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Palindromic Triangulars
Factorization Records Full Listing The Subsets Most Beautiful square penta hexa hepta octa nona

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

An Animated Example

Triangular numbers are defined and calculated by this extraordinary intricate and excruciatingly complex formula.
So, this line is for experts only

So far this compilation counts 195 palindromic triangulars.

Triangular numbers, whether palindromic or not, can only end with the following digits 0, 1, 3, 5, 6 or 8.
Alas, my palindromes may not have leading 0's! So the zero option must not be investigated.

There exist no palindromic triangulars of length  12, 30, 40, 47, 48, 50, 56, 60.

A B C →  2 3 1 

A B C →  2 1 0 

A B C →  2 5 3 

Every hexagonal number is a triangular number.

A good source for such statements is “The Book of Numbers”
by John H. Conway and Richard K. Guy.
Click on the image on the left for more background about the book.
The book can be ordered at www.amazon.com.

The set of hexagonal numbers is a subset of the triangular numbers.
In fact every other triangular number T_n is a hexagonal number,
with H_n = T_2n–1. So in practice this means that whenever you have
an odd basenumber of a triangular (it doesn't matter if it is palindromic
or not) add 1 and divide by 2 and you'll have your basenumber of
a hexagonal number.
For instance Base_T = 3185 gives (3185+1)/2 or Base_H = 1593.
Both basenumbers yield resp. 5073705 ! (See T₂₀ and H₁₃ in the list
of palindromic triangulars and hexagonals.)

The best way to get a 'structural' insight as how to imagine tetrahedrals is to visit these sites :

One can visualize a triangular number as the surface value of an equilateral triangle.
Stick two triangles together so that they form a rectangle and you'll immediately understand the above formula.

1 2 3 base

Explaining a triangular number in everyday language :
it's the sum of the consecutive integers beginning from 1 up to base.
So, the basenumber is the highest number you add to the sum.

Example (see above figure) : 1 + 2 + 3 + (base = 4) gives the triangular number 10, the total of all stars.

The triangular numbers (1,3,6,10,15,...) can be found in the famous Pascal's Triangle.
They appear in the third row like this :

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1

Bright S. Jaja (email) a sophomore student from Madison, Wisconsin
remarked [ September 28, 1999 ] the following interesting relation between
the five first rows of the Pascal's Triangle and the powers of 11.

110 = 1
111 = 11
112 = 121
113 = 1331
114 = 14641

José Antônio Fabiano Mendes (email) from Rio de Janeiro, Brazil
counters the above observation [ January 26, 2000 ].

Jaja's remark will always be true if we disregard the carry-1 rule, namely:
1 4 6 4 1 x 11 = 1 5 10 10 5 1
1 5 10 10 5 1 x 11 = 1 6 15 20 15 6 1
and so on ...
Here we are not using the base 10 but an infinite base, so to speak.

Marc Jacobs (email) from Burbank, California [ November 9, 2001 ]
thinks that the 11ⁿ pattern does continue past 11⁴ in Pascal's Triangle,
but in a convoluted manner, not quite like José's:

11⁵ = 161051 -- the 5^th row says: 1 5 10 10 5 1

now here's where the trickery comes in...

start by keeping the last three digits: 051

add the next digits together in pairs, from right to left:

1 + 0 (from the two 10s) = 1

1 + 5 (from 10) = 6

and the first number: 1

string it all together -- 161051 !

11⁶ = 1771561 -- the 6^th row says: 1 6 15 20 15 6 1

start by keeping the last three digits: 561

add the next digits together in pairs, from right to left:

1 + 0 = 1

2 + 5 = 7

1 + 6 = 7

and the first number: 1

string it all together -- 1771561

11⁷ = 19487171 -- the 7^th row says: 1 7 21 35 35 21 7 1

start by keeping the last three digits: 171

2 + 5 = 7

3 + 5 = 8

3 + 1 = 4

2 + 7 = 9

and the first number: 1

string it all together -- 19487171

Now it gets a bit trickier - it involves carrying:
11⁸ = 214358881 -- the 8^th row says: 1 8 28 56 70 56 28 8 1

start by keeping the last three digits: 881

now work from right to left, adding pairs and carrying:

2 + 6 = 8

5 + 0 = 5

7 + 6 = 13 (keep 3, carry 1 to next pair)

5 + 8 = 13 + 1 = 14 (keep 4, carry 1 to next pair)

2 + 8 = 10 + 1 = 11 (keep 1, carry 1)

the first 1 + the carried 1 = 2

string it all together -- 214358881

3 digit numbers in the triangle mean we add them as 2 digit numbers:
11⁹ = 2357947691 -- the 9^th row says: 1 9 36 84 126 126 84 36 9 1

start by keeping the last three digits: 691

3 + 4 = 7

8 + 6 = 14 (keep 4, carry 1)

12 + 6 = 18 + 1 = 19 (keep 9, carry 1)

12 + 4 = 16 + 1 = 17 (keep 7, carry 1)

8 + 6 = 14 + 1 = 15 (keep 5, carry 1)

3 + 9 = 12 + 1 = 13 (keep 3, carry 1)

the first 1 + the carried 1 = 2

string it all together -- 2357947691

Good luck with higher numbers!
But I don't see any reason why this shouldn't continue in this manner.

Sloane's A001110 gives the first numbers that are both Triangular and Square.

 Check out also Sloane's entries A001108 and A001109 to discover more about the topic.

 Relationship between Triangular and Square Numbers

 Square and Triangular numbers

 From Eric Weisstein's Math Encyclopedia : Triangular Square Number

Sloane's A014979 gives the first numbers that are both Triangular and Pentagonal.

Sloane's A046194 gives the first numbers that are both Triangular and Heptagonal.

Sloane's A046183 gives the first numbers that are both Triangular and Octagonal.

Sloane's A048909 gives the first numbers that are both Triangular and Nonagonal.

Another interesting entry from Sloane's table is :

Michael Mann (email)
found an interesting way to construct the first few triangular numbers...

If you write how many 3-digit numbers have 0,1,2,3,4,... as a sum of their digits,
you will get the sequence of the triangular numbers.
This means any number which can be represented using three decimal positions
(say 10 = 010), i.e. those that have ⩽ 3 digits.

Consider these numbers :

how many have 0 as a sum of their digits -- 1 (0)
sum 1 -- 3 numbers (1, 10, 100)
sum 2 -- 6 (2, 20, 200, 11, 110, 101)
sum 3 -- 10 (3, 30, 300, 21, 12, 102, 201, 111, 120, 210)
sum 4 -- 15 (4, 40, 400, 31, 13, 130, 310, 301, 103, 202, 22, 220, 211, 212, 122)
sum 5 -- 21 (32, 23, 203, 302, ...)
etc.

It is just easier to count the numbers with particular sum of the digits by representing
each number in 3 decimal positions : say, for sum 3 :

C(3, 1) //those which have only 3 in them, 3, 30, 300
C(3, 1)*C(2, 1) //those which have one 2 and one 1 in them 210, 021, 120 etc.
C(3, 3) //those that have 3 ones in them

It looks like the sequence produced is a part of triangular numbers sequence.
I don't know what the exact explanation for it is. If you have one couldn't you share it ?

Michael Mann's own reply [ July 31, 2000 ]
Michael thought more about it himself in the following days
and realised that his conjecture is true only for Sum ⩽ 9.
Here is his proof why it is true only for the small values : for Sum ⩽ 9.

The number of ⩽ 3 digit number corresponding to each value of the Sum is the
number of ways to put Sum non-distinguishable items into 3 distinguishable
boxes where each box can hold 0 ⩽ n ⩽ Sum items.
= C(n+2, 2) = C(n+2, n) = (n+1) + n*(n-1)/2 -- a triangular number.

Normal basenumbers and their palindromic triangular numbers two examples:

In some cases the basenumbers are themselves palindromic! two examples:

Can you think of more beautiful and mind-boggling numbers...
As it turns out, these numbers are extremely rare and hard to find.
Please inform me if you know more palindromic triangulars
or whether you're about to challenge me by trying to extend the list.
You'll find my e-mail address at the bottom of this page.

Click here to display the full listing of palindromic triangulars. Click here to display the subsets of palindromic triangulars.

I found FOUR YEARS IN A ROW in the following palindromic triangulars !
Will these be the most prolific years of my life ?
[122] 1873574437207991455541997027344753781
[73] 15199896744769899151
[116] 5952926739999190550919999376292595
[40] 6874200024786

The 'smallest' triangular number with FOUR digit 8's in its decimal expansion is palindromic !
[16] 828828
[See Sloane A036525]

The sum of the first thirteen triangle basenumbers is itself palindromic
1 + 2 + 3 + 10 + 11 + 18 + 34 + 36 + 77 + 109 + 132 + 173 + 363 = 969

The last palindromic basenumber 363 can be expressed as the sum of consecutive powers of the base 3
[14] 363 = 3¹ + 3² + 3³ + 3⁴ + 3⁵

All the EVEN palindromic triangulars carry the factor 11 in their genes.
This number 11 is itself a palindromic basenumber, refer to [6].
And 11 is the only existing palindromic prime with an 'even' number of digits.

Dividing into groups of three :
[106] 353.520.620.692.923 equals A.B.C.D.E
( A + B + C + D + E ) = 3108
Just a number like any other, where it not for the property that 3108 is the sum of seven repdigital palindromes :
111 + 222 + 333 + 444 + 555 + 666 + 777 = 7 x 444
Creating “a triplet of duo's ” is another way to split this one up :
(35)(35)_(206)(206)_(92)(92)__(3)
By the way summing up all three pairs generates the number of the beast nl. 666.
I found that the beast's number pops up in more than one palindromic triangular (see for instance [9]). So be warned...

[31] 179.158 equalling A.B
( A x B ) = 179 x 158 = 28282. Palindromes creep up in many unexpecting ways !

Here's the INTEL triangular love affair :
[54] 681909070909186
[55] 683727232727386
[56] 684866959668486    ¿ So, what happened with the 286 ?

Four is a rewarding number when dealing with palindromic triangulars as e.g. when groupings of four are summed up in :
[121] 300721668093919607706919390866127003
Nine groups of four looks like : 3007 + 2166 + 8093 + 9196 + (0)770 + 6919 + 3908 + 6612 + 7003
And yes, after adding them together... you've guessed it : a palindrome shows up : 47674
Consider its basenumber for a moment : 775.527.779.120.670.322
where the extraction of the middle digits of the six trio's reveals this nice undulating pattern 7.2.7.2.7.2
Add up the numbercouples surrounding these middle digits :
7_5 + 5_7 + 7_9 + 1_0 + 6_0 + 3_2 = 313 again a palindromic number.

The next triangular consists only of the odd digits 1, 3, 5 and 9 :
[48] 539593131395935
The next triangular consists only of the even digits 0, 2, 6 and 8 :
[61] 8208268228628028

There are a few beautiful basenumbers composed only of the digits 3, 4, 5 and 6.
They seem to grow like crystals.
[4] 3
[9] 36
[14] 363
[35] 365436
[50] 34456434
[53] 36545436
[99] 36543454565436
[131] 3654345456545434563 - Most beautiful palindromic triangular

This palindromic basenumber [71] 2664444662 has exactly three palindromic factors !
And above all, it only needs the digits 1 and 2 to show them !
2
11
121111121

Starting numbers predicting the end... that must be the end !
[74] 11151642876
( 1 + 1 + 1 + 5 ) ( 1 + 6 ) ( 4 + 2 ) = (8) (7) (6) = ..876.

1. A068641 Smallest n-digit palindromic triangular number. If no such number exists then 0. - Amarnath Murthy
2. A068642 Index of the smallest n-digit palindromic triangular number. If no such number exists then 0. - Amarnath Murthy
3. A068643 Largest n-digit palindromic triangular number. If no such number exists then 0. - Amarnath Murthy
4. A068644 Index of the largest n-digit palindromic triangular number. If no such number exists then 0. - Amarnath Murthy

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Patrick De Geest - Belgium - Short Bio - Some Pictures
E-mail address : pdg@worldofnumbers.com