More Palindromic Products of Integer Sequences

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More Palindromic Products of Integer Sequences
Reversal Products

Introduction

Palindromic numbers are numbers which read the same from

left to right (forwards) as from the right to left (backwards)

Here are a few random examples : 525, 3883, 946101649

Palindromic Products of Integers Sequences are defined and calculated by this extraordinary intricate and excruciatingly complex formula. So, this line is for experts only :

(S1) x (S2) x (S3) x ...

As you can see from a quick glance at this page, many sequences just took a fresh start.
So, there is your chance to contribute. You may add to existing topics or you may even send
in totally new topics. It's up to you!

Allow me to show you one of my most favourite palindromic numbers 323323
Why this number ?
Well, mainly because it can be expressed as the result of some remarkable and unrelated operations.
Please follow WONplate 52 where more items are added as time progresses.

323323	7 x 11 x 13 x 17 x 19
	1⁷ + 2² + 3⁸ + 4⁹ + 5⁵ + 6⁶ + 7¹ + 8⁴ + 9³
	323323 is a substring of its base 4 representation 1032{323323}.
	The string 323323 was found at position 985589 counting from the first digit after the decimal point of . [ Pi-Search Page ]

The first is as the product of Five consecutive primes !
The second as a sum of the nine digits 1 through 9 [ 123456789 ] each raised to a power.
The different exponents thereby forming a combination of these same nine digits ! [ 728956143 ]
The third and the fourth are self-explanatory...
Furthermore by adding the squares of the digits of my palindrome we find :

3² + 2² + 3² + 3² + 2² + 3² = 44
The left or right half of my palindrome is also a palindrome : 323

The product of these Two consecutive palindromes

11 x 22 equals 242.
242 also is the sum of four consecutive integers namely

59 + 60 + 61 + 62
and is also the sum of four consecutive palindromes :

44 + 55 + 66 + 77

Enjoyable palindromic outcomes with Threefold sequence (n) x (n+2) x (n+4)

35 x 37 x 39 = 50505
35 + 37 + 39 = 111

Here's a bonus operation : 50505 / 111 = 455

Find two numbers which differ by 22 so that when each number x is multiplied with (x+1)
they form a palindromic number pair.
Answer by Salil Palkar & Nishant Redkar, Mumbai, India:
Source : http://www.mindsport.org/archives/03 01 01.htm (dead link now)

Regarding the problem of the palindromic number pair, the
two numbers are 5291 and 5313. This is how me and my
friend cracked it: To have x(x+1) as a palindrome number x
must have 1, 2, 3, 6, 7 or 8 in the units place. Since the two
numbers differ by 22, units place digit should differ by 2 — ie,
it can be 3 and 1 in the unit places of the two numbers or 8
and 6. We find for x(x+1) we have 2 in the units place for
numbers ending with 3, 1, 6, 8. Now a palindrome has 2 in
the units place if it starts with 2. Also since the first digit of
the palindrome is 2 then first digit of x has to be 1, 4 or 5. On
trying further there are 32 pairs possible but on trying on the
calculator we were left with only 16 pairs for 4 or 3 digit
numbers. On analysing further we were left with only one
pair and it is 5291 and 5313. For x = 5291, x(x+1) =
27999972 and for x = 5313, x(x+1) = 28233282.

1051 & 1061

a curious pair of consecutive primes

Have you noticed the following with Two consecutive primes

1051 x 1061 = 1115111
1051 + 1061 = 2112
1051 ² + 1061 ² = 2230322

All three operations yield a palindromic number !

Note that we can easily transform these numbers into palindromic primes by simply inserting a zero.
1051 and 1061 becomes 10501 and 10601 respectively.
Both these numbers can be expressed as the sum of three consecutive primes :

10501 = 3491 + 3499 + 3511

10601 = 3529 + 3533 + 3539

Inserting one more zero makes another interesting product, Hugo Sánchez reported
from Caracas Venezuela [March 17, 1999 ].

100501 x 100601 = 10110501101 ( ps.: 100601 is composite.)
Hugo played around with the primes 1051 and 1061 and found more palindromes by inserting various operation signs.

(1051) + (10x61) = 1661

(10^5+1) x (10^6+1) = 100001100001

The two consecutive primes 1051 and 1061 share the following property (Sloane A052033).
All the base b representations (b < 10) with expansions interpreted as decimal numbers are composite.

1051₁₀
1387₉ = 19 * 73
2033₈ = 19 * 107
3031₇ = 7 * 433
4511₆ = 13 * 347
13201₅ = 43 * 307
100123₄ = 59 * 1697
1102221₃ = 3 * 3 * 3 * 40823
10000011011₂ = 19 * 20771 * 25339

1061₁₀
1408₉ = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 11
2045₈ = 5 * 409
3044₇ = 2 * 2 * 761
4525₆ = 5 * 5 * 181
13221₅ = 3 * 3 * 13 * 113
100211₄ = 23 * 4357
1110022₃ = 2 * 199 * 2789
10000100101₂ = 613 * 4027 * 4051

If 1049 is included the property holds also for the three consecutive primes 1049, 1051 and 1061 !
See Prime Curios! 1049

[July 16, 2001 ]
If we are to find the smallest prime to be appended to 1051 or 1061 so that
the concatenation is a square then the following beautiful solutions pop up :

1051 & 8758721 yielding the square 10518758721.
Note that this square is 102561²

The real gem comes with prime 1061 though !
1061 & 10601 yielding the square 106110601.
The appended prime is the same as the original prime with only an extra zero inserted.
Note that this square is 10301² and that 10601 and 10301 are both palindromic primes !

1051² equals 5! + 0! + 6! + 9! + 7! + 7! + 9! + 7! + 9!
Both 1051 and 506977979 are prime !
Source : Prime Curios! 506977979 by G. L. Honaker, Jr.

Index Nr	Base Integer Sequence
Index Nr	Palindromic Product of Integer Sequences	Length

	Fivefold sequence (n) x (n+2) x (n+4) x ...
1	31 x 33 x 35 x 37 x 39
1	51.666.615	8

	Five consecutive primes
1	7 x 11 x 13 x 17 x 19
1	323.323	6

	Fourfold sequence (n) x (n+2) x (n+4) x (n+6)
2	12 x 14 x 16 x 18
2	48.384	5
1	7 x 9 x 11 x 13
1	9.009	4

	Four consecutive primes
1	5 x 7 x 11 x 13
1	5.005	4

	Three consecutive palindromic primes
1	7 x 11 x 101
1	7.777	4
0	[1 is not considered as a prime] 1 x 2 x 3
0	6	1

	Three consecutive palindromes
Note that all palindromes (except the first one) can be constructed by using pattern expansion. Example of one infinite pattern by repeatedly inserting more zero's : 1356531 1030506050301 1003005006005003001 etc. The length of the numbers is the sequence 7, 13, 19, 25, ... [each time +6]. Make a sequence starting with group (1,3,4). Apply now to each term the operation (+6,+6,+6) and repeat the process with the new group. We get : 1,3,4, 7,9,10, 13,15,16, 19,21,22, 25,27,28, 31,33,34, ... (Sloane's A029739) These sequential numbers starting from the second group match exactly the number of digits of the following palindromes which are the product of three consecutive palindromes !
13	1.000.000.001 x 1.000.110.001 x 1.000.220.001
13	1.000.330.027.200.660.027.200.330.001	28
12	999.979.999 x 999.989.999 x 999.999.999
12	999.969.997.200.060.002.799.969.999	27
11	100.000.001 x 100.010.001 x 100.020.001
11	1.000.300.050.006.000.500.030.001	25
10	10.000.001 x 10.011.001 x 10.022.001
10	1.003.302.720.660.272.033.001	22
9	9.997.999 x 9.998.999 x 9.999.999
9	999.699.720.060.027.996.999	21
8	1.000.001 x 1.001.001 x 1.002.001
8	1.003.005.006.005.003.001	19
7	100.001 x 101.101 x 102.201
7	1.033.272.662.723.301	16
6	99.799 x 99.899 x 99.999
6	996.972.060.279.699	15
5	10.001 x 10.101 x 10.201
5	1.030.506.050.301	13
4	1.001 x 1.111 x 1.221
4	1.357.887.531	10
3	979 x 989 x 999
3	967.262.769	9
2	101 x 111 x 121
2	1.356.531	7
1	1 x 2 x 3
1	6	1

	Three consecutive primes
1	7 x 11 x 13
1	1.001	4

	Threefold sequence (n) x (n+2) x (n+4)
2	202 x 204 x 206
2	8.488.848	7
1	35 x 37 x 39
1	50.505	5

	Three consecutives (n) x (n+1) x (n+2)
2	77 x 78 x 79
2	474.474	6
1	1 x 2 x 3
1	6	1
	Record found on May 9, 1997
	Two consecutive primes Searched exhaustively upto length 20. Following prime must be greater than 10.000.000.000 !
One can also find these rare numbers in Sloane's table : %N Palindromes that are product of two consecutive primes. under A028979. %N Product of prime and following prime is palindromic. under A028888. %N Product of prime and previous prime is palindromic. under A028978.
6	Prime Curios! 1.934.063 x 1.934.071
6	3.740.615.160.473	13
5	1.051 x 1.061
5	1.115.111	7
4	191 x 193
4	36.863	5
3	17 x 19
3	323	3
2	7 x 11
2	77	2
1	2 x 3
1	6	1

	Two consecutive palindromic primes
7	111.010.111 x 111.020.111
7	12.324.354.845.342.321	17
6	100.111.001 x 100.131.001
6	10.024.214.741.242.001	17
5	100.060.001 x 100.111.001
5	10.017.106.860.171.001	17
4	101 x 131
4	13.231	5
3	11 x 101
3	1.111	4
2	7 x 11
2	77	2
1	2 x 3
1	6	1
0	[1 is not considered as a prime] 1 x 2
0	2	1

	Two consecutives (n) x (n+1)

	Two consecutive palindromes This rather boring series is very easy to continue... I stopped with 1111000001111 x 1111001001111 which gives 1234322113468643112234321
11	999 x 1.001
11	999.999	6
10	202 x 212
10	42.824	5
9	121 x 131
9	15.851	5
8	111 x 121
8	13.431	5
7	101 x 111
7	11.211	5
6	99 x 101
6	9.999	4
5	77 x 88
5	6.776	4
4	11 x 22
4	242	3
3	9 x 11
3	99	2
2	2 x 3
2	6	1
1	1 x 2
1	2	1

	Twofold sequence (n) x (n+2)

Contributions

Chad Davis is the first contributor to this page.

Hugo Sánchez found some interesting palindromes- goto topic.

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