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Search: id:A082907
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| A082907 |
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A function applied to entries of Pascal-triangle as follows: C[n,j] is replaced by GCD[2^n,C[n,j]], providing largest power of 2 dividing C[n,j]. |
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+0
3
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| 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 2, 4, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 4, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 4, 8, 2, 8, 4, 8, 1, 1, 1, 4, 4, 2, 2, 4, 4, 1, 1, 1, 2, 1, 8, 2, 4, 2, 8, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 4, 2, 4, 1, 8, 4, 8, 1, 4, 2, 4, 1, 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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If N is a power of 2, then then the first N rows are invariant under all 6 symmetries of an equilateral triangle. - Paul Boddington (psb(AT)maths.warwick.ac.uk), Dec 17 2003
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FORMULA
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T[n, j] = a[n]/(a[j]*a[n-j]) where a[n]=A060818[n]. T[n, j] = (b[j]*b[n-j])/b[n] where b[n]=A001316[n] (Gould's sequence). - Paul Boddington (psb(AT)maths.warwick.ac.uk), Dec 17 2003
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EXAMPLE
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Triangle read by rows:
1,
1,1,
1,2,1,
1,1,1,1,
1,4,2,4,1,
1,1,2,2,1,1,
1,2,1,4,1,2,1,
1,1,1,1,1,1,1,1,
1,8,4,8,2,8,4,8,1,
1,1,4,4,2,2,4,4,1,1,
n=-1+2^k: such rows consist of 1-s since all C[n,j]-s are odd.
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MATHEMATICA
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Flatten[Table[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}], {n, 0, 25}], 1]
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CROSSREFS
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Cf. A000005, A007318, A000079.
Sequence in context: A091255 A135303 A036065 this_sequence A146532 A119335 A155869
Adjacent sequences: A082904 A082905 A082906 this_sequence A082908 A082909 A082910
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KEYWORD
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nonn,tabl
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Apr 23 2003
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