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.

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

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