# Permutation identities similar to $(7901234568 / 9876543210) \cdot 1234567890 = 0987654312$

It is well known that $9876543210/1234567890 = 109739369/13717421 = 8.0000000729...$

Recently I posted at

http://list.seqfan.eu/pipermail/seqfan/2012-October/010235.html

my observation that exactly the same ratio also could be expressed via another pair of , particularly as $7901234568 / 987654312 = 109739369/13717421 = 8.0000000729...$

and that as a result

$(7901234568 / 987654321) * 123456789 = 987654312$

which could be re-written as

$(7901234568 / 9876543210) * 1234567890 = 0987654312$

The above re-write (with slightly artificial but still valid pre-pending right hand side of the initial arithmetic identity with leftmost $0$) makes it to be the integer arithmetic identity, where all members are some specific permutations of all decimal base digits $1,2,...,8,9,0$ (with no duplicates)

So all four numbers in above expression (two pairs yielding same ratio) are base 10 permutations.

Is there only one such integer arithmetic identity involving permutations (with no duplicates) of all decimal base digits $1,2 \dots ,8,9,0$ (for base $10$) or there are more identities of the similar type?

The semi-brute force computational method to find the answer for this question is to construct list of all unique permutations (in ascending value order) and then to generate all possible products between elements, belonging to first half of the list (with smaller values), and those, belonging to the second half of the list (with bigger values), and then to look for such products, which are equal one to another in value.

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In any base other than $2$, there is $0x * y0=x0 * 0y$ where $x$ and $y$ are different permutations of the non-zero digits. This doesn't feel like what you are looking for. –  Ross Millikan Oct 16 '12 at 22:55
Yes I am looking for the cases, where all permutations involved are fully (more than being different by the position of 0 like in your case) unique. –  Alex Oct 18 '12 at 13:20

R. J. Cano (in private email communications with me) expressed his intuitive prediction that in the base 10 more pairs (he generously named them PovolotskyPairs) of such permutations could be found, that is such pairs of distinct digits permutations, which yield the very same ratio 109739369/13717421 = 8.0000000729... .
Also R. J. Cano expressed his prediction that permutation pairs with similar ratio properties (but not with same ratio value as in the base 10) also could be found for other bases.
Later R. J. Cano informed me that after spending one day he (using 2 PARI programs, which he wrote) produced the following results for such pairs of distinct digits permutations found for all bases "n" in the range from 2 to 10 (outputted in decimal, that is in the base 10 format).
The program searched for the maximum number of pairs with the smallest, less than "n", but wherever is possible greater than 1 ratio.

Here are results of his program augmented with my analysis:

? assumedBase=2; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {1,1}; 2) {2,2};

Ratio 1

Apparently in this case (base 2) there are no pairs with the ratio greater than 1.

In the base 2 notation
1) {01, 01}; 2) {10, 10};

Sorted in increasing order (and divided by d=1) Multiplication factors (to get base 2 PovolotskyPairs from the ratio):

k(m) = {1, 2}

? ? assumedBase=3; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {7,5}; 2) {21,15};

Ratio 7/5 = 1.4

In the base 3 notation
1) {021,012}; 2) {210,120};

Sorted in increasing order (and divided by d=1) Multiplication factors (to get base 3 PovolotskyPairs from the ratio):

k(m) = {1, 3}

? ? assumedBase=4; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {114,54}; 2) {57,27}; 3) {228,108};

Ratio 19/9 = 2.111...

In the base 4 notation
1) {1302, 0312}; 2) {0321,0123}; 3) {3210,1230};

Sorted in increasing order (and divided by d=3) Multiplication factors (to get base 4 PovolotskyPairs from the ratio):

k(m) = {1, 2, 4}

? ? assumedBase=5; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {1758,582}; 2) {586,194}; 3) {2930,970};

Ratio 293/97=3.0206...

In the base 5 notation
1) {24013, 04312}; 2) {04321,01234}; 3) {43210,12340)

Sorted in increasing order (and divided by d=2) Multiplication factors (to get base 5 PovolotskyPairs from the ratio):

k(m) = {1, 3, 5}

which are described by the formula

k(m) = 2*m -1 for m=1,2,3

? ? assumedBase=6; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {29860,7460}; 2) {22395,5595}; 3) {14930,3730}; 4) {7465,1865}; 5) {44790,11190};

Ratio 1493/373=4.00268...

In the base 6 notation
1) {350124,054312}; 2) {251403, 041523}; 3) {153042,025134}; 4) {054321,012345}; 5) {543210,123450};

Sorted in increasing order (and divided by d=5) Multiplication factors (to get base 6 PovolotskyPairs from the ratio):

k(m) = {1, 2, 3, 4, 6}

? ? assumedBase=7; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {571905,114375}; 2) {114381,22875}; 3) {800667,160125};

Ratio 38127/7625=5.000262...

In the base 7 notation
1) {4601235,0654312}; 2) {0654321,0123456}; 3) {6543210,1234560}

Sorted in increasing order (and divided by d=3) Multiplication factors (to get base 7 PovolotskyPairs from the ratio):

k(m) = {1, 5, 7)

? assumedBase=8; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {12326118,2054346}; 2) {10271765,1711955}; 3) {8217412,1369564}; 4) {6163059,1027173}; 5) {4108706,684782}; 6) {2054353,342391}; 7) {16434824,2739128};

Ratio 293479/48913 = 6.000020...

In the base 8 notation
1) {57012346, 07654312}; 2) {47136025, 06417523}; 3){37261504,05162734};4) {27405163,03726145}; 5) {17530642,02471356}; 6) {07654321,01234567}; 7) {76543210,12345670};

Sorted in increasing order (and divided by d=7) Multiplication factors (to get base 8 PovolotskyPairs from the ratio):

k(m) = {1, 2, 3, 4, 5, 6, 8}

? assumedBase=9; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {296618812,42374108}; 2) {211870580,30267220}; 3) {127122348,18160332}; 4) {42374116,6053444}; 5) {381367044,54480996};

Ratio = 10593529/1513361=7.00000132...

In the base 9 notation
1) {680123457,087654312}; 2) {482603715, 062851734}; 3) {285174063,037148256}; 4) {087654321,012345678}; 5) {876543210,123456780};

Sorted in increasing order (and divided by d=4) Multiplication factors (to get base 9 PovolotskyPairs from the ratio):

k(m) = {1, 3, 5, 7, 9}

which are described by the formula

k(m) = 2*m - 1 for m=1,2,3,4,5

? assumedBase=10; ? reportPovolotskyPairs() Loading permutations... Performing calculations... Reporting the pairs: 1) {7901234568,0987654312}; 2) {6913580247,0864197523}; 3) {4938271605,0617283945}; 4) {3950617284,0493827156}; 5) {1975308642,0246913578}; 6) {0987654321,0123456789}; 7) {9876543210,1234567890};

Ratio = 109739369/13717421 = 8.0000000729...

Sorted in increasing order (and divided by d=9) Multiplication factors (to get base 10 PovolotskyPairs from the ratio):

k(m) = { 2, 4, 5, 7, 8, 10}

which are described by the formula

k(m) = (3 + (-1)^m + 6*m)/4 for m=1,2,3,4,5,6

Analyzing above R. J. Cano results - it is interesting to observe the seven facts:

1) The number of "PovolotskyPairs" is:

{2,2,3,3,5,3,7,5,7,...} for n>=1 where n=r-1 where r is the base radix

Judging by above sequence it appears that the number of "PovolotskyPairs" is related to phi, which is the Euler totient function - see A039649, A039650, A214288 in the OEIS.

2) The ratio increases in its value as base grows

3) Starting from base 3 and going to base 10, the ratio is getting closer to exact integer value

4) The sequence of gcd dividers d for covered by calculations range of bases

r = {2, 3, 4, 5, 6, 7, 8, 9, 10}

is d = {1, 1, 3, 2, 5, 3, 7, 4, 9}

which is described by the formula

d(r) = -((-3 + (-1)^(r-1))*(r-1))/4 for r >= 2

which gives d(r) sequence continuation as:

d = {1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, 17, 9, 19, 10, 21, 11, 23, 12, ...}

which is A026741 in OEIS

5) I sent (over email) to OEIS superseeker the sequence consisting of numerators of obtained ratios and received following reply:

Report on [ 1,7,19,293,1493,38127,293479,10593529,109739369]:

    SUCCESS


Transformation T006 gave a match with the following sequences (up to a limit of 10):

%I A051847

%S A051847 1,19,1493,293479,109739369,66987982331,60693710471869,

%T A051847 76519827268721103,128138108936443028945,275176672984400058317539,

%U A051847 737345594135016860806925221,2411620538399461719230688945719

%N A051847 Bisection of A051846, divided by the term position.

%F A051847 a(n) = A051846(2n-1)/(2n-1)

%A A051847 Antti Karttunen Dec 13 1999

List of transformations used:

T006 elements of odd index in the sequence

More detailed analysis revealed that the sequence of ratio's numerators is described as:

a(n) = -4*((n-1)*(n+1)^(n+1)+1)/(((-1)^n-3)*n^3) for n=1 ... infinity

it could also be expressed as:

a(n) = -4*A051846(n)/((-3 + (-1)^n)*n) for n=1 ... infinity

First 30 terms are:

{1, 7, 19, 293, 1493, 38127, 293479, 10593529, 109739369, 5135610071, 66987982331, 3856048810781, 60693710471869, 4149140360751583, 76519827268721103, 6058888636862818097, 128138108936443028945, 11533996620790579909159, 275176672984400058317539, 27752288337333416976560341, 737345594135016860806925221, 82360974454696873163308732367, 2411620538399461719230688945719, 295545481092344912381290837570473, 9455799675198193629383691706982713, 1261454506836133847144542645138683127, 43783142861585517778127873524741134539, 6316283357812746048940357306645174509389, 236375074773470498995448767887162245274509, 36667114799331395762941381321682972438910911}

This sequence is submitted into OEIS as A221740

6) Also I sent (over email) to OEIS superseeker the sequence consisting of denominators of obtained ratios and received following reply:

Report on [ 1,5,9,97,373,7625,48913,1513361,13717421]:

    SUCCESS


Transformation T006 gave a match with the following sequences :

%I A051848

%S A051848 1,9,373,48913,13717421,6698798233,5057809205989,5465701947765793,

%T A051848 8008631808527689309,15287592943577781017641,

%U A051848 36867279706750843040346261,109619115381793714510485861169

%N A051848 Bisection of A023811, divided by the term position.

%F A051848 a(n) = A023811[2n-1]/(2n-1)

%Y A051848 Cf. A051847.

%A A051848 Antti Karttunen Dec 13 1999

More detailed analysis revealed that the sequence of ratio's denominators is described as:

a(n) = -4*(((n+1)^(n+1)-(n+1))/((n+1)-1)^2-1)/((-3+(-1)^n)*n) for n=1 ... infinity

it could also be expressed as:

a(n) = -4*A023811(n+1)/((-3 + (-1)^n)*n) for n=1 ... infinity

First 30 terms are:

{1, 5, 9, 97, 373, 7625, 48913, 1513361, 13717421, 570623341, 6698798233, 350549891889, 5057809205989, 319164643134737, 5465701947765793, 403925909124187873, 8008631808527689309, 678470389458269406421, 15287592943577781017641, 1460646754596495630345281, 36867279706750843040346261, 3921951164509374912538511065, 109619115381793714510485861169, 12849803525754126625273514676977, 393991653133258067890987154457613, 50458180273445353885781705805547325, 1683967033137904529927995135566966713, 233936420659731335145939159505376833681, 8441966956195374964123170281684365902661, 1264383268942461922860047631782171463410721}

This sequence is submitted into OEIS as A221741

7) As a result the first 30 ratios are:

{1, 7/5, 19/9, 293/97, 1493/373, 38127/7625, 293479/48913, 10593529/1513361, 109739369/13717421, 5135610071/570623341, 66987982331/6698798233, 3856048810781/350549891889, 60693710471869/5057809205989, 4149140360751583/319164643134737, 76519827268721103/5465701947765793, 6058888636862818097/403925909124187873, 128138108936443028945/8008631808527689309, 11533996620790579909159/678470389458269406421, 275176672984400058317539/15287592943577781017641, 27752288337333416976560341/1460646754596495630345281, 737345594135016860806925221/36867279706750843040346261, 82360974454696873163308732367/3921951164509374912538511065, 2411620538399461719230688945719/109619115381793714510485861169, 295545481092344912381290837570473/12849803525754126625273514676977, 9455799675198193629383691706982713/393991653133258067890987154457613, 1261454506836133847144542645138683127/50458180273445353885781705805547325, 43783142861585517778127873524741134539/1683967033137904529927995135566966713, 6316283357812746048940357306645174509389/233936420659731335145939159505376833681, 236375074773470498995448767887162245274509/8441966956195374964123170281684365902661, 36667114799331395762941381321682972438910911/1264383268942461922860047631782171463410721}

The 30th ratio in decimals is:

29.00000000000000000000000000000000000000000158...

What is left undone is analytical proof of my following conjecture:

Every complete set of all distinct unique possible digits permutations (with no duplicates) in any numerical base (radix) r contains some prime number of permutations pairs (PovolotskyPairs), which have the same specific and unique ratio value, defined as:

(n^2 (n+1)^n-(n+1)^n+1) / (-n^2+n (n+1)^n+(n+1)^n-n-1)

for n=1 ... infinity where n = r - 1

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Although both of the observations are intuitive, they are certainly very interesting! Is there any chance that you could put the code on pastebin for us to look at? –  Eric Stucky Jan 20 '13 at 23:25
R. J. Cano merged those two mentioned above PARI/GP programs into one and posted it on oeis.org/w/images/b/b8/PovolotskyPairs.gp.txt –  Alex Jan 22 '13 at 21:01
I also submitted A212958 to OEIS –  Alex Feb 21 '13 at 20:11
In base 4, you show 3 pairs that give the ratio $19/9$. But aren't there 6 pairs that give the ratio 4? In base 4, $1230/0123,1320/0132,2130/0213,2310/0231,3120/0312,3210/0321$ all give 4. Similarly, in, say, base 10, you get 362880 pairs that give ratio 10. –  Gerry Myerson Jul 15 at 7:07
@Gerry Myerson - The program searched for the pairs with the smallest (but wherever is possible greater than 1 ) ratio. I added this condition in the text of the answer. Thanks! –  Alex Jul 15 at 12:49