

A080572


Number of ordered pairs (i,j), 0 <= i,j < n, for which (i & j) is nonzero, where & is the bitwise AND operator.


0



0, 1, 2, 7, 8, 15, 24, 37, 38, 49, 62, 81, 98, 121, 146, 175, 176, 195, 216, 247, 272, 307, 344, 387, 420, 463, 508, 559, 608, 663, 720, 781, 782, 817, 854, 909, 950, 1009, 1070, 1141, 1190, 1257, 1326, 1405, 1478, 1561, 1646, 1737, 1802, 1885, 1970, 2065, 2154
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OFFSET

1,3


COMMENTS

Conjectured to be less than or equal to lcs(n) (see sequence A063437). The value of a(2^n) is that given in Stinson and van Rees and the value of a(2^n1) is that given in Fu, Fu and Liao. This function gives an easy way to generate these two constructions.


REFERENCES

C. Fu, H. Fu and W. Liao, A new construction for a critical set in special Latin squares, Proceedings of the Twentysixth Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, Florida, 1995), Congressus Numerantium, Vol. 110 (1995), pp. 161166.
D. R. Stinson and G. H. J. van Rees, Some large critical sets, Proceedings of the Eleventh Manitoba Conference on Numerical Mathematics and Computing (Winnipeg, Manitoba, 1981), Congressus Numerantium, Vol. 34 (1982), pp. 441456.


LINKS

Table of n, a(n) for n=1..53.
R. Bean, Three problems on partial Latin squares, Problem 418 (BCC19,2), Discrete Math., 293 (2005), 314315.
J. M. Dover, On two OEIS conjectures, arXiv:1606.08033 [math.CO], 2016.


FORMULA

a(2^n)=4^n3^n; a(2^n+1)=4^n3^n+1; a(2^n1)=4^n3^n2^(n+1)+3
a(0)=a(1)=0, a(2n) = 3a(n)+n^2, a(2n+1) = a(n)+2a(n+1)+n^21. This was proved by Jeremy Dover.  Ralf Stephan, Dec 08 2004


CROSSREFS

Cf. A063437.
Sequence in context: A064293 A063729 A023178 * A263602 A162664 A032689
Adjacent sequences: A080569 A080570 A080571 * A080573 A080574 A080575


KEYWORD

easy,nonn


AUTHOR

Richard Bean (rwb(AT)eskimo.com), Feb 22 2003


STATUS

approved



