|
| |
|
|
A098756
|
|
Smallest available integer which fits into the repeating pattern 9876543210.
|
|
1
|
|
|
|
9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 98, 76, 54, 32, 10, 987, 65, 43, 210, 9876, 543, 2109, 87, 654, 3210, 98765, 432, 109, 876, 5432, 1098, 765, 43210, 987654, 32109, 8765, 432109, 87654, 321098, 7654, 3210987, 6543, 21098, 76543, 210987, 65432, 10987, 6543210
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
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. - Hagen von Eitzen, Aug 16 2009
|
|
|
LINKS
|
|
|
|
FORMULA
|
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
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
|
|
|
EXAMPLE
|
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.
|
|
|
PROG
|
(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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
base,easy,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
