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).
From Geoffrey Critzer, Dec 15 2011: (Start)
a(n) = 12*a(n-1) - 20*a(n-2).
O.g.f.: 1/(1-10*x) - 1/(1-2*x). (End)
a(n) = f(n,0) where 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
E.g.f.: exp(2*x)*(exp(8*x) - 1). - Elmo R. Oliveira, Mar 26 2025
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
(SageMath) [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
More terms from Elmo R. Oliveira, Mar 26 2025
STATUS
approved
