A089398 -id:A089398 - cp's OEIS frontend

A089400 a(n) = m - A089398(2^m - n) for m>=n.

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

Offset: 0

Paul D. Hanna, Oct 30 2003

nonn

A089398(n) = n-th column sum of binary digits of k*2^(k-1), where summation is over k>=1, without carrying between columns.

			a(6)=4 since 7 - A089398(2^7 - 6) = 7 - 3 = 4.

Cf. A089398, A089401.

a(2^k)=1 (for k>1), a(2^k+j)=1+a(j) (for 2^k-k>j>=0), a(2^k-j)=1+A089401(j) (for k>j>0).

A089401 a(n) = m - A089398(2^m + n) for m>=n.

Original entry on oeis.org

1, 1, 3, 2, 4, 5, 6, 5, 7, 8, 11, 9, 11, 12, 13, 12, 14, 15, 18, 18, 19, 20, 21, 20, 22, 23, 26, 24, 26, 27, 28, 27, 29, 30, 33, 33, 36, 36, 37, 36, 38, 39, 42, 40, 42, 43, 44, 43, 45, 46, 49, 49, 50, 51, 52, 51, 53, 54, 57, 55, 57, 58, 59, 58, 60, 61, 64, 64, 67, 69, 69, 68, 70

Offset: 1

Paul D. Hanna, Oct 30 2003

nonn

A089398(n) = n-th column sum of binary digits of k*2^(k-1), where summation is over all k>=1, without carrying from columns sums that may exceed 2.

Row sums of triangular arrays in A103582 and in A103583. - Philippe Deléham, Apr 04 2005

			a(6)=5 since 7 - A089398(2^7 + 6) = 7 - 2 = 5.

Cf. A089398, A089400.

f[n_] := Block[{lg = Floor[Log[2, n]] + 1}, Sum[ Join[ Reverse[ IntegerDigits[n - i + 1, 2]], {0}][[i]], {i, lg}]]; Table[n - f[2^n + n] + 2, {n, 0, 72}] (* Robert G. Wilson v, Mar 29 2005 *)

a(n)=n/2+1/2*sum(k=1,n,(-1)^floor((n-k)/2^(k-1))) \\ Benoit Cloitre

{a(n)=if(n<=0,0,m=floor(log(n)/log(2)); if(n-2^m<=m,n-m+a(n-2^m),2^m-1+a(n-2^m)))} \\ Paul D. Hanna, Mar 28 2005

a(n) = n/2+1/2*sum(k=1, n, (-1)^floor((n-k)/2^(k-1))). - Benoit Cloitre, Mar 28 2005

Let a(0)=0; when n - 2^[log_2(n)] <= [log_2(n)] then a(n) = a(n - 2^[log_2(n)]) + n - [log_2(n)], else a(n) = a(n - 2^[log_2(n)]) + 2^[log_2(n)] - 1. Thus a(2^m) = 2^m - m for all m>=0; for 0<=k<=m: a(2^m + k) = a(k) + 2^m + k - m; for mPaul D. Hanna, Mar 28 2005

More terms from Benoit Cloitre and Robert G. Wilson v, Mar 28 2005

A089399 a(n) satisfies: 2^a(n)+1 = sum(k=1,n, A089398(k)*2^(k-1)) for n>2, with a(1)=a(2)=0.

Original entry on oeis.org

0, 0, 3, 4, 5, 6, 8, 9, 10, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 51, 52, 53, 54, 55, 57, 58, 59, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 69, 70, 71, 72, 73

Offset: 1

Paul D. Hanna, Oct 30 2003

nonn

A089398(n) = n-th column sum of binary digits of k*2^(k-1), where summation is over all k>=1, without carrying from columns sums that may exceed 2.

			a(7)=8 since 2^8+1=257=(1)+(0)2+(2)2^2+(1)2^3+(1)2^4+(1)2^5+(3)2^6,
and A089398 begins: {1,0,2,1,1,1,3,2,2,0,3,2,2,2,4,3,3,1,...}.

Cf. A089398.

A103581 A102371 written in base 2.

Original entry on oeis.org

1, 10, 111, 1100, 11101, 111110, 1111011, 11111000, 111111001, 1111111010, 11111111111, 111111110100, 1111111110101, 11111111110110, 111111111110011, 1111111111110000, 11111111111110001, 111111111111110010

Offset: 1

Philippe Deléham, Mar 23 2005

The number of zeros in the n-th term appears to match A089398. - Benoit Cloitre, Mar 24 2005

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.

Cf. A007088, A102371, A103583, A103582.

a(n) = A007088(A102371(n)). - Michel Marcus, May 08 2020

More terms from Benoit Cloitre, Mar 24 2005

A105034 Binary equivalents of A105033.

Original entry on oeis.org

0, 1, 0, 11, 10, 1, 100, 111, 110, 101, 0, 1011, 1010, 1001, 1100, 1111, 1110, 1101, 1000, 11, 10010, 10001, 10100, 10111, 10110, 10101, 10000, 11011, 11010, 11001, 11100, 11111, 11110, 11101, 11000, 10011, 10, 100001, 100100, 100111, 100110

Offset: 0

N. J. A. Sloane, Apr 04 2005

Number of 1's in a(n) is A089398(n). - Philippe Deléham, Apr 05 2005.

The version 0, 01, 000, 0011, 00010, 000001, ... is obtained by interchanging 0 and 1 in A103581: 1, 10, 111, 1100, 11101, 111110, .... - Philippe Deléham, Apr 07 2005

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.

Cf. A105033, A102370.

Cf. triangular array in A103589.

More terms from Benoit Cloitre, Apr 04 2005

A103175 A001787 written in base 2.

Original entry on oeis.org

0, 1, 100, 1100, 100000, 1010000, 11000000, 111000000, 10000000000, 100100000000, 1010000000000, 10110000000000, 110000000000000, 1101000000000000, 11100000000000000, 111100000000000000

Offset: 0

Philippe Deléham, Mar 30 2005

Cf. A089398.

More terms from Erich Friedman, Aug 08 2005

cp's OEIS Frontend

A089400 a(n) = m - A089398(2^m - n) for m>=n.

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A089401 a(n) = m - A089398(2^m + n) for m>=n.

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A089399 a(n) satisfies: 2^a(n)+1 = sum(k=1,n, A089398(k)*2^(k-1)) for n>2, with a(1)=a(2)=0.

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A103581 A102371 written in base 2.

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A105034 Binary equivalents of A105033.

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A103175 A001787 written in base 2.

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