Download 106. Comments on `Triangular numbers and perfect squares` Let the

106. Comments on ‘Triangular numbers and perfect squares’
Let the triangular numbers {Tn } be given by Tn = n(n + 1)/2 for all integers n. Then
the {Tn } form a bilateral series,
· · · 10, 6, 3, 1, 0, 0, 1, 3, 6, 10, · · · .
(1) The authors of [1] find all squares that are sums of three consecutive triangular numbers.
If we let
s−1 = 64, s0 = 1, s1 = 4,
and continue the sequence in both directions using the recurrence
sn+3 − 99sn+2 + 99sn+1 − sn = 0,
we obtain the bilateral sequence
· · · 611524, 6241, 64, 1, 4, 361, 35344, · · · ,
precisely those squares that are sums of three consecutive triangular numbers.
The generating function for the “right hand sequence” 4, 361, · · · is
4 − 35q + q 2
1 − 99q + 99q 2 − q 3
while the generating function for the “left hand sequence” 1, 64, · · · is
1 − 35q + 4q 2
.
1 − 99q + 99q 2 − q 3
(2) The generating function for the squares that are sums of four consecutive triangular
numbers is
4(1 − 10q + q 2 )
.
1 − 35q + 35q 2 − q 3
(3) There are no squares that are sums of five consecutive triangular numbers, since the
sum of five consecutive triangular numbers is divisible by 5 but not by 25.
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(4) There are no squares that are the sum of six consecutive triangular numbers, since the
sum of six consecutive triangular numbers is 2 modulo 3, and squares are 0 or 1 modulo
3.
(5) There are no squares that are the sum of seven consecutive triangular numbers, since
the sum of seven consecutive triangular numbers is divisible by 7 but not by 49.
(6) There is just one square, 36, that is the sum of eight consecutive triangular numbers,
36 = 15 + 10 + 6 + 3 + 1 + 0 + 0 + 1 = 1 + 0 + 0 + 1 + 3 + 6 + 10 + 15.
(7) There are no squares that are the sum of nine consecutive triangular numbers, since
the sum of nine consecutive triangular numbers is 3 modulo 9, so is divisible by 3 but not
by 9.
(8) There are no squares that are the sum of ten consecutive triangular numbers, since
the sum of ten consecutive triangular numbers is divisible by 5 but not by 25.
(9) Just when you thought there weren’t going to be any, the squares that are sums of 11
consecutive triangular numbers are given by
s−1 = 14318656, s0 = 121, s1 = 302500
or by
s−1 = 296356225, s0 = 1936, s1 = 14641
and (in both directions)
sn+3 − 155235sn+2 + 155235sn+1 − sn = 0.
The four generating functions are
121(1 − 36899q + 2500q 2)
121(2500 − 36899q + q 2 )
,
,
2
3
1 − 155235q + 155235q − q
1 − 155235q + 155235q 2 − q 3
121(121 − 34535q + 16q 2 )
121(16 − 34535q + 121q 2 )
and
.
2
3
1 − 155235q + 155235q − q
1 − 155235q + 155235q 2 − q 3
(10) There are no squares that are the sum of 12 consecutive triangular numbers, since
the sum of 12 consecutive triangular numbers is congruent to 22, 28 or 46 modulo 48, and
squares are not.
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(11) The squares that are sums of 13 consecutive triangular numbers are given by
s−1 = 1249924, s0 = 169, s1 = 43264
or by
s−1 = 6558721, s0 = 676, s1 = 8281
and in both directions
sn+3 − 10403sn+2 + 10403sn+1 − sn = 0.
The four generating functions are
169(1 − 3007q + 256q 2 )
169(256 − 3007q + q 2 )
,
,
1 − 10403q + 10403q 2 − q 3 1 − 10403q + 10403q 2 − q 3
169(4 − 2803q + 49q 2 )
169(49 − 2803q + 4q 2 )
and
.
1 − 10403q + 10403q 2 − q 3
1 − 10403q + 10403q 2 − q 3
I would like to acknowledge the help given me by my colleague Peter Brown, who
performed certain calculations following Chrystal’s Algebra [2, pp. 479–485].
I have found that 14, 15, 16, 17, 18, 19, 20, 21 and 24 consecutive triangular numbers
are “out”, while 22, 23 and 25 are “in”.
References
[1] Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares, Math. Gaz.
86 (November 2002) pp. 423–431.
[2] Chrystal, Algebra, An elementary text–book for the higher classes of secondary schools
and for colleges, Part II, seventh edition, Chelsea, New York, 1964.