A107221 Numbers such that sum of digits is 7 and product of digits is 8.

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

Table of n, a(n) for n=1..10.

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

Adjacent sequences: A107218 A107219 A107220 * A107222 A107223 A107224

Keyword

nonn,base,fini,full

Author

Zak Seidov, May 13 2005

Status

approved