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A107221
Numbers such that sum of digits is 7 and product of digits is 8.
0
124, 142, 214, 241, 412, 421, 1222, 2122, 2212, 2221
OFFSET
1,1
COMMENTS
Sequence is finite.
There are no other terms. Proof. From the fact that the product of the digits is 8, we can conclude that each number contains either three '2's or a '2' and a '4' or a single '8' together with arbitrarily many '1's. In the first case there has to be exactly one '1', otherwise the sum wouldn't be 7. This gives the numbers 1222, 2122, 2212, 2221. In the second case, as well, there can be only one '1', this yields the numbers 124, 142, 214, 241, 412, 421. The third case is not possible because 8, by itself, is already bigger than 7. All those terms are listed, hence the sequence is complete. - Stefan Steinerberger, Jun 07 2007
LINKS
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
CROSSREFS
Sequence in context: A080537 A030492 A031203 * A300325 A056085 A129010
KEYWORD
nonn,base,fini,full
AUTHOR
Zak Seidov, May 13 2005
STATUS
approved