A060458 Maximum value seen in the final n decimal digits of 2^j for all values of j.
8, 96, 992, 9984, 99968, 999936, 9999872, 99999744, 999999488, 9999998976, 99999997952, 999999995904, 9999999991808, 99999999983616, 999999999967232, 9999999999934464, 99999999999868928, 999999999999737856, 9999999999999475712, 99999999999998951424, 999999999999997902848
Offset: 1
Examples
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.
Links
Programs
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Magma
[10^n-2^n : n in [1..20]]; // Wesley Ivan Hurt, Sep 25 2014
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Maple
A060458:=n->10^n-2^n: seq(A060458(n), n=1..20); # Wesley Ivan Hurt, Sep 25 2014
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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*) -
PARI
a(n)=sum(j=0,n-1,2^(3*n-2*j)*binomial(n,j)) \\ R. J. Cano, May 15 2014
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PARI
A060458(n)=(5^n-1)<
M. F. Hasler, Oct 31 2014 -
Sage
[10^n - 2^n for n in range(1,19)] # Zerinvary Lajos, Jun 05 2009
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
Extensions
Edited by M. F. Hasler, Oct 31 2014
More terms from Elmo R. Oliveira, Mar 26 2025
Comments