|
%I #31 Jan 17 2020 14:07:41
%S 1,2,3,4,5,6,7,24,64,134,205,463,660,661,40663,42710,42711,60007,
%T 62047,636703,3352072,3352272,3451473,4217603,7755336,16450603,
%U 63717005,233173324,3115653067,4577203604,61777450236,147402312024
%N Base-8 Armstrong or narcissistic numbers, written in base 8.
%C Whenever a term ends in 0, then a(n+1) = a(n) + 1 is also a term. Like the other single-digit terms, zero would satisfy the definition (n = Sum_{i=1..k} d[i]^k when d[1..k] are the base-8 digits of n), but here only positive numbers are considered. - _M. F. Hasler_, Nov 18 2019
%H Joseph Myers, <a href="/A010351/b010351.txt">Table of n, a(n) for n = 1..62</a> (the full list of terms, from Winter)
%H Gordon L. Miller and Mary T. Whalen, <a href="https://www.fq.math.ca/Scanned/30-3/miller.pdf">Armstrong Numbers: 153 = 1^3 + 5^3 + 3^3</a>, Fibonacci Quarterly, 30-3 (1992), 221-224.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NarcissisticNumber.html">Narcissistic Number</a>
%H D. T. Winter, <a href="http://web.archive.org/web/20100109234250/http://ftp.cwi.nl:80/dik/Armstrong">Table of Armstrong Numbers</a> (latest backup on web.archive.org from Jan. 2010; page no longer available), published not later than Aug. 2003.
%e 432 = 660_8 (= 6*8^2 + 6*8^1 + 0*8^0), and 6^3 + 6^3 + 0^3 = 432, therefore 660 is in the sequence. It's easy to see that 432 + 1 then also satisfies the equation, as for any term that is a multiple of 8. - _M. F. Hasler_, Nov 21 2019
%o (PARI) [fromdigits(digits(n,8))|n<-A010354] \\ _M. F. Hasler_, Nov 18 2019
%Y Cf. A010354 (a(n) written in base 10).
%Y In other bases: A010343 (base 4), A010345 (base 5), A010347 (base 6), A010349 (base 7), A010352 (base 9), A005188 (base 10).
%K base,fini,full,nonn
%O 1,2
%A _N. J. A. Sloane_
%E Edited by _Joseph Myers_, Jun 28 2009
|
