When I use the term ninedigital in these articles I always refer to a strictly zeroless pandigital (digits from 1 to 9 each appearing just once).
Fifth Page
Topic 5.13 [ July 15, 2015 ]
Finding a ninedigital as a substring in the decimal expansion
of that same ninedigital raised to a power p
Here I am looking for ninedigitals raised to a power p so that the same
ninedigital pops up as a substring in the decimal expansion of that number.
For that purpose I use UBASIC. One limitation here is that for the largest
ninedigital an overflow occurs when p is greater than 119. Someone who
is equipped with better tools can raise that exponent to higher values and,
no doubt, will certainly find much more solutions. Good hunting! P@rick.
See WONplate 195 for the pandigital version of this topic.
In both cases the expansion ends with our ninedigital !!
In fact I detected a pattern here as for all exponents +50 (starting with 1) it occurs.
1, 51, 101, 151, 201, ...
Can this phenomenon be explained mathematically ?
Yes, as Alexandru Petrescu does with an elegant proof hereunder.
Proof
By Alexandru Petrescu (PhD in applied mathematics) [ April 17, 2022 ]
Let a = 365781249. Condition is: 109 | (a51a) or 109 | a(a501). But gcd(a,109) = 1 so 109 | (a501).
Factorization: a501 = (a1)(a+1)(a4a3+a2a+1)(a4+a3+a2+a+1)(a20a15+a10a5+1)(a20+a15+a10+a5+1)
a1 = 365781248 = 28 x 7 x 17 x 12007 (1)
a+1 = 365781250 = 2 x 57 x 2341 (2)
For any number b having the units digit equal to 9 we have:
b2k = 1 (mod 10) and b2k+1 = 9 (mod 10)
So: a4a3+a2a+1 = 19+19+1 = 5 (mod 10) (3)
a20a15+a10a5+1 = 19+19+1 = 5 (mod 10) (4)
From (1)-(4) we have 29 x 59 = 109 | (a501).
Generally 109 | (a50p+1a) because a50p1 = (a50)p1 = (a501)(....)
The vertical bar "|" stands for 'divides'
Pari/gp has a simple command for producing the 6 terms of the factorization of our polynomial
(14:00) gp > factor(a^50-1)
%1 =
[ a - 1 1]
[ a + 1 1]
[ a^4 - a^3 + a^2 - a + 1 1]
[ a^4 + a^3 + a^2 + a + 1 1]
[a^20 - a^15 + a^10 - a^5 + 1 1]
[a^20 + a^15 + a^10 + a^5 + 1 1]
Statistics and curios for the ninedigital variation
In total there are 65795 ninedigitals that can be expressed as sums of two squares in 1 or more ways.
This is around 18% of the 9! possible ninedigitals. Here is the distribution list :
In total there are exactly 100 ninedigitals expressible as a sum of two squares whereby
the concatenation of its basenumbers forms a ninedigital (77 in total) or a pandigital (23 in total).
As a coincidence there are also 23 ninedigitals whereby the basenumbers A and B are the same.
They show up on the right side of the table.
One ninedigital stands out from the rest namely 317928645 because it is the
only one that has more than one solution. Beware this is a unique case!
This couple are the first two from an eightfold (*_8) solution for this ninedigital.
Note: this curio was first observed by Peter Kogel already in 2005.
317928645 =
↗ ↘
25382 + 176492
29432 + 175862
Here is the complete list of all the hundred ninedigital solutions.
What is under construction is the list regarding the pandigitals. B.S. Rangaswamy sent me already one example
to wet your appetite. See at the bottom of the table.
1
175236849 = 34952 + 127682
612859437 = 137492 + 205892
143752968 = 2 * 84782
2
179684325 = 36542 + 128972
618542937 = 136592 + 207842
145897362 = 2 * 85412
3
189237465 = 45362 + 129872
689537412 = 109742 + 238562
162973458 = 2 * 90272
4
197485632 = 53762 + 129842
756928314 = 149672 + 230852
164275938 = 2 * 90632
5
218367945 = 54692 + 137282
783691245 = 185732 + 209462
178945362 = 2 * 94592
6
231649785 = 65282 + 137492
786425193 = 159482 + 230672
183974562 = 2 * 95912
7
234169785 = 76592 + 132482
814593672 = 158942 + 237062
219367458 = 2 * 104732
8
234718965 = 87932 + 125462
817439625 = 137492 + 250682
346581792 = 2 * 131642
9
234971685 = 56792 + 142382
824691537 = 103592 + 267842
423579618 = 2 * 145532
10
237916845 = 89732 + 125462
841769325 = 170582 + 234692
453968712 = 2 * 150662
11
238176549 = 69452 + 137822
867154293 = 169532 + 240782
461593728 = 2 * 151922
12
238971465 = 76592 + 134282
874932516 = 174962 + 238502
497638152 = 2 * 157742
13
243956817 = 27842 + 153692
891746325 = 137852 + 264902
537198642 = 2 * 163892
14
249813657 = 82592 + 134762
912645873 = 164972 + 253082
571963842 = 2 * 169112
15
251649873 = 53672 + 149282
916782345 = 185072 + 239642
618534792 = 2 * 175862
16
256847193 = 49682 + 152372
921847653 = 160982 + 257432
637459218 = 2 * 178532
17
268371954 = 69752 + 148232
925781634 = 143972 + 268052
639174258 = 2 * 178772
18
283974165 = 43592 + 162782
927814653 = 157982 + 260432
654279138 = 2 * 180872
19
286753194 = 82952 + 147632
934687125 = 143702 + 269852
654713298 = 2 * 180932
20
312897645 = 89342 + 152672
943725186 = 187052 + 243692
765421938 = 2 * 195632
21
317928645 = 25382 + 176492 = 29432 + 175862
948571236 = 136802 + 275942
913524768 = 2 * 213722
22
326785149 = 49652 + 173822
972354861 = 174692 + 258302
943256178 = 2 * 217172
23
328459617 = 54362 + 172892
973416285 = 174062 + 258932
958431762 = 2 * 218912
24
341978265 = 64592 + 173282
25
345172689 = 87452 + 163922
26
346297185 = 79532 + 168242
27
362794185 = 54962 + 182372
28
365984721 = 24362 + 189752
29
368529417 = 53762 + 184292
30
369471825 = 59642 + 182732
31
374921865 = 83522 + 174692
32
378129645 = 83492 + 175622
33
385267914 = 23672 + 194852
34
413629578 = 64532 + 192872
35
435869712 = 78242 + 193562
36
436521978 = 96272 + 185432
37
478192653 = 46982 + 213572
38
481379265 = 48572 + 213962
39
489731265 = 56972 + 213842
40
497163825 = 48962 + 217532
41
497231865 = 59762 + 214832
42
514926873 = 57632 + 219482
43
523687194 = 63452 + 219872
44
523861794 = 78632 + 214952
45
532978641 = 87962 + 213452
46
592876413 = 61982 + 235472
47
598314762 = 64712 + 235892
48
613259874 = 18752 + 246932
49
613478925 = 74582 + 236192
50
614829357 = 78692 + 235142
51
619423785 = 15962 + 248372
52
631548297 = 87962 + 235412
53
635127849 = 91682 + 234752
54
639218457 = 73562 + 241892
55
648912537 = 38642 + 251792
56
651429378 = 68132 + 245972
57
674921853 = 41972 + 256382
58
682143597 = 61892 + 253742
59
694725138 = 96872 + 245132
60
712839546 = 97352 + 248612
61
726389145 = 89762 + 254132
62
726439185 = 71642 + 259832
63
729563481 = 39842 + 267152
64
758394621 = 43952 + 271862
65
789631245 = 97412 + 263582
66
812596473 = 56132 + 279482
67
817453962 = 65492 + 278312
68
823415697 = 35162 + 284792
69
825697314 = 39152 + 284672
70
839147625 = 91562 + 274832
71
846352197 = 73592 + 281462
72
865972341 = 38462 + 291752
73
872156394 = 16352 + 294872
74
914367285 = 64712 + 295382
75
931427586 = 68312 + 297452
76
934816725 = 74312 + 296582
77
951246738 = 85172 + 296432
Another ninedigital that stands out from the rest is 934167285 because it is the only
one that can be written as a sum of two squares in two different ways such that both
their basenumbers forms a ninedigital when multiplied together.
A nice unique case!
934167285 =
↗ ↘
95132 + 290462 and 9513 * 29046 = 276314598
206012 + 225782 and 20601 * 22578 = 465129378
In total there are 219 ninedigitals expressible in this way.
The smallest is 248635917 = 106142 + 116612 and 10614 * 11661 = 123769854
The largest is 987431562 = 215192 + 228992 and 21519 * 22899 = 492763581
Statistics and curios for the pandigital variation
In total there are 568801 pandigitals that can be expressed as sums of two squares in 1 or more ways.
This is around 17,416 % of the 9*9! possible pandigitals. Here is the distribution list :
One lovely pandigital already popped up on my screen 1073982645
It combines in an elegant way the two numberformats i.e. ninedigital and pandigital !
This couple are the middle two from a fourfold (*_4) solution for this pandigital.
1073982645 =
↗ ↘
89462 + 315272
198542 + 260732
Then three more pandigitals with double solutions popped up later on !
They are fully pandigital solutions throughout. Enjoy them!
7095281346 =
↗ ↘
207452 + 816392
398612 + 742052
7125094386 =
↗ ↘
312692 + 784052
497312 + 682052
7465102389 =
↗ ↘
369422 + 781052
469832 + 725102
Two pandigitals that stand out from the rest are 5921803476 and 8097452136 because
they are the only ones that can be written as a sum of two squares such that both their
basenumbers form a pandigital when concatenated as well as when multiplied together.
A nice couplet !
While tracing for long(er) chains of pandigitals I stumbled across the following unique and
unexpectedly double expression. A thing of beauty !
Note that A and B are in descending order here. I didn't find a case whereby A and B are ascending.
7983056241 = 75129 2 + 48360 2
and the concatenation of A | B gives us
7512948360 = 75306 2 + 42918 2
a similar all_pandigital expression !
Can you find longer chains of pandigital expressions using perhaps other operations than concatenation ?
More subcategories
All the pandigitals (41) equal to 2 x A2 are A =
22887, 23124, 24957, 25941, 26409, 26733, 27276, 29685, 31389, 35367,
39036, 39147, 39432, 39702, 40293, 41997, 42843, 43059, 44922, 45258,
45624, 46464, 49059, 50889, 53568, 54354, 57321, 59268, 59727, 60984,
61098, 61611, 61866, 62634, 65436, 68823, 68982, 69087, 69696, 69732,
69798.
Note that two of them (see underlined) are palindromic !
All the pandigitals (46) equal to (A)2 + (AReversed)2 are A (smallest) =
10716, 12804, 14496, 14967, 16053, 18126, 18528, 19317, 20493, 21423, 21792,
22839, 23205, 23286, 23544, 24267, 26058, 27324, 28557, 28563, 30597, 32325,
32838, 33105, 37824, 41676, 41718, 41736, 42378, 43497, 43725, 45018, 46464,
47217, 49245, 49305, 51327, 52866, 53436, 54456, 55296, 56247, 60927, 63666,
67137, 69696.
Note that the same two palindromes appear again !
The most beautiful one however seems to me the next one
8493716052 = 636662 + 666362
When the number of the beast gets involved...
There are 2187 pandigitals that can be expressed as (A)2 + (BAnagram of A)2
The reversals of above paragraph are evidently included in this total.
0001 1023849765 = 223862 + 228632 smallest
0002 1027935648 = 223082 + 230282 second
0003 1027968345 = 176282 + 267812 third
- - -
0666 3921846057 = 107162 + 617012 indexed by the number of the beast yields a reversal
1666 7914563208 = 260582 + 850622 indexed by a near number of the beast yields also a reversal
- - -
2185 9864025713 = 339722 + 933272 third last
2186 9872350146 = 154892 + 981452 penultimate
2187 9876135240 = 657422 + 745262 largest
The following is special as the five lowest and five highest digits of the pandigital are nicely separated
1082 56789_04132 = 507542 + 557042 curio
There are four pandigitals whose second half is also an anagram of A or B
0371 26904_57813 = 358172 + 375182 curio
1062 54903_21768 = 187622 + 716822 curio
1175 60294_17538 = 518372 + 578132 curio
1865 85764_92130 = 209312 + 902132 curio
These two solutions are also noteworthy
0957 5047168932 = 246662 + 666242 curio with number of the beast
1731 8140532769 = 158882 + 888152 curio not with number of the beast but same structure
Topic 5.11 [ April 27, 2014 ]
Order out of chaos using the ninedigits
Topic 5.8 [ December 30, 2010 ]
From a posting to [SeqFan]
by Eric Angelini
"As an afterthought, here's one I like (because of
the symmetry in the operations) that was appropriate
for the countdown on Friday night { can you find out the exact year? }:
10 + 9 * 8 * 7 / 6 * 5 * 4 + 321
Happy New Year! "
Topic 5.7 [ April 2008 ]
Fractions using the same digits as their decimal representation
A webpage by Christian Boyer
The above link came from a reply in the SeqFan mailing list where
Alexander R. Povolotsky's topic was about approximating Pi
using just nine- and pandigitals. Here is his *best* combination !
689725314(0) / 219546387(0) = 3.141592642...
Alexander R. Povolotsky [ October 8, 2022 ] writes
Since then the better approximations were found per
Using each number (1-9 EXACTLY ONCE) can you make 2 distinct 9 digits numbers,
so the quotient of the two numbers is as close to Pi as possible?
429751836 / 136794258 = 3.14159265369164852899... (pi + 1.01855e-10)
467895213 / 148935672 = 3.14159265350479621759... (pi - 8.49969e-11)
Pi = 3.1415926535897932384626433832795028842
I conjecture that the higher the pandigital number base then the better and better Pi approximations
can be found and that in the base where n=infinity the actual Pi number could be achieved...
Someone with good computer programming skills perhaps could check whether my conjecture is true or not.
Topic 5.6 [ March 31, 2008 ]
Blending palindromes with nine- & pandigitals using multiplication by 9
by B.S. Rangaswamy
Topic 5.4 [ October 26, 2005 ]
From Palindromic Squares to Pandigitals
264 is a very interesting number since it is the 12th basenumber of a palindromic square.
The square itself is this nice palindrome 69696.
Note the presence of the number of the Beast ! 69696 .
Did you know that when we power up 264 with two exponents and add them up
that we arrive at a pandigital number... in two different ways !
2643 + 2644 = 4875932160 2644 + 2644 = 9715064832
The second equation can be written as a palindromic expression itself using 69696squared and then doubled
2 * 69696 ^ 2
(There exists another 5-digit palindrome with this property. Can you discover it ?)
B.S. Rangaswamy [ August 11, 2006 ] was enthused by this presentation
Topic 5.1 [ September 4, 2005 ]
Generating Pandigitals from Palindromes through Fibonacci iteration
by B.S. Rangaswamy
" I got inspired by your presentation of the derivation of 68 ninedigit numbers (with all numerals from 1 to 9)
from palindromes through Fibonacci iteration. I have developed it further by arriving at a dozen 10 digit numbers
(with all numerals from 0 to 9) from 2 to 10 digit palindromes. Some of the 10 digit numbers arrived at
together with their mother palindromes are :
It is interesting to note that 6300036 leads to pandigital 1467908532, which matches identically
with each stage of iteration of 630036 to 146798532. I came across another curio as well :
630036 146798532
6300036 1467908532
9530359 123894675
95300359 1238904675
Following closely resembling palindromes lead to closely matching pandigitals :
592070295 2960351478
952070259 4760351298
This task began at my son's residence in Florida US and was completed at Bangalore India.
I was thrilled at the discovery of each of these pandigitals.
I am grateful to you for your encouragement and guidance in this venture.
B.S.Rangaswamy "
In total there are 117 palindromes that yield pandigital numbers
The smallest one is 75257 and the largest one is 4376006734 ¬
1. Typical generation of pandigitals (10 digits) from palindromes through Fibonacci iteration