|Journal of Integer Sequences, Vol. 2 (1999), Article 99.1.7|
Abstract: We introduce a generalization of the Connell Sequence which relies on two parameters: the modulus m and the successive row length difference r. We show its relationship with polygonal numbers, examine its limiting behavior, and find an expression for the general term.
1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, ...
in the On-Line
Encyclopedia of Integer Sequences), in which the first odd number
is followed by the next two even numbers, which in turn are followed by
the next three odd numbers, and so on.
Lakhtakia and Pickover  arranged this sequence as a concatenation of finite subsequences,
|Subsequence Number :||Subsequence :|
|3||5, 7, 9|
|4||10, 12, 14, 16|
where the nth subsequence contains
elements, the last of which is n2. So if we let
the nth element of the Connell sequence then
|c( Tn ) = n2 ,||(1)|
= n (n + 1)/ 2
denotes the nth triangular number.
Lakhtakia and Pickover  used (1) to show that
|lim n c(n) / n = 2 ,||(2)|
thus explaining the limiting behavior of the Connell sequence. The same
authors remarked that (2) could have been obtained directly from
the formula for c(n) given by Connell 
|c(n) = 2n - Floor( ( 1 + Sqrt( 8n - 7 ) ) / 2 ) .||(3)|
Stevens  defined a generalized Connell k-sequence Ck for integers k >= 2 (whereby the classical Connell sequence becomes the special case C2 ). In this paper we further generalize the Connell sequence by affixing another parameter onto Stevens's sequence Ck.
|Definition 1: Let m >= 2 and r >=
be integers. We denote by Cm,r(n) the
term of the generalized Connell sequence with parameters m
, or, simply, the Connell (
m , r )-sequence . The sequence is defined as follows:
1. The sequence is formed by concatenating subsequences S1, S2, ... , each of finite length.
2. The subsequence S1 consists of the element 1.
3. If the nth subsequence Sn ends with the element e , then the (n+1)th subsequence Sn+1 begins with the element e+1 .
4. If Sn consists of t elements, then Sn+1 consists of t + r elements.
5. Each subsequence is nondecreasing, and the difference between
two consecutive elements in the same subsequence is m .
The sequence Ck of Stevens
is the sequence C k,1 .
When (m , r) = ( 3 , 2 )
|2||2, 5, 8|
|3||9, 12, 15, 18, 21|
|4||22, 25, 28, 31, 34, 37, 40|
The final elements 1, 8, 21, 40, ... in the subsequences appear to be the octagonal numbers, En = n (3n-2 ). The nth subsequence Sn contains exactly 2n - 1 elements, and from the identity 1 + 3 + ... + (2n-1) = n2 we obtain
C3,2( n2 ) = En .
Just as the Connell sequence relates triangular numbers to squares (see (1)), the sequence C3,2 relates squares to octagonal numbers. Triangular numbers, squares, and octagonal numbers are all examples of polygonal numbers.
|Definition 2: For integers k >= 3,
the nth k-gonal number
We shall demonstrate the relationship between generalized Connell sequences and polygonal numbers. Consider the sequence Cm,r . By induction (see conditions 2 and 4 in Definition 1), Sn contains exactly ( n - 1 ) r + 1 elements. Therefore to reach the end of nth subsequence Sn , we must count exactly
| S1| + | S2| + ... + | Sn | = Pr+2( n )
elements of the sequence Cm,r . Hence
the last element of Sn is Cm,r
) ) . Thus Sn+1 begins (see condition
3 of Definition 1) with the element Cm,r
) ) + 1 , and, since | Sn+1 | = nr
+ 1, it ends (see condition 5 of Definition 1) with
the element Cm,r ( Pr+2( n ) ) + 1 + m ( nr + 1 ) . But this last element is also expressible as Cm,r ( Pr+2( n + 1 ) ) . Therefore, assuming the induction hypothesis Cm,r ( Pr+2( n ) ) = Pmr+2( n ) , we obtain
Cm,r ( Pr+2( n + 1 ) ) = Pmr+2( n ) + 1 + m ( nr + 1 ) = Pmr+2( n + 1 ) ,
and hence by induction we have for all positive integers n,
(1) is the special case of (4) when ( m
, r ) = ( 2 , 1 ). As we remarked in Section 2,
C3,2( P4( n ) ) = P8( n) .
Another example is given by C3,1 (A033292)
|3||6, 9, 12|
|4||13, 16, 19, 22|
Here the pentagonal numbers P5( n ) = n ( 3n - 1 ) / 2 at the end of each subsequence.
It is interesting to note that, since all elements of Sn are congruent to n ( mod m ), we obtain the following property of polygonal numbers: Pmr+2( n) is congruent to n ( mod m ).
Cm,r ( n ) = Cm,r ( Pr+2(
) + i )
= Cm,r( Pr+2( j ) ) + 1 + ( i - 1 ) m
= Pmr+2( j ) + 1 + ( i - 1 ) m.
Thus, since 1 <= i <= 1
+ rj and n = Pr+2(
) + i , we have A <= Cm,r
) / n <= B , where
A = ( Pmr+2( j ) + 1 ) / ( Pr+2( j ) + 1 + rj ) ,
B = ( Pmr+2( j ) + 1 + jmr ) / ( Pr+2( j ) + 1 ) .
Recalling Definition 2, it is a simple matter to verify that A
and B both converge to the limit m
as j tends toward infinity. Therefore
T( n ) = mn - Cm,r ( n ) .
We assume n > 1 and write n = Pr+2( j ) + i exactly as in Section 4. Then after some algebra we find that T( n ) = ( j +1 )( m - 1 ), so j + 1 = T( n ) / ( m - 1 ). Since n >= Pr+2( j) + 1 ,
j + 1 <= ( 3r - 2 + Sqrt( 8r ( n -1 ) + ( r - 2 )2 ) ) / 2k ,
T( n ) = ( m - 1 ) Floor( ( 3r - 2 + Sqrt( 8r ( n -1 ) + ( r - 2 )2 ) ) / 2k ) ,
 Amer. Math. Monthly, 66, no. 8 (October, 1959), 724. Elementary Problem E1382.
 Amer. Math. Monthly, 67, no. 4 (April, 1960), 380. Solution to Elementary Problem E1382.
 Lakhtakia, A. & Pickover, C. The Connell sequence, J. Recreational Math., v. 25, no. 2 (1993), 90-92.
 Stevens, G. A Connell-like
sequence, J. of Integer Sequences, v.1, Article
(Concerned with sequences A001614 , A033292 , A045928 , A045929 , A045930 .)
Received February 9 1999; published in Journal of Integer Sequences, March 16 1999.