**Palindromic Products of Integers Sequences** are defined and calculated by this extraordinary intricate and excruciatingly complex formula.

So, this line is for experts only _{}

As you can see from a quick glance at this page, many sequences just took a fresh start.

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

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 |
---|---|

1^{7} + 2^{2} + 3^{8} + 4^{9} + 5^{5} + 6^{6} + 7^{1} + 8^{4} + 9^{3}
| |

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 different exponents thereby forming a combination of these same nine digits ! [ 728956143 ]

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

35 x 37 x 39 =Here's a bonus operation :50505

35 + 37 + 39 =111

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 are5291and5313. 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) =

27999972and for x = 5313, x(x+1) =28233282.

1051 & 1061 |
---|

Have you noticed the following with **Two consecutive primes**

1051 x 1061 =All three operations yield a palindromic number !1115111

1051 + 1061 =2112

1051^{2}+ 1061^{2}=2230322

Notethat we can easily transform these numbers into palindromic primes by simply inserting a zero.

1051 and 1061 becomes10501and10601respectively.

Both these numbers can be expressed as the sum of three consecutive primes :

10501= 3491 + 3499 + 3511

10601= 3529 + 3533 + 3539

from Caracas Venezuela [

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__.

1051_{ 10}1387 _{ 9} = 19 * 732033 _{ 8} = 19 * 1073031 _{ 7} = 7 * 4334511 _{ 6} = 13 * 34713201 _{ 5} = 43 * 307100123 _{ 4} = 59 * 16971102221 _{ 3} = 3 * 3 * 3 * 4082310000011011 _{ 2} = 19 * 20771 * 25339 |
1061_{ 10}1408 _{ 9} = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 112045 _{ 8} = 5 * 4093044 _{ 7} = 2 * 2 * 7614525 _{ 6} = 5 * 5 * 18113221 _{ 5} = 3 * 3 * 13 * 113100211 _{ 4} = 23 * 43571110022 _{ 3} = 2 * 199 * 278910000100101 _{ 2} = 613 * 4027 * 4051 |

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 **10**25**61**^{2}

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 **10301**^{2} and that **10601** and **10301** are both palindromic primes !

1051^{2} 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 | Length |
---|---|---|

Palindromic Product of Integer Sequences | Length | |

Fivefold sequence (n) x (n+2) x (n+4) x (n+6) x (n+8) Searched up to basenumber 100.000.000 | ||

1 | 31 x 33 x 35 x 37 x 39 | 2 |

51.666.615 | 8 | |

Five consecutive primes | ||

1 | 7 x 11 x 13 x 17 x 19 | 1 - 2 |

323.323 | 6 | |

Fourfold sequence (n) x (n+2) x (n+4) x (n+6) Searched up to basenumber 100.000.000 | ||

2 | 12 x 14 x 16 x 18 | 2 |

48.384 | 5 | |

1 | 7 x 9 x 11 x 13 | 1 - 2 |

9.009 | 4 | |

Four consecutive primes | ||

1 | 5 x 7 x 11 x 13 | 1 - 2 |

5.005 | 4 | |

Three consecutive palindromic primes | ||

1 | 7 x 11 x 101 | 1 - 2 - 3 |

7.777 | 4 | |

0 | [1 is not considered as a prime] 1 x 2 x 3 | 1 |

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 :
030506050301003005006005003001 etc.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 | 10 |

1.000.330.027.200.660.027.200.330.001 | 28 | |

12 | 999.979.999 x 999.989.999 x 999.999.999 | 9 |

999.969.997.200.060.002.799.969.999 | 27 | |

11 | 100.000.001 x 100.010.001 x 100.020.001 | 9 |

1.000.300.050.006.000.500.030.001 | 25 | |

10 | 10.000.001 x 10.011.001 x 10.022.001 | 8 |

1.003.302.720.660.272.033.001 | 22 | |

9 | 9.997.999 x 9.998.999 x 9.999.999 | 7 |

999.699.720.060.027.996.999 | 21 | |

8 | 1.000.001 x 1.001.001 x 1.002.001 | 7 |

1.003.005.006.005.003.001 | 19 | |

7 | 100.001 x 101.101 x 102.201 | 6 |

1.033.272.662.723.301 | 16 | |

6 | 99.799 x 99.899 x 99.999 | 5 |

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 | 3 |

1.356.531 | 7 | |

1 | 1 x 2 x 3 | 2 |

6 | 1 | |

Three consecutive primes | ||

1 | 7 x 11 x 13 | 1 - 2 |

1.001 | 4 | |

Threefold sequence (n) x (n+2) x (n+4) Searched up to basenumber 100.000.000 | ||

2 | 202 x 204 x 206 | 3 |

8.488.848 | 7 | |

1 | 35 x 37 x 39 | 2 |

50.505 | 5 | |

Three consecutives (n) x (n+1) x (n+2) Searched up to basenumber 100.000.000 | ||

2 | 77 x 78 x 79 | 2 |

474.474 | 6 | |

1 | 1 x 2 x 3 | 1 |

6 | 1 | |

Record 7 found by Jan van Delden on [ May 20, 2011 ] (source)Record 6 found by PDG on [ May 9, 1997 ] | ||

Two consecutive primes Searched exhaustively up to length 20. | ||

| ||

7 | 13.422.495.703 x 13.422.495.727 | 11 |

180.163.391.219.193.361.081 | 21 | |

6 | Prime Curios! 1.934.063 x 1.934.071 | 7 |

3.740.615.160.473 | 13 | |

5 | 1.051 x 1.061 | 4 |

1.115.111 | 7 | |

4 | 191 x 193 | 3 |

36.863 | 5 | |

3 | 17 x 19 | 2 |

323 | 3 | |

2 | 7 x 11 | 1 - 2 |

77 | 2 | |

1 | 2 x 3 | 1 |

6 | 1 | |

Two consecutive palindromic primes | ||

7 | 111.010.111 x 111.020.111 | 9 |

12.324.354.845.342.321 | 17 | |

6 | 100.111.001 x 100.131.001 | 9 |

10.024.214.741.242.001 | 17 | |

5 | 100.060.001 x 100.111.001 | 9 |

10.017.106.860.171.001 | 17 | |

4 | 101 x 131 | 3 |

13.231 | 5 | |

3 | 11 x 101 | 2 - 3 |

1.111 | 4 | |

2 | 7 x 11 | 1 - 2 |

77 | 2 | |

1 | 2 x 3 | 1 |

6 | 1 | |

0 | [1 is not considered as a prime] 1 x 2 | 1 |

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 | 3 - 4 |

999.999 | 6 | |

10 | 202 x 212 | 3 |

42.824 | 5 | |

9 | 121 x 131 | 3 |

15.851 | 5 | |

8 | 111 x 121 | 3 |

13.431 | 5 | |

7 | 101 x 111 | 3 |

11.211 | 5 | |

6 | 99 x 101 | 2 - 3 |

9.999 | 4 | |

5 | 77 x 88 | 2 |

6.776 | 4 | |

4 | 11 x 22 | 2 |

242 | 3 | |

3 | 9 x 11 | 1 - 2 |

99 | 2 | |

2 | 2 x 3 | 1 |

6 | 1 | |

1 | 1 x 2 | 1 |

2 | 1 | |

Twofold sequence (n) x (n+2) |

Chad Davis (email) is the first contributor to this page.

Hugo Sánchez (email) found some interesting palindromes - go to topic.

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E-mail address : pdg@worldofnumbers.com