Triangular numbers are defined and calculated by this extraordinary intricate and excruciatingly complex formula. So, this line is for experts only _{}
base x ( base + 1 ) ------------------------ or ( base ^{2} + base ) / 2 2
The best way to get a 'structural' insight as how to imagine tetrahedrals is to visit these sites :
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 Worthwhile links : Eric Weisstein's page Pascal's Triangle Math Forum
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.
11^{0} = 1 11^{1} = 11 11^{2} = 121 11^{3} = 1331 11^{4} = 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 11^{n} pattern does continue past 11^{4} in Pascal's Triangle, but in a convoluted manner, not quite like José's:
11^{5} = 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 ! 11^{6} = 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 11^{7} = 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: 11^{8} = 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: 11^{9} = 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.
Good luck with higher numbers! But I don't see any reason why this shouldn't continue in this manner.
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'.
Here is a formula that generates these numbers which are both triangular and square. It is written in the UBASIC syntax : 10 for N=1 to 9 20 X=ceil(((17+12*sqrt(2))^N+(17-12*sqrt(2))^N-2)/32) 30 print N,X 40 next N
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
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.
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.
Highlighting out extra characteristics or relationships about the PALINDROMIC TRIANGULARS is a way to give these numbers added value. Some can be quite obvious as others can be more elaborate or even far-fetched. There's almost no limit in what you can find out. While extending the list with the next higher palindromic triangular may be very difficult to accomplish, searching for extra features in the existing ones is something that we all could embark on. “ The only limitation is one's imagination ” as the saying goes. To give you an idea I'll show you some results of my own humble investigations. If you consult the full listing of the palindromic triangulars you'll see a column with the hypertext word 'Info'. This will lead you to a new page displaying additional information about the referenced triangulars. Things like prime factors, number-lengths, discovery dates, features, comments... can be found there. My first idea was to incorporate them also in the table, but that would have lead to an overloaded table difficult to manage. Some random excerpts :
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 = 3^{1} + 3^{2} + 3^{3} + 3^{4} + 3^{5}
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.
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