A212814 a(n) = number of integers k >= 7 such that A212813(k) = n.

1, 3, 11, 2632

Offset: 0

Author

N. J. A. Sloane, May 30 2012. I added Hans Havermann's comment May 31 2012

Keywords

• nonn

Comments

The next term may be very large, see A212815.

Comment from Hans Havermann, Sequence Fans Mailing List, May 31 2012: The 11 numbers k for which A212813(k)=2 are 9, 11, 14, 20, 24, 27, 28, 40, 45, 48, 54. Empirically, it appears that 2632 is the sum of the number of prime partitions (A000607) of the eleven numbers 8, 10, 13, 19, 23, 26, 27, 39, 44, 47, 53. I hesitate turning this into a conjecture only because the 3 numbers k for which A212813(k)=1 are 7, 10, 12 and the sum of the number of prime partitions of the three numbers 6, 9, 11 is twelve, not eleven (the extra partition being, I think, 2+2+2).

Examples

			The 11 numbers k for which A212813(k)=2 are 9, 11, 14, 20, 24, 27, 28, 40, 45, 48, 54 (see A212816).
		

References

• Bellamy, O. S.; Cadogan, C. C. Subsets of positive integers: their cardinality and maximality properties. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 167--178, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561043 (82b:10006)

Links

• Hans Havermann, Conjecture regarding A212814(4)

Crossrefs

Cf. A036288, A212813, A212815, A212816, A212908, A212909.