A098756 - OEIS

%I #10 Jun 10 2018 04:22:02

%S 9,8,7,6,5,4,3,2,1,0,98,76,54,32,10,987,65,43,210,9876,543,2109,87,

%T 654,3210,98765,432,109,876,5432,1098,765,43210,987654,32109,8765,

%U 432109,87654,321098,7654,3210987,6543,21098,76543,210987,65432,10987,6543210

%N Smallest available integer which fits into the repeating pattern 9876543210.

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

%C 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. - _Hagen von Eitzen_, Aug 16 2009

%F 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_, Jan 04 2005

%F 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. - _Hagen von Eitzen_, Aug 16 2009

%e 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.

%o (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); \\ _Hagen von Eitzen_, Aug 16 2009

%K base,easy,nonn

%O 1,1

%A _Eric Angelini_, Oct 01 2004

%E Corrected and extended by _Sam Alexander_, Jan 04 2005

%E More terms from _Hagen von Eitzen_, Aug 16 2009