|
| |
|
|
A056888
|
|
a(n) = number of k such that sum of digits of 9^k is 9n.
|
|
1
|
|
|
|
2, 3, 2, 0, 4, 1, 3, 1, 1, 5, 2, 2, 3, 1, 0, 3, 6, 2, 3, 0, 0, 4, 1, 3, 1, 4, 1, 1, 0, 1, 3, 2, 3, 5, 1, 1, 3, 3, 2, 5, 0, 3, 3, 1, 1, 3, 2, 2, 0, 2, 1, 5, 2, 1, 1, 1, 1, 3, 4, 5, 1, 0, 1, 3, 2, 1, 2, 4, 5, 1, 1, 2, 1, 0, 1, 2, 4, 1, 2, 5, 0, 2, 4, 3, 2, 2, 1, 2, 2, 2, 0, 2, 3, 2, 1, 5, 1, 0, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Proposed by Mark Sapir, Math. Dept., Vanderbilt University, who remarks (August 2000) that he can prove that a(n) is always finite and that a(1) = 2.
Values of a(n) for n>1 computed numerically by Michael Kleber, Sep 02 2000 and David W. Wilson, Sep 06 2000.
All terms except the first are only conjectures. For the theorem that a(n) is always finite, see Senge-Straus and Stewart. - N. J. A. Sloane, Jan 06 2011
|
|
|
REFERENCES
|
H. G. Senge and E. G. Straus, PV-numbers and sets of multiplicity, Periodica Math. Hungar., 3 (1971), 93-100.
C. L. Stewart, On the representation of an integer in two different bases, J. Reine Angew. Math., 319 (1980), 63-72.
|
|
|
LINKS
|
Table of n, a(n) for n=1..99.
|
|
|
EXAMPLE
|
There are two powers of 9 with digit-sum 9, namely 9 and 81, so a(1) = 2.
|
|
|
CROSSREFS
|
Cf. A065999.
Sequence in context: A165192 A104771 A307688 * A286297 A339451 A111182
Adjacent sequences: A056885 A056886 A056887 * A056889 A056890 A056891
|
|
|
KEYWORD
|
nonn,base
|
|
|
AUTHOR
|
N. J. A. Sloane, Sep 05 2000
|
|
|
STATUS
|
approved
|
| |
|
|
