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%I #28 Jan 17 2024 10:54:35
%S 1,2,3,4,5,6,7,8,41,50,126,127,468,469,1824,8052,8295,9857,1198372,
%T 3357009,3357010,6287267,156608073,156608074,403584750,403584751,
%U 586638974,3302332571,42256814922,42256814923,114842637961,155896317510,552468844242,552468844243,647871937482,686031429775
%N Base-9 Armstrong or narcissistic numbers (written in base 10).
%C From _M. F. Hasler_, Nov 20 2019: (Start)
%C 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 9 digits of n), but here only positive numbers are considered.
%C Terms a(n+1) = a(n) + 1 (n = 11, 13, 20, 23, 25, 29, 33, 48, 51, 57) correspond to solutions a(n) that are multiples of 9, in which case a(n) + 1 is also a solution. (End)
%H Joseph Myers, <a href="/A010353/b010353.txt">Table of n, a(n) for n = 1..58</a> (the full list of terms, from Winter)
%H René-Louis Clerc, <a href="https://hal.science/hal-04376934">Perfect r-narcissistic numbers in any base</a>, hal-04376934, 2024.
%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://ftp.cwi.nl/dik/Armstrong">Table of Armstrong Numbers</a>
%e 126 = 150_9 (= 1*9^2 + 5*9^1 + 0*9^0) = 1^3 + 5^3 + 0^3. It is easy to see that 126 + 1 then also satisfies this relation, as for all other terms that are multiples of 9. - _M. F. Hasler_, Nov 20 2019
%t Select[Range[9^7], # == Total[IntegerDigits[#, 9]^IntegerLength[#, 9]] &] (* _Michael De Vlieger_, Jan 17 2024 *)
%o (PARI) select( {is_A010353(n)=n==vecsum([d^#n|d<-n=digits(n,9)])}, [0..10^4]) \\ This gives only terms < 10^6, for illustration of is_A010353(). - _M. F. Hasler_, Nov 20 2019
%Y Cf. A010352 (a(n) written in base 9).
%Y In other bases: A010344 (base 4), A010346 (base 5), A010348 (base 6), A010350 (base 7), A010354 (base 8), A005188 (base 10), A161948 (base 11), A161949 (base 12), A161950 (base 13), A161951 (base 14), A161952 (base 15), A161953 (base 16).
%K base,fini,full,nonn
%O 1,2
%A _N. J. A. Sloane_
%E Edited by _Joseph Myers_, Jun 28 2009
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