A159780 Inner product of the binary representation of n and its reverse.

0, 1, 0, 2, 0, 2, 1, 3, 0, 2, 0, 2, 0, 2, 2, 4, 0, 2, 0, 2, 1, 3, 1, 3, 0, 2, 2, 4, 1, 3, 3, 5, 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, 2, 4, 2, 4, 0, 2, 2, 4, 0, 2, 2, 4, 0, 2, 2, 4, 2, 4, 4, 6, 0, 2, 0, 2, 0, 2, 0, 2, 1, 3, 1, 3, 1, 3, 1, 3, 0, 2, 0, 2, 2, 4, 2, 4, 1, 3, 1, 3, 3, 5, 3, 5, 0, 2, 2, 4, 0, 2, 2, 4, 1

Offset: 0

Author

T. D. Noe, Apr 22 2009

Keywords

• nonn
• base

Comments

a(n) gives the number of 1's that coincide in the binary representation of n and its reverse. For the n in A140900, we have a(n)=0. The number k first appears at n=2^k-1.

Also central terms and right edge of the triangle in A173920: a(n)=A173920(2*n,n)=A173920(n,n). [From Reinhard Zumkeller, Mar 04 2010]

a(A000225(n)) = n and a(m) < n for m < A000225(n). [Reinhard Zumkeller, Oct 21 2011]

a(n) = sum(A030308(n,k)*A030308(n,A070939(n)-1-k): k = 0..A070939(n)-1). - Reinhard Zumkeller, Mar 10 2013

Examples

			14 is represented by the binary vector (1,1,1,0). The reverse is (0,1,1,1). The inner product is 1*0+1*1+1*1+0*1 = 2. Hence a(14) = 2.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A216176.

Programs

Haskell

a159780 n = sum $ zipWith (*) bs $ reverse bs
   where bs = a030308_row n
-- Reinhard Zumkeller, Mar 10 2013, Oct 21 2011

Mathematica

Table[d=IntegerDigits[n,2]; d.Reverse[d], {n,0,1023}]