A006063 A card-arranging problem: values of n such that there exists a permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a cube for every i.

7, 19, 26, 37, 44, 56, 63, 66, 68, 80, 82, 85, 87, 98, 100, 103, 105, 110, 112, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 135, 147, 149, 150, 151, 152, 155, 156, 159, 171, 173, 174, 175, 176, 177, 178, 179

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn

Comments

Apparently Gardner (1975) quotes Papaikonomou as showing that there can be at most one solution for a given n. However, this is incorrect: see A096680 for n values with more than one such permutation. - Ray Chandler

For any n, the number of permutations is permanent(m), where the n X n matrix m is defined m(i,j) = 1 or 0, depending on whether i+j is a cube or not. Hence, n is in this sequence if permanent(m) > 0.

References

• M. Gardner, Mathematical Games column, Scientific American, Mar 1975.
• M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 81.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A095986 (for squares), A096680.

Formula

Conjecture: a(n) = n + 124 for n >= 173, i.e. there is such a permutation for every n >= 173. Verified for 173 <= n <= 1000. - Robert Israel, Aug 28 2018

Extensions

Entry revised Jul 18 2004 based on comments from Franklin T. Adams-Watters

a(8) and later terms from Ray Chandler, Jul 26 2004