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

Offset: 0

Author

Richard Bean, Feb 22 2003

Keywords

• easy
• nonn

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^n-1) is that given in Fu, Fu and Liao. This function gives an easy way to generate these two constructions.

From Gus Wiseman, Mar 30 2019: (Start)

Also the number of ordered pairs of positive integers up to n with at least one binary carry. A binary carry of two positive integers is an overlap of the positions of 1's in their reversed binary expansion. For example, the a(2) = 1 through a(6) = 15 ordered pairs are:

(1,1) (1,1) (1,1) (1,1) (1,1)

(2,2) (1,3) (1,3) (1,3)

(2,2) (2,2) (1,5)

(2,3) (2,3) (2,2)

(3,1) (3,1) (2,3)

(3,2) (3,2) (3,1)

(3,3) (3,3) (3,2)

(4,4) (3,3)

(3,5)

(4,4)

(4,5)

(5,1)

(5,3)

(5,4)

(5,5)

(End)

a(n) is also the number of even elements in the n X n symmetric Pascal matrix. - Stefano Spezia, Nov 14 2022

References

• C. Fu, H. Fu and W. Liao, A new construction for a critical set in special Latin squares, Proceedings of the Twenty-sixth Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, Florida, 1995), Congressus Numerantium, Vol. 110 (1995), pp. 161-166.
• 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. 441-456.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000
• R. Bean, Three problems on partial Latin squares, Problem 418 (BCC19,2), Discrete Math., 293 (2005), 314-315.
• J. M. Dover, On two OEIS conjectures, arXiv:1606.08033 [math.CO], 2016.
• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 29.

Crossrefs

Cf. A063437.

Cf. A000120, A005061, A006218, A050315, A155609, A247935, A267610, A267700.

Cf. A325096, A325098, A325102, A325103, A325104, A325106, A325124.

Programs

Maple

f:=proc(n) option remember; local t;
if n <= 1 then 0
elif (n mod 2) =  0 then 3*f(n/2)+(n/2)^2
else t:=(n-1)/2; f(t)+2*f(t+1)+t^2-1; fi; end;
[seq(f(n),n=0..100)]; # N. J. A. Sloane, Jul 01 2017

Mathematica

a[0] = a[1] = 0; a[n_] := a[n] = If[EvenQ[n], 3*a[n/2] + n^2/4, 2*a[(n-1)/2 + 1] + a[(n-1)/2] + (1/4)*(n-1)^2 - 1];
Array[a, 60, 0] (* Jean-François Alcover, Dec 09 2017, from Dover's formula *)
Table[Length[Select[Tuples[Range[n-1],2],Intersection[Position[Reverse[IntegerDigits[#[[1]],2]],1],Position[Reverse[IntegerDigits[#[[2]],2]],1]]!={}&]],{n,0,20}] (* Gus Wiseman, Mar 30 2019 *)

Formula

a(2^n) = 4^n-3^n = A005061(n); a(2^n+1) = 4^n-3^n+1 = A155609(n); a(2^n-1) = 4^n-3^n-2^(n+1)+3.

a(0)=a(1)=0, a(2n) = 3a(n)+n^2, a(2n+1) = a(n)+2a(n+1)+n^2-1. This was proved by Jeremy Dover. - Ralf Stephan, Dec 08 2004

a(n) = (A325104(n) - n)/2. - Gus Wiseman, Mar 30 2019