A168157 Number of 0's in the matrix whose lines are the binary expansion of the first n primes.

1, 1, 4, 4, 9, 10, 19, 21, 22, 23, 23, 37, 40, 42, 43, 45, 46, 47, 69, 72, 76, 78, 81, 84, 88, 91, 93, 95, 97, 100, 100, 136, 141, 145, 149, 152, 155, 159, 162, 165, 168, 171, 172, 177, 181, 184, 187, 188, 191, 194, 197, 198, 201, 202, 263, 268, 273, 277, 282, 287

Offset: 1

Author

M. F. Hasler, Nov 21 2009

Keywords

• base
• nonn

Comments

The matrix is to be taken of minimal size, i.e., have n lines and the number of columns needed to write the n-th prime in the last line, A035100(n). Otherwise said, there is no zero column except for n=1 (prime(1) = 2 = 10[2] in binary).

The number of zeros in the last line of the matrix is given by A035103(n).

One has a(n)=a(n-1) iff n = A059305(k) for some k, i.e. prime(n) is a Mersenne prime A000668(k) = A000225(A000043(k)).

If prime(n)=2^2^k+1 is a Fermat prime (A019434), n>2, then one has a(n)=a(n-1)+n-1+2^k-1.

More generally, the "big jumps" a(n+1) > a(n)+n happen whenever a column is added, i.e. when prime(n) = A014234(k) <=> prime(n+1) = A104080(k) for some k,n>1.

Examples

			a(4)=4 is the number of zeros in the matrix [010] /* = 2 in binary */ [011] /* = 3 in binary */ [101] /* = 5 in binary */ [111] /* = 7 in binary */
		

Programs

PARI

A168157(n)=n*#binary(prime(n))-sum(i=1,n,norml2(binary(prime(i))))

Formula

a(n)=n*A035100(n)-A095375(n).