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 A060458 Maximal value seen in the final n decimal digits of 2^j for all values of j. 5
 8, 96, 992, 9984, 99968, 999936, 9999872, 99999744, 999999488, 9999998976, 99999997952, 999999995904, 9999999991808, 99999999983616, 999999999967232, 9999999999934464, 99999999999868928, 999999999999737856 (list; graph; refs; listen; history; text; internal format)
 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 Index entries for linear recurrences with constant coefficients, signature (12,-20). 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 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 A060458:=n->10^n-2^n: seq(A060458(n), n=1..20); # Wesley Ivan Hurt, Sep 25 2014 MATHEMATICA RecurrenceTable[{a[n] == 12 a[n - 1] - 20 a[n - 2], a == 0, a == 8}, a[n], {n, 1, 20}]  (* Geoffrey Critzer, Dec 15 2011*) PROG (Sage) [10^n - 2^n for n in xrange(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)<

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Last modified November 22 03:43 EST 2019. Contains 329388 sequences. (Running on oeis4.)