that I like to share with you :

93² = 8649
9933² = 98664489
999333² = 998666444889
99993333² = 9998666644448889
9999933333² = 99998666664444488889

Each square belongs to the general classification P₂(4.n) with n = 1, 2, 3, 4, 5, etc.

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This one was sent in by Jeff Heleen where he noticed that the root consists of only 2 digits

333303² = 111090889809
33333003² = 1111089088998009
3333330003² = 11111088908899980009

Each square belongs to the general classification P₂(4.n) with n = 3, 4, 5, etc.

33² = 1089 could belong to the pattern also if the condition of the root having 2 distinct digits is dropped.
But since I haven't a solution for P₂(4.2) of this kind, the pattern's smoothness is lost!

Another neat arrangement of digits is for this member of P₂(4.5):
3600180036² = 1296.1296.2916.1296.1296
Periods used here only to separate out the interesting stuff.

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Here is my collection of such numbers with palindromic squareroots.
0² = 0
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
33² = 1089
44² = 1936
55² = 3025
66² = 4356
88² = 7744
99² = 9801
181² = 32761
191² = 36481
232² = 53824
252² = 63504
272² = 73984
282² = 79524
292² = 85264
323² = 104329
353² = 124609
616² = 379456
626² = 391876
686² = 470596
717² = 514089
737² = 543169
757² = 573049
777² = 603729
797² = 635209
929² = 863041
1441² = 2076481
1991² = 3964081
2552² = 6512704
2882² = 8305924
2992² = 8952064
4554² = 20738916
7557² = 57108249
7777² = 60481729
10401² = 108180801
28582² = 816930724
32423² = 1051250929
35853² = 1285437609    pandigital
40104² = 1608330816
50405² = 2540664025
50705² = 2570997025
84648² = 7165283904    pandigital
97779² = 9560732841    pandigital

Jeff Heleen (email) from New Hampshire, USA.

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( © All rights reserved ) - Last modified : December 13, 2021.

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Patrick De Geest - Belgium - Short Bio - Some Pictures
E-mail address : pdg@worldofnumbers.com

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