A132674 a(1)=1, a(n) = 10*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

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

Offset

1,2

Comments

Also: a(1)=1, a(n) = maximal positive integer < a(n-1) not yet in the sequence, if it exists, 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 permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

Links

Table of n, a(n) for n=1..69.

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).

Crossrefs

For parameters p=2 to p=9 see A132666 - A132673.

For a similar recurrence rule concerning Fibonacci and Lucas numbers see A132664 and A132665.

Sequence in context: A143473 A055121 A249591 * A090293 A164732 A070562

Adjacent sequences: A132671 A132672 A132673 * A132675 A132676 A132677

Keyword

nonn

Author

Hieronymus Fischer, Aug 24 2007, Sep 15 2007

Status

approved