SEQUENCES OF CONSECUTIVE n-NIVEN NUMBERS*
H. G. Grundmae
Department of Mathematics, Bryn Mawr College, Bryn Mawr, PA 19010
e-mail: grundman@brynmawr.edu(Submitted August 1992)

A Niven number [2] is a positive integer that is divisible by the sum of its digits. In 1982,
Kennedy [3] showed that there do not exist sequences of more than 21 consecutive Niven
numbers. In 1992, Cooper & Kennedy [1] improved this result by proving that there does notexist a sequence of more than 20 consecutive Niven numbers. They also proved that this bound is
the best possible by producing an infinite family of sequences of 20 consecutive Niven numbers.
For any positive integer n > 2, define an ft-Niven number to be a positive integer that isdivisible by the sum of the digits in its base n expansion. This paper examines the maximal possible lengths of sequences of consecutive ft-Niven numbers. The main result is given in the following theorem.
Main Theorem: For any n > 2, there does not exist a sequence of more than In consecutive«-Niven numbers.
Note that this result is entirely general and gives Cooper & Kennedy's result as a special case.
Fix any n > 2. The following notation will be used throughout this paper. A number written
in base n will be subscripted with (n). For example, 12 = 22 (5) • When a string of (nonboldface)variables is subscripted, it is assumed that each variable represents a single digit in the given base.
For example, y(w) represents a number that can be expressed in two base n digits, 0<ij <n<(Note that / = 0 is allowed.) A boldface variable in such a string represents a (possibly empty)
string of digits in the given base. For example, a22 (5) represents a number congruent to 12
modulo 25. Let s(a) be the sum of the digits in the string a.
Lemma 1: Suppose thatis a sequence of n consecutive «-Niven numbers. Then n divides s(a).
Proof: Let s = 5(a). The base n digit sums of the numbers a0 ( w ) ,..., a(/? - l) (w) are $, s + 1,..., s + n-l. Exactly one of the digit sums is divisible by n. The corresponding w-Niven number
must also be divisible by n and thus must be a0 (w) . Hence, n divides the digit sum of a0 ( w ) , i.e., ndivides s.
Lemma 2: The n + l consecutive numbers a00 ( w ) ,..., a!0 (w) are not all ^-Niven numbers.
Proof: Suppose to the contrary. Since w-Niven numbers are by definition positive,- s =5(a) > 0. Further, by Lemma 1, n divides s. Thus, n < s. The base n digit sum of both a01 (w) and* This material is based on work supported by the National Science Foundation under Grant No. DMS9115349. The Government has certain rights in this material.
174[MAYSEQUENCES OF CONSECUTIVE Yl NIVEN NUMBERS

alO(w) is s +1. Since s +1 divides each, it must divide their difference, w - 1 . So s +Contradiction.
Lemma 3: Ifi^n-1numbers.

l<n-l<s.

and s(a) + / > 0, then m(n- l)(w) and a(z 4- l)(w-2) (w) are not both ^-Niven

Proof: Let 5 = 5(a). The base ^ digit sum of both &i(n-\){n) and a(z + l)(«-2) (w) is5+i + n -1. If both areft-Nivennumbers, then s + i + n-l must also divide their difference, n-\.
But s+i>0 implies that $ + i + n-l>n-l.
Contradiction.
Theorem 4: If ay(/?) is the first term in a sequence of length at least 2n of consecutive ft-Niven
numbers, then i~n-\ and7 = 0.
Proof: Let a//(w) be the first term in such a sequence. Let s = s(a). Suppose i^n-\.

Thena(i + 1)0(B), a(i + l)l (n) , . . , a ( / + l)(w - l) (;i)
is a subsequence of consecutive n-Niven numbers. By Lemma 1, n divides s + z + l, and sos +1 > 0. Further, both a/(« - l)(w) and a(/ +1)(« - 2)(77) are «-Niven numbers. But this is impossible by Lemma 3.
Similarly, if i-n-1 and7 ^ 0, then the sequence contains the subsequence(a + l)00 (B) ,...,(a + l)10
which cannot all be ??-Niven numbers by Lemma 2.
We now prove the Main Theorem as an easy corollary to Theorem 4.
Proof of the Main Theorem: Suppose xl5 x 2 ,... is a sequence of more than 2n consecutive
w-Niven numbers. By Theorem 4, both xx and x2 end in zero when written in base n. This isclearly impossible.
It is not known whether this bound is the best possible. The earlier results show that it is thebest possible for n = 10 and computer calculations have verified that it is optimal for a number ofother small values of n. A general proof, however, that applies to all values of n has yet to befound.
Conjecture 5: For each n > 2, there exists a sequence of consecutive ^-Niven numbers of length
In.
REFERENCES
1.

Curtis Cooper & Robert E. Kennedy. "On Consecutive Niven Numbers." The Fibonacci
Quarterfy.31.2 (1993): 146-51.
2. R. Kennedy, T. Goodman, & C. Best. "Mathematical Discovery and Niven Numbers." The
MATYC Journal 14 (1980):21-25.
3. Robert E. Kennedy. "Digital Sums, Niven Numbers, and Natural Density." Crux Mathematicorum 8 (1982): 129-33.
AMS Classification Numbers: 11A07, 11A63•••••••••1994]

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