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%I #23 Nov 05 2020 06:45:46
%S 1,2,3,4,5,6,7,8,9,10,61,72,126,370,855,1161,1216,1280,10657,16841,
%T 16842,17864,17865,36949,36950,63684,66324,71217,90120,99594,99595,
%U 141424,157383,1165098,1165099,5611015,11959539,46478562,203821954,210315331,397800208,826098079,1308772162,1399714480
%N Base-11 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 11 digits of n), but here only positive numbers are considered.
%C Terms a(n+1) = a(n) + 1 (n = 20, 22, 24, 30, 34, 56, 67, 57, 195, ...) correspond to solutions a(n) that are multiples of 11, in which case a(n) + 1 is also a solution. (End)
%H Joseph Myers, <a href="/A161948/b161948.txt">Table of n, a(n) for n = 1..134</a> (the full list of terms, from Winter)
%H Henk Koppelaar and Peyman Nasehpour, <a href="https://arxiv.org/abs/2008.08187">On Hardy's Apology Numbers</a>, arXiv:2008.08187 [math.NT], 2020.
%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 16841 = 11720_11 (= 1*11^4 + 1*11^3 + 7*11^2 + 2*11^1 + 0*11^0) = 1^5 + 1^5 + 7^5 + 2^5 + 0^5. It's easy to see that 16841 + 1 then also satisfies this relation, as for all terms that are multiples of 11. - _M. F. Hasler_, Nov 20 2019
%t Select[Range[10^7], # == Total[IntegerDigits[#, 11]^IntegerLength[#, 11]] &] (* _Michael De Vlieger_, Nov 04 2020 *)
%o (PARI) select( {is_A161948(n)=n==vecsum([d^#n|d<-n=digits(n,11)])}, [0..10^5]) \\ This gives only terms < 10^5, for illustration of is_A161948(). - _M. F. Hasler_, Nov 20 2019
%Y In other bases: A010344 (base 4), A010346 (base 5), A010348 (base 6), A010350 (base 7), A010354 (base 8), A010353 (base 9), A005188 (base 10), A161949 (base 12), A161950 (base 13), A161951 (base 14), A161952 (base 15), A161953 (base 16).
%K base,fini,full,nonn
%O 1,2
%A _Joseph Myers_, Jun 22 2009
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