A060460 Consider the final n decimal digits of 2^j for all values of j. They are periodic. Sequence gives position (or phase) of the maximal value seen in these n digits.
3, 12, 53, 254, 1255, 6256, 31257, 156258, 781259, 3906260, 19531261, 97656262, 488281263, 2441406264, 12207031265, 61035156266, 305175781267, 1525878906268, 7629394531269, 38146972656270, 190734863281271
Offset: 1
Examples
a(2) = 5*3-(3+4*0) = 15-3 = 12, etc...
For n=2, the last 2 digits of powers of 2 have the period {2,4,8,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,4,8,16,32} displayed in A000855. The maximum is 96 and it occurs at 2^12=4096. So a(2)=12.
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
- Index entries for sequences related to final digits of numbers
- Index entries for linear recurrences with constant coefficients, signature (7, -11, 5).
Programs
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Mathematica
nxt[{n_,a_,b_}]:={n+1,b,5b-(3+4(n-1))}; NestList[nxt,{2,3,12},20][[All,2]] (* or *) Table[2*5^(n-1)+n,{n,30}] (* or *) LinearRecurrence[{7,-11,5},{3,12,53},30] (* Harvey P. Dale, Aug 01 2021 *)
Formula
a(1) = 3, a(n) = 5*a(n-1)-(3+4*(n-2)).
a(n) = a(n) = 2*5^(n-1) + n.
G.f.: (-3 + 9 x - 2 x^2)/((-1 + x)^2 (-1 + 5 x)) - Harvey P. Dale, Aug 01 2021
Extensions
Offset 1 (and formulas adapted) from Michel Marcus, Mar 25 2020
Comments