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Search: id:A119387
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| A119387 |
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a(n) = number of binary digits (1's and nonleading 0's) which remain unchanged in their positions when n and (n+1) are written in binary. |
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+0
1
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| 0, 0, 1, 0, 2, 1, 2, 0, 3, 2, 3, 1, 3, 2, 3, 0, 4, 3, 4, 2, 4, 3, 4, 1, 4, 3, 4, 2, 4, 3, 4, 0, 5, 4, 5, 3, 5, 4, 5, 2, 5, 4, 5, 3, 5, 4, 5, 1, 5, 4, 5, 3, 5, 4, 5, 2, 5, 4, 5, 3, 5, 4, 5, 0, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 2, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 6, 5, 6, 4, 6, 5, 6, 3, 6
(list; graph; listen)
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OFFSET
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0,5
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FORMULA
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a(n) = A048881(n) + A086784(n+1). (A048881(n) is the number of 1's which remain unchanged between binary n and (n+1). A086784(n+1) is the number of nonleading 0's which remain unchanged between binary n and (n+1).)
a(A000225(n))=0. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 29 2006
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EXAMPLE
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9 in binary is 1001. 10 (decimal) is 1010 in binary. 2 binary digits remain unchanged (the leftmost two digits) between 1001 and 1010. So a(9) = 2.
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PROGRAM
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(C) #include <stdio.h> #define NMAX 200 int sameD(int a, int b) { int resul=0 ; while(a>0 && b >0) { if( (a &1) == (b & 1)) resul++ ; a >>= 1 ; b >>= 1 ; } return resul ; } int main(int argc, char*argv[]) { for(int n=0; n<NMAX; n++) printf("%d, ", sameD(n, n+1)) ; return 0 ; } - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 29 2006
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CROSSREFS
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Cf. A048881, A086784.
Sequence in context: A127969 A081733 A102587 this_sequence A055941 A068076 A138498
Adjacent sequences: A119384 A119385 A119386 this_sequence A119388 A119389 A119390
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KEYWORD
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easy,nonn
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AUTHOR
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Leroy Quet (qq-quet(AT)mindspring.com), Jul 26 2006
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EXTENSIONS
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More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 29 2006
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