1,1

a(n) must be chosen so its rightmost digit is not 1, with the exception a(n)=1.

If n > 14 then a(n+81) is a(n) augmented by a full ten digit period (suitably rotated). Hence for n > 14 we have a(n+81) = 10^10 * a(n) + c(n mod 81). For c one easily finds c(0) = 5432109876, c(1) = 4321098765, c(2) = c(0), c(3) = 3210987654, c(4) = c(5) = c(1), c(6) = c(7) = c(3), c(8) = 2109876543 etc. [From Hagen von Eitzen (math(AT)von-eitzen.de), Aug 16 2009]

To calculate the n-th term (n>10), find the greatest i<n such that first digit of a(i)=(last digit of a(n-1))-1 mod 10; then a(n) is a(i) augmented with the next digit (or next 2 digits if next digit is 1) - Sam Alexander (amnalexander(AT)yahoo.com), Jan 04 2005

G.f.: P(x)/((1-x^81)*(1-10^10*x^81)) with a degree 176 polynomial P(x) = 9*x + 8*x^2 + 7*x^3 + 6*x^4 + ... + 760000000000*x^174 + 520000000000*x^175 + 320000000000*x^176. [From Hagen von Eitzen (math(AT)von-eitzen.de), Aug 16 2009]

To find the next term after 5432, we look for the most recent term beginning with 1 (1=2-1), which is 109. Augment 109 to get 1098.

(PARI) M=vector(9, x, x); for(n=1, 10, print(n, " ", 10-n)); s=9; for(n=11, 1000, d=M[s]%10; M[s]=10*M[s]+d-1; if(d==0, M[s]+=10); if(d==2, M[s]*=10); print(n, " ", M[s]); s=(M[s]-1)%10); [From Hagen von Eitzen (math(AT)von-eitzen.de), Aug 16 2009]

Sequence in context: A109910 A084019 A061601 this_sequence A112454 A138531 A030076

Adjacent sequences: A098753 A098754 A098755 this_sequence A098757 A098758 A098759

base,easy,nonn

Eric Angelini (eric.angelini(AT)kntv.be), Oct 01 2004

Corrected and extended by Sam Alexander (amnalexander(AT)yahoo.com), Jan 04 2005

More terms from Hagen von Eitzen (math(AT)von-eitzen.de), Aug 16 2009