OFFSET
1,1
COMMENTS
Consider the final n decimal digits of 2^j for all values of j. They are periodic. Sequence gives maximal value seen in these n digits.
With f(n)=a(n+1)-a(n), the difference f(n)-a(n) is always 8*10^n meaning that a(n) becomes its own "first differences" sequence when each term is prefixed a digit '8'. For higher order differences, the prefix 8 becomes: 8*10^n*sum_{k=0..m-1} 9^k where m is the order. - R. J. Cano, May 11 2014
LINKS
FORMULA
a(n) = 10^n-2^n = 2^n*(5^n-1).
a(n) = 12*a(n-1) - 20*a(n-2); O.g.f.:1/(1-10x)-1/(1-2x). - Geoffrey Critzer, Dec 15 2011
a(n) = f(n,0); Given f(x,y)=sum{j=0..x+y-1}(2^(3*x-2*j)*binomial(x,j)). - R. J. Cano, May 15 2014
a(n) = 2^(n+2)*A003463(n). - R. J. Cano, Sep 25 2014
a(n) = 8*A016134(n-1). - R. J. Mathar, Mar 10 2022
EXAMPLE
Maximum of the last 4 digits of powers of 2 is 9984=10000-16. It occurs at 2^254. 2^254=289480223.....01978282409984 (with 77 digits, last 4 ones are ...9984). The period length of the last-4-digit segment is A005054(4)=500. For n=4 period: amplitude=9984, phase=254.
MAPLE
MATHEMATICA
RecurrenceTable[{a[n] == 12 a[n - 1] - 20 a[n - 2], a[0] == 0, a[1] == 8}, a[n], {n, 1, 20}] (* Geoffrey Critzer, Dec 15 2011*)
PROG
(Sage) [10^n - 2^n for n in range(1, 19)] # Zerinvary Lajos, Jun 05 2009
(PARI) a(n)=sum(j=0, n-1, 2^(3*n-2*j)*binomial(n, j)) \\ R. J. Cano, May 15 2014
(Magma) [10^n-2^n : n in [1..20]]; // Wesley Ivan Hurt, Sep 25 2014
(PARI) A060458(n)=(5^n-1)<<n \\ M. F. Hasler, Oct 31 2014
CROSSREFS
KEYWORD
base,nonn,easy
AUTHOR
Labos Elemer, Apr 09 2001
EXTENSIONS
Edited by M. F. Hasler, Oct 31 2014
STATUS
approved