Journal of Integer Sequences, Vol. 1 (1998), Article 98.1.4 |

Department of Mathematics

Hartwick College, Oneonta, NY 13820

Email address: StevensG@Hartwick.edu

We introduce a generalization of the Connell
Sequence, generated by using groups of *q*
terms each a multiple of *q* (rather than just being even or odd as
*q* is), and find an expression for the general term.

**INTRODUCTION.**

The Connell Sequence
1, 2, 4, 5, 7, 9, 10, 12, 14, 16, ...
(A001614
in the *On-Line
Encyclopedia of Integer Sequences*)
is generated as follows:

In 1959 Ian Connell [1] posed the problem of proving that this sequence has general term:

In this article we consider the following generalization of the Connell Sequence.

Definition: Let {S}
be the Connell-like sequence generated as follows:
_{n}Take the first multiple of one, the next two multiples of two, the next three multiples of three, the next four multiples of four, ..., the next q multiples of q, etc. |

The resulting sequence
(now A033291)
is_{ } ^{ }

**THE DISCOVERY.**

Let *q* denote the multipliers
used in the sections of the general sequence so that *q* runs through
all positive integer values and is used as a multiplier *q* times.
Note that the subscript of the last term of the section which uses the
multiplier *q* will be the *q ^{th}* triangular number,

Thus we can define a sequence related to *S _{n}* by

This sequence uses an increment of 1 for three terms, then 2 for three
terms, then 3 for three terms, and so forth. If we ignore the three
term groupings and think about a sequence that increments by 1, then by
2, then by 3, etc., it would have to look basically like ½ *q*^{2}.
To get the groupings of three, we can divide this by 3 and use the floor
function to obtain Floor(*q*^{2} / 6) which produces
the sequence 0, 0, 1, 2, 4, ..., which is not quite right. A slight
adjustment allows us to write_{ }

This result, however, has been obtained by observing the first few terms
of some infinite sequences. It is conceivable that the pattern displayed
by the first twenty terms of the sequence *v _{q}* might change
at some point. (Using a computer program to check up to a

**THE PROOF.**

Theorem. Let {S}
be the sequence defined above and let _{n}q = _{n}q(n)
= Ceiling (½ (-1 + sqrt(1 + 8n)))
so that q is the multiplier used in the section of the
sequence containing the term _{n}S. Let _{n}v
= Floor(_{q}q(q + 1) / 6)
for q >= 1.
Then S = _{n}q - _{n}nq_{n}v_{q}_{(n)}
for all n >= 1. |

**Proof **(by induction on *n*).
For *n* = 1, *q*_{1} = 1 and *S _{n}*
=

Assume the theorem is true for

and so

as required.

If *S _{n}* and

Now

*q _{n}*

= (

= (

This difference will thus be at most *q* + 1 if (*q* + 1)*v _{q}*

Let

We consider six cases, the possibilities for *q* mod 6, and compute
(omitting the arithmetic details) the exact value of *f*(*q*)
in each case:

In each case, since *k* >= 0, we find *f*(*q*) >= 0, so
that the difference between *S _{n}* and the proposed expression
for

**ADDENDUM.**

After reading an initial version of this paper, a colleague, Gerry Hunsberger, suggested another way of generalizing the Connell Sequence. Instead of thinking of the original seqence in terms of even and odd numbers, it is also possible to think of it as "one integer congruent to 1 mod 2, the next two integers congruent to 2 mod 2, the next three integers congruent to 3 mod 2, the next four integers congruent to 4 mod 2, and so forth. This still produces the Connell Sequence but allows for a nice generalization by changing the modulus.

Definition. The Connell k-Sequence, C,
is defined by taking one integer congruent to 1 mod _{k}k, the next
two integers congruent to 2 mod k, the next three integers congruent
to 3 mod k, the next four integers congruent to 4 mod k,
and so forth. |

For example, the Connell 3-Sequence would be 1, 2, 5, 6, 9, 12, 13, 16, 19, 22, 23, 26,... (now A033292) and the 8-Sequence would be 1, 2, 10, 11, 19, 27, 28, 36, 44, 52, 53, 61, ... (now A033293).

Within each grouping of terms, the terms increase by an amount *k*,
and when changing from one grouping to the next, the increment is 1. The
groupings change after each trianular number. Thus we can consider the
*n ^{th}* term as being obtained by adding

n) = kn - (k - 1)Floor(½
(-1+sqrt(8n+1))). |

**References.**
**[1]** *American Mathematical Monthly*,
**66** (1959), 724. Elementary Problem E1382.
**[2]** *American Mathematical Monthly*,
**67** (1960), 380.
Solution to Elementary Problem E1382.
**[3]**
Lakhtakia, A., V. K. Varadan, & V. V. Varadan.
Connell arrays. *Archiv für Elektronik und*
* Übertragungstechnik*, **42**, (3) (1988), 186-189.
**[4]** Lakhtakia, A. & Clifford Pickover.
The Connell sequence. *J. Recreational Math.*,
**25** (1993), 90-92.

Received Jan. 30, 1998; published in Journal of Integer Sequences, April 26, 1998.

Return to