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A132674
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a(1)=1, a(n)=10*a(n-1) if the minimal natural number not encountered so far is greater than a(n-1), else a(n)=a(n-1)-1.
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0
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1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 110, 109, 108, 107, 106, 105, 104, 103, 102, 101, 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Also: a(1)=1, a(n)=maximal positive number <a(n-1) not encountered so far, if existing, else a(n)=10*a(n-1).
Also: a(1)=1, a(n)=a(n-1)-1, if a(n-1)-1>0 and has not been encountered so far, else a(n)=10*a(n-1).
A reordering of the natural numbers. The sequence is self-inverse, in that a(a(n))=n.
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FORMULA
| The following formulas are given for a general parameter p>2 considering the recurrence rule above (i.e. a(n)=p*a(n-1)...; p=10 for this sequence).
G.f.: g(x)=(x(1-2x)/(1-x)+px^2*f'(x^((2p-1)/(p-1)))+((2p-1)/p^2)*(f'(x^(1/(p-1)))-px-1)/(1-x) where f(x)=sum{k>=0, x^(p^k)} and f'(z)=derivative of f(x) at x=z.
a(n)=((3p-1)*p^(r/2)-p-1)/(p-1)-n if both, r and s are even, else a(n)=((p^2+2p-1)*p^((s-1)/2)-p-1)/(p-1)-n, where r=ceiling(2*log_p(((p-1)n+p)/(2p-1))) and s=ceiling(2*log_p(((p-1)n+p)/p)-1).
a(n)=(p^floor(1+(k+1)/2)+(2p-1)*p^floor(k/2)-p-1)/(p-1)-n, where k=r if r is odd, else k=s (with respect to r and s above; formally, k=((r+s)-(r-s)*(-1)^r)/2).
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CROSSREFS
| For parameters p=2 to p=9 see A132666 -132673.
For a similar recurrence rule concerning Fibonacci and Lucas numbers see A132664 and A132665.
Sequence in context: A178914 A143473 A055121 * A164732 A070562 A070641
Adjacent sequences: A132671 A132672 A132673 * A132675 A132676 A132677
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KEYWORD
| nonn
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AUTHOR
| Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 24 2007, Sep 15 2007
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