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A053697 a(n+1)=a(n)+a^(n), where the addition is in base 11 and where a^(n) is obtained from a(n) by replacing each digit with its multiplicative inverse modulo 11. Zero digits, if any, are deleted. 0
1, 2, 8, 14, 27, 94, 137, 284, 947, 1384, 2847, 9484, 13847, 28484, 94847, 138484, 284847, 948484, 1384847, 2848484, 9484847, 13848484, 28484847, 94848484, 138484847, 284848484, 948484847, 1384848484, 2848484847, 9484848484 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Conjecture. For any positive integer a(1), the sequence generated according to the above rule eventually cycles through the forms a(k)=[1][4^a][3][(84)^b],..., a(k+6)=[1][4^a][3][(84)^(b+1)], or through a(k)=[1][5^a][4][(84)^b],..., a(k+6)=[1][5^a][4][(84)^(b+1)], for nonnegative integers a and b. The sequence listed above, with a(1)=1, is an example of the first type.

LINKS

Table of n, a(n) for n=1..30.

FORMULA

Conjectures from Colin Barker, Feb 15 2016: (Start)

a(n) = a(n-2) + 10*a(n-3) - 10*a(n-5) for n>7.

G.f.: x*(1+2*x+7*x^2+2*x^3-x^4+10*x^5-10*x^6+10*x^8) / ((1-x)*(1+x)*(1-10*x^3)). (End)

From Robert Israel, Feb 15 2016: (Start)

For k >= 1:

a(6*k-4) = 2*10^(2*k-2)  + 84*(10^(2*k-2)-1)/99.

a(6*k-3) = 94*10^(2*k-3) + 84*(10^(2*k-3)-10)/99 + 7.

a(6*k-2) = 13*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.

a(6*k-1) = 2*10^(2*k-1)  + 84*(10^(2*k-1)-10)/99 + 7.

a(6*k)   = 94*10^(2*k-2) + 84*(10^(2*k-2)-1)/99.

a(6*k+1) = 13*10^(2*k-1) + 84*(10^(2*k-1)-10)/99 + 7.

Colin Barker's conjectures follow from these. (End)

CROSSREFS

Sequence in context: A324785 A173974 A056677 * A092346 A196227 A166045

Adjacent sequences:  A053694 A053695 A053696 * A053698 A053699 A053700

KEYWORD

nonn,base,easy

AUTHOR

John W. Layman, Feb 14 2000

STATUS

approved

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Last modified April 4 21:43 EDT 2020. Contains 333238 sequences. (Running on oeis4.)