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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
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
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
Sequence in context: A307340 A084019 A061601 * A267193 A112454 A304621
KEYWORD
base,easy,nonn
AUTHOR
Eric Angelini, Oct 01 2004
EXTENSIONS
Corrected and extended by Sam Alexander, Jan 04 2005
More terms from Hagen von Eitzen, Aug 16 2009
STATUS
approved