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 A008952 Leading digit of 2^n. 15
 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 6, 1, 2, 5, 1, 2, 4, 8, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 2, 5, 1, 2, 4, 9, 1, 3, 7, 1, 3, 6, 1, 2, 4, 9, 1, 3, 7, 1, 3, 6, 1, 2, 4, 9, 1, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Statistically, sequence obeys Benford's law, i.e. digit d occurs with probability log_10(1 + 1/d); thus 1 appears about 6.6 times more often than 9. - Lekraj Beedassy, May 04 2005 The most significant digits of the n-th powers of 2 are not cyclic and in the first 1000000 terms, 1 appears 301030 times, 2 appears 176093, 3 appears 124937, 4 appears 96911, 5 appears 79182, 6 appears 66947, 7 appears 57990, 8 appears 51154 and 9 appears 45756 times. - Robert G. Wilson v, Feb 03 2008 In fact the sequence follows Benford's law precisely by the equidistribution theorem. - Charles R Greathouse IV, Oct 11 2015 LINKS Robert G. Wilson v, Table of n, a(n) for n = 0..100000. Brady Haran and Dmitry Kleinbock, Powers of 2, Numberphile video (2015). More footage. Jaap Spies, A Bit of Math, The Art of Problem Solving, Jaap Spies Publishers (2019). Wikipedia, Benford's law. Wikipedia, Zipf's law. FORMULA a(n) = [2^n / 10^([log_10(2^n)])] = [2^n / 10^([n*log_10(2)])]. a(n) = A000030(A000079(n)). - Omar E. Pol, Jul 04 2019 MAPLE a:= n-> parse(""||(2^n)[1]): seq(a(n), n=0..100); # Alois P. Heinz, Aug 06 2021 MATHEMATICA a[n_] := First@ IntegerDigits[2^n]; Array[a, 105, 0] (* Robert G. Wilson v, Feb 03 2008 and corrected Nov 24 2014 *) PROG (PARI) a(n)=digits(2^n)[1] \\ Charles R Greathouse IV, Oct 11 2015 (Python) def A008952(n): return int(str(1<

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Last modified November 30 12:40 EST 2022. Contains 358441 sequences. (Running on oeis4.)