cp's OEIS Frontend

No other terms <= 200000. - Harvey P. Dale, Dec 16 2010

No other terms <= 1320000. - Robert G. Wilson v, Dec 18 2010

There are almost certainly no further terms.

Almost certainly there are no further terms.

Comments from Donovan Johnson on the computation of this sequence, Dec 05 2010 (Start):

The number of digits of 13^k is approximately 1.114*k, so I defined an array d() that is a little bigger than 1.114 times the maximum k value to be checked. The elements of d() each are the value of a single digit of the decimal expansion of 13^k with d(1) being the least significant digit.

It's easier to see how the program works if I start with k = 2.

For k = 1, d(2) would have been set to 1 and d(1) would have been set to 3.

k = 2:

x = 13*d(1) = 13*3 = 39

y = 39\10 = 3 (integer division)

x-y*10 = 39-30 = 9, d(1) is set to 9

x = 13*d(2)+y = 13*1+3 = 16, y is the carry from previous digit

y = 16\10 = 1

x-y*10 = 16-10 = 6, d(2) is set to 6

x = 13*d(3)+y = 13*0+1 = 1, y is the carry from previous digit

y = 1\10 = 0

x-y*10 = 1-0 = 1, d(3) is set to 1

These steps would of course be inside a loop and that loop would be inside a k loop. A pointer to the most significant digit increases usually by one and sometimes by two for each successive k value checked. The number of steps of the inner loop is the size of the pointer. A scan is done from the first element to the pointer element to get the digit sum.

(End)

No other terms < 3*10^6. - Donovan Johnson, Dec 07 2010

From Donovan Johnson, Dec 03 2010: (Start)

To generate the additional terms I used PFGW.exe to get the decimal expansion for each number of the form 167^n (n <= 50000). Then I wrote a program in powerbasic to read the pfgw.out file and get the digit sums.

The digit sum is 10 times the n value for terms a(5) to a(56). (End)

I believe that this sequence is finite. - N. J. A. Sloane, Dec 05 2010

Conjecture: the sequence is finite, since (sum of the digits of 6^k)/k -> log_10(6)*4.5 ~ 3.50168 as k->infinity (this is also a conjecture). - Robert Gerbicz, May 08 2008

Next term, if it exists, exceeds 100000. - Sean A. Irvine, Apr 05 2010

The keyword "hard" refers to the difficulty of finding, or disproving the existence of, the next term. - N. J. A. Sloane, Nov 09 2024

The set of these numbers appears to be finite, and probably 2781 is its largest element.

The motivation for this sequence is the study of the behavior of the sum of digits of powers of a given number. Statistically, sumdigits(n^k) ~ 4.5*log_10(n')*k (where n' = n without trailing 0's), but typically fluctuations of some percent persist up to large values of k. (Cf. the graph of sequences n^k cited in the cross-references.)

The ratio of 11/10 is somewhat arbitrary, but larger ratios of the simple form (1 + 1/m) yield quite small subsets of this sequence (for m=2 the only element is 118, for m=3 the set is {31, 86, 118}, for m=1 it is empty), and smaller ratios yield much larger (possibly infinite?) sets. Also, the condition can be written sumdigits(N^k)-N > N/10, and 10 is the base we are using.

To compute the sequence A247889 we would like to have a rule telling us when we can stop the search for an exponent. It appears that sumdigits(N^k) >= 2*N is a limit that works for all N; the present sequence gives counterexamples to the (r.h.s.) limit of 1.1*N. The above comment mentioned the counterexamples {118} resp. {31, 86, 118}) for limits N*3/2 and N*4/3.

118 is exceptional in the sense that it appears to be the only number m for which the smallest k such that sumdigits(m^k) = m occurs after the smallest k such that sumdigits(m^k) > m*3/2. If this last limit is decreased to m*4/3, then 31 and 86 also have this property. It appears that no number has this property if the limit is increased to 2m, see also A247889.

It is also remarkable that many values in the sequence are repeated (19, 55, 64, 109, 190, 154, 280, 289, 334 (3 times), 379, 406, 424, ...), while most other numbers never appear.