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Search: id:A054898
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| A054898 |
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Sum_{k>0} floor(n/9^k). |
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3
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| 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11
(list; graph; listen)
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OFFSET
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0,19
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COMMENT
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Different from the highest power of 9 dividing n!.
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FORMULA
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floor[n/9] + floor[n/81] + floor[n/729] + floor[n/6561] + ....
a(n)=(n-A053830(n))/8
Recurrence: a(n)=floor(n/9)+a(floor(n/9)); a(9*n)=n+a(n); a(n*9^m)=n*(9^m-1)/8+a(n). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
a(k*9^m)=k*(9^m-1)/8, for 0<=k<9, m>=0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
Asymptotic behavior: a(n)=n/8+O(log(n)), a(n+1)-a(n)=O(log(n)); this follows from the inequalities below. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
a(n)<=(n-1)/8; equality holds for powers of 9. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
a(n)>=(n-8)/8-floor(log_9(n)); equality holds for n=9^m-1, m>0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
lim inf (n/8-a(n))=1/8, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
lim sup (n/8-log_9(n)-a(n))=0, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
lim sup (a(n+1)-a(n)-log_9(n))=0, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
G.f.: g(x)=sum{k>0, x^(9^k)/(1-x^(9^k))}/(1-x). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 14 2007
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MATHEMATICA
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Table[t = 0; p = 9; While[s = Floor[n/p]; t = t + s; s > 0, p *= 9]; t, {n, 0, 100} ]
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CROSSREFS
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Cf. A011371 and A054861 for analogues involving powers of 2 and 3.
Cf. A054899, A067080, A098844, A132033.
Sequence in context: A111856 A111857 A133879 this_sequence A167383 A167661 A102682
Adjacent sequences: A054895 A054896 A054897 this_sequence A054899 A054900 A054901
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KEYWORD
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nonn
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AUTHOR
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Henry Bottomley (se16(AT)btinternet.com), May 23 2000
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