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[ December 5, 2021 ] [ Last update November 4, 2022 ]
Ascending  meta_ninedigital  and  octamanco  prime factors
in the list of palindromic squares and how to program them
Patrick De Geest

Browsing the list of the factorization of the palindromic squares
PLAIN TEXT SQUARES

I noticed that a few of them are  meta_ninedigital . What does that mean ?
Well it means that all the nine digits are present except for the zero !
And that some of these nine digits are repeated but always grouped together.
Without these repetitions we have only a pure ninedigital number that is
always divisible by 9 and thus cannot be a prime factor.
Moreover the digits are in ascending order (from 1 up to 9) !
Every binary square root or basenumber begins with 10...
Here are these ascending meta_ninedigital primes (AM9) from the list

562 10100101110100101 {BL=17} [102012042434245292542434240210201] {PL=33}
< 3^2 * 1122233456677789 >
574 10101100100110101 {BL=17} [102032223232444292444232322230201] {PL=33}
< 3^2 * 1122344455567789 >
774 1000011011101100001 {BL=19} [1000022022323444349434443232202200001] {PL=37}
< 3^2 * 111112334566788889 >
884 1011100001000011101 {BL=19} [1022323212022222449442222202123232201] {PL=37}
< 3^2 * 112344444555556789 >
1153 100010010111010010001 {BL=21} [10002002122404324432923442340422120020001] {PL=41}
< 3^2 * 11112223345667778889 >
1155 100010011010110010001 {BL=21} [10002002302242325422922452324220320020001] {PL=41}
< 3^2 * 11112223445567778889 >
1221 100110010010010011001 {BL=21} [10022014104204304602920640340240141022001] {PL=41}
< 3^2 * 11123334445556667889 >
1255 101001000111000100101 {BL=21} [10201202023422242232923224222432020210201] {PL=41}
< 3^2 * 11222333345666677789 >
1279 101100000111000001101 {BL=21} [10221210022444200234943200244422001212201] {PL=41}
< 3^2 * 11233333345666666789 >
1648 10000110100100101100001 {BL=23} [100002202014124054042292240450421410202200001] {PL=45}
< 3^2 * 1111123344455566788889 >
1686 10001011000100011010001 {BL=23} [100020221024121422422292224224121420122020001] {PL=45}
< 3^2 * 1111223444455556778889 >
1799 10100100001110000100101 {BL=23} [102012020032422224023292320422224230020210201] {PL=45}
< 3^2 * 1122233333456666677789 >
1807 10100110000100001100101 {BL=23} [102012222014120044222292222440021410222210201] {PL=45}
< 3^2 * 1122234444455555677789 >
1822 10101100000100000110101 {BL=23} [102032221212020222224292422222020212122230201] {PL=45}
< 3^2 * 1122344444455555567789 >
1830 10110000100100100001101 {BL=23} [102212102024024032042292240230420420201212201] {PL=45}
< 3^2 * 1123333344455566666789 >
2290 1000010011001001100100001 {BL=25} [1000020022102222341244029204421432222012200200001] {PL=49}
< 3^2 * 3541 * 27961 * 1122233456677789 > = < 3^2 * 111112223444555677788889 >
2300 1000010101001001010100001 {BL=25} [1000020202104032241422409042241422304012020200001] {PL=49}
< 3^2 *111112233444555667788889 >
2330 1000100001101011000010001 {BL=25} [1000200012202242203432229222343022422022100020001] {PL=49}
< 3^2 * 111122222344556777778889 >
2511 1010010001001001000100101 {BL=25} [1020120202122042041224209024221402402212020210201] {PL=49}
< 3^2 * 112223333444555666677789 >
2542 1011000001001001000001101 {BL=25} [1022121002024024023004229224003204204202001212201] {PL=49}
< 3^2 * 112333333444555666666789 >

I find the appearance of these factors (starting high up in the list)
very striking ! Multiply them by 9 and you end up with
palindromic numbers of the Binary Square Root family !

Some formal notation

1122233456677789 could be written as
1(2)2(3)3(2)4(1)5(1)6(2)7(3)8(1)9(1)

11233333345666666789 could be written as
1(2)2(1)3(6)4(1)5(1)6(6)7(1)8(1)9(1)

Generally with variables
1(A)2(B)3(C)4(D)5(E)6(F)7(G)8(H)9(I)

 Now the question ! 
Who can send me a program that prints out these AM9 meta_ninedigital
primes from smallest (length > 9) to largest (your choice how large).
Preferably I want a Pari/gp script. But other languages
are welcome too (C++, etc.).

I wonder if these curious primes are plentiful or rare.
What percentage will give a palindromic squareroot
when multiplied by 9 or even another multiplicand ?

With some trial and error I found these small
meta_ninedigital primes :  1234555556789 {L=13} E=5
and  1112234567789 {L=13} A=3, B=2, G=2...
Not sure if there are smaller ones !

Here is a not too long (!) meta_ninedigital PRP example
1(9240)2(1)3(1)4(1)5(1)6(1)7(1)8(1)9(1)


Browsing the list of the factorization of the palindromic squares
PLAIN TEXT SQUARES

I also noticed that a few of them are  octamanco . What does that mean ?
Well it means that eight digits are present except for the zero and the eight !
And that some of these eight digits are repeated but always grouped
together and can become a prime factor.
Moreover the digits are in ascending order (from 1 up to 9) !
Every binary square root or basenumber begins with 11...
Here are these ascending octamanco primes (OM8) from the list

605 11001001110010011 {BL=17} [121022025422441494144224520220121] {PL=33}
< 3^2 * 1222333456667779 >
613 11010001110001011 {BL=17} [121220124442223494322244421022121] {PL=33}
< 3^2 * 1223333456666779 >
909 1100011001001100011 {BL=19} [1210024202323442049402443232024200121] {PL=37}
< 3^2 * 122223444555677779 >
1350 110110000010000011011 {BL=21} [12124212102202202424942420220220121242121] {PL=41}
< 3^2 * 12234444445555556779 >
1359 111000010010010000111 {BL=21} [12321002222222320224942202322222220012321] {PL=41}
< 3^2 * 12333334445556666679 >
1889 11001000001110000010011 {BL=23} [121022001024422220221494122022224420100220121] {PL=45}
< 3^2 * 1222333333456666667779 >
2568 1100000001101011000000011 {BL=25} [1210000002422224201212249422121024222242000000121] {PL=49}
< 3^2 * 122222222344556777777779 >
2627 1100101000001000001010011 {BL=25} [1210222210203200204222229222224020023020122220121] {PL=49}
< 3^2 * 122233444444555555667779 >
2647 1101100000001000000011011 {BL=25} [1212421210002202200024249424200022022000121242121] {PL=49}
< 3^2 * 122344444444555555556779 >
2656 1110000001001001000000111 {BL=25} [1232100002222222221002249422001222222222000012321] {PL=49}
< 3^2 * 123333333444555666666679 >

Some formal notation

1222333456667779 could be written as
1(1)2(3)3(3)4(1)5(1)6(3)7(3)9(1)

Generally with variables
1(A)2(B)3(C)4(D)5(E)6(F)7(G)9(H)

 Now the question ! 
Who can send me a program that prints out these OM8 octamanco
primes from smallest to largest (your choice how large).
Preferably I want a Pari/gp script. But other languages
are welcome too (C++, etc.).
[ December 10, 2021 ]
Alexandru Petrescu is the first person to send me some results.
He lists the smallest prime meta_ninedigitals (AM9) of lengths 11, 12 & 13
with the condition that maximum three presences of every digit occur.

[Length] + [Meta_9digital primes]
11	12234567899
11	12344567789
11	12345678899



12	112344567899
12	112345667789
12	122333456789
12	123334556789
12	123344567789
12	123345566789
12	123345667889
12	123444566789
12	123445666789
12	123445667899
12	123445677899
12	123455678999
12	123456777889
12	123456788899
 
13	1112234567789
13	1112334567899
13	1122345567899
13	1122345677899
13	1123344566789
13	1123345556789
13	1123345677899
13	1123445667889
13	1123455667889
13	1222334567789
13	1222344556789
13	1223334567899
13	1223344567889
13	1223345667889
13	1223345678999
13	1223444567899
13	1223445566789
13	1223455678999
13	1223456667899
13	1233345677899
13	1233455667889
13	1233455677789
13	1233456778999
13	1233456788899
13	1234455566789
13	1234455677789
13	1234455677899
13	1234456677899
13	1234456777889
13	1234555666789
13	1234555667789
13	1234555678999
13	1234556667899
13	1234556788999
13	1234566777889

[ December 11, 2021 ]
Alexandru Petrescu (email)
proposes besides these ascending meta_ninedigitals
the use of descending meta_ninedigitals as well.
With the jumbled variation (see ninedig7.htm)
we have our trio AM9, DM9 & JM9.

Ascending meta_ninedigital (AM9) numbers
numbers of form 1(A)2(B)3(C)4(D)5(E)6(F)7(G)8(H)9(I)

Descending meta_ninedigital (DM9) numbers
numbers of form 9(A)8(B)7(C)6(D)5(E)4(F)3(G)2(H)1(I)

Jumbled meta_ninedigital (JM9) numbers
numbers of form a(A)b(B)c(C)d(D)e(E)f(F)g(G)h(H)i(I)
with {a,b,c,d,e,f,g,h,i} ∈ {1,2,3,4,5,6,7,8,9}

 Problem 1. 
Find the smallest AM9 (respectively DM9) primes.
Observation: the length of such a number is minimum 10,
because every ninedigital number is always divisible by 9.

(AM9) (12234567899, 12344567789, 12345678899) have length 11.

(DM9) 9876543211 The greatest ninedigital suffixed by 1. Has length 10.

 Problem 2. 
Find the smallest AM9 (respectively DM9) primes,
with property A<B<C<D<E<F<G<H<I.

(AM9) 1223334444555556666667777777888888888999999999999 has length 49.
numbers of form 1(1)2(2)3(3)4(4)5(5)6(6)7(7)8(9)9(12)

(DM9) 9887776666555554444443333333222222222111111111111111 has length 52.
numbers of form 9(1)8(2)7(3)6(4)5(5)4(6)3(7)2(9)1(15)

In the run-up to his Pari/gp program Alexandru observed that the AM9
meta_ninedigitals all end with only one 9 !
Otherwise the last two digits of 9n (a palindrome) are 91,
so the first two digits must be 19 and this is impossible.

If n = [ 1(A)2(B)3(C)4(D)5(D)6(C)7(B)8(A–1)9(1) ] then

9n = [ 10(A–1)10(B–1)10(C–1)10(D–1)10(D–1)10(C–1)10(B–1)10(A–1)1 ]
(index is related to the number of 0’s.)

Here is his proof (in a broader framework!)


Decimal representation of n using repunit formula is:

n = 9 + 10 * 8 * 10A-1 – 1
9		 + 10A * 7 * 10B – 1
9		 + 10A+B * 6 * 10C – 1
9		 + 10A+B+C * 5 * 10D – 1
9		 + 10A+B+C+D * 4 * 10D – 1
9		 + 10A+B+C+2D * 3 * 10C – 1
9		 + 10A+B+2C+2D * 2 * 10B – 1
9		 + 10A+2B+2C+2D * 10A – 1
9

We calculate 9n by iteration; summing the first two summands,
then this result with the third, and so on.

a1 = 9 *{9 + 10 * 8 * 10A-1 – 1
9		 } = 81 + 80 * (10A–1 – 1) = 1 + 8 * 10A
a2 = a1 + 7 * 10A * (10B – 1) = 1 + 10A + 7 * 10A+B
a3 = a2 + 6 * 10A+B * (10C – 1) = 1 + 10A + 10A+B + 6 * 10A+B+C
a4 = a3 + 5 * 10A+B+C * (10D – 1) = 1 + 10A + 10A+B + 10A+B+C + 5 * 10A+B+C+D
a5 = a4 + 4 * 10A+B+C+D * (10D – 1) = 1 + 10A + 10A+B + 10A+B+C + 10A+B+C+D + 4 * 10A+B+C+2D
a6 = a5 + 3 * 10A+B+C+2D * (10C – 1) = 1 + 10A + 10A+B + 10A+B+C + 10A+B+C+D + 10A+B+C+2D + 3 * 10A+B+2C+2D
a7 = a6 + 2 * 10A+B+2C+2D * (10B – 1) = 1 + 10A + 10A+B + 10A+B+C + 10A+B+C+D + 10A+B+C+2D + 10A+B+2C+2D + 2 * 10A+2B+2C+2D
a8 = a7 + 10A+2B+2C+2D * (10A – 1) = 1 + 10A + 10A+B + 10A+B+C + 10A+B+C+D + 10A+B+C+2D + 10A+B+2C+2D + 10A+2B+2C+2D + 102A+2B+2C+2D

So 9n = a8


And finally the long awaited Pari/gp code

which is surprisingly short!


s=0;n=11;for(a=2,n,for(b=1,n,for(c=1,n,for(d=1,n,p=2*(a+b+c+d)+1;\
x=vector(p);x[1]=1;x[a+1]=1;x[a+b+1]=1;x[a+b+c+1]=1;x[a+b+c+d+1]=1;\
x[a+b+c+2*d+1]=1;x[a+b+2*c+2*d+1]=1;x[a+2*b+2*c+2*d+1]=1;x[p]=1;\
y=fromdigits(x);z=y/9;t=y^2;d1=digits(t);if(isprime(z)==1,s++;\
write("c:/pari/am9pal.txt",s,",",a,",",b,",",c,",",d,",",z))))));\
printf("%.4g",100*s/(n^3*(n-1)));print("%")

or in a slightly more readable layout


default(parisizemax,"1G") ;
{
s=0;n=11;
for(a=2,n,
   for(b=1,n,
      for(c=1,n,
         for(d=1,n,
            p=2*(a+b+c+d)+1;
            x=vector(p);
            x[1]=1;
            x[a+1]=1;
            x[a+b+1]=1;
            x[a+b+c+1]=1;
            x[a+b+c+d+1]=1;
            x[a+b+c+2*d+1]=1;
            x[a+b+2*c+2*d+1]=1;
            x[a+2*b+2*c+2*d+1]=1;
            x[p]=1;
            y=fromdigits(x);
            z=y/9;
            t=y^2;
            d1=digits(t);
            if(isprime(z)==1,s++;
		write("c:/pari/am9pal.txt",s,",",a,",",b,",",c,",",d,",",z));
         )
      )
   )
);
printf("%.4g",100*s/(n^3*(n-1)));print("%");
}


A000211 Prime Curios! Prime Puzzle
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E-mail address : pdg@worldofnumbers.com