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Consecutive Triangles (Posted on 2009-11-29) Difficulty: 4 of 5
A, B, C and D are triangular numbers.

A, B and C are always consecutive while D is their sum.

Determine (and explain as best as possible1) how such sets of values are distributed across the number system.

1. This can be explained in terms of a single variable expression.

See The Solution Submitted by brianjn    
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Solution computer-aided solution | Comment 3 of 8 |

DECLARE FUNCTION isTri# (t#)
DECLARE FUNCTION tr# (n#)
DEFDBL A-Z
tr1 = 1
tr2 = 3
tr3 = 6
adder = 4
sum = tr1 + tr2 + tr3
DO
 IF isTri(sum) THEN PRINT tr1; tr2; tr3, isTri(tr1); isTri(tr2); isTri(tr3), sum; isTri(sum): PRINT
 tr4 = tr3 + adder: adder = adder + 1
 sum = sum - tr1 + tr4
 tr1 = tr2: tr2 = tr3: tr3 = tr4
LOOP

FUNCTION isTri (t)
 n = INT(SQR(t * 2))
 np = n + 1
 IF n * np = 2 * t THEN isTri = n:  ELSE isTri = 0
END FUNCTION

FUNCTION tr (n)
 tr = n * (n + 1) / 2
END FUNCTION

finds                                           Among triangular numbers
                                                    Ordinal of
A  B  C                                               A  B  C                          D  Ord of D
1  3  6                                               1  2  3                         10  4
36  45  55                                            8  9  10                       136  16
595  630  666                                        34  35  36                     1891  61
                                                                           
8646  8778  8911                                    131  132  133                  26335  229
121771  122265  122760                              493  494  495                 366796  856
1701090  1702935  1704781                          1844  1845  1846              5108806  3196
23711941  23718828  23725716                       6886  6887  6888             71156485  11929
330334956  330360660  330386365                   25703  25704  25705          991081981  44521
4601234485  4601330415  4601426346                95929  95930  95931        13803991246  166156
64087907136  64088265153  64088623171            358016  358017  358018     192264795460  620104
892633045591  892634381730  892635717870        1336138  1336139 1336140   2677903145191  2314261
12432788092530  12432793079070  12432798065611  4986539  4986540 4986541  37298379237211  8636941

The ordinal triangular number of A, follows Sloane's A082840, which has a comment by our own brianjn, based upon this problem. The ordinal position of D is given by A133161.

The formula given in Sloane for the ordinal of A (i.e., the nth triangular number), is:

With a=2+sqrt(3), b=2-sqrt(3)
a(n)=-3/2+(1/12)(a-2b+5)a^n+(1/12)(b-2a+5)b^n.

To test that out:

DEFDBL A-Z
CLS
a = 2 + SQR(3): b = 2 - SQR(3)
FOR n = 1 TO 10
 ans = -3 / 2 + (1 / 12) * (a - 2 * b + 5) * a ^ n + (1 / 12) * (b - 2 * a + 5) * b ^ n
 PRINT n, ans; TAB(38); INT(ans + .000000001#)
NEXT

finds:

1             .9999999999999999      1
2             7.999999999999999      8
3             34                     34
4             131                    131
5             492.9999999999999      493
6             1844                   1844
7             6885.999999999998      6886
8             25702.99999999999      25703
9             95928.99999999997      95929
10            358015.9999999999      358016

The successive 9's show the rounded value at the right is the actual value.


  Posted by Charlie on 2009-11-29 19:04:46
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