A161953 Base-16 Armstrong or narcissistic numbers (written in base 10).
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 342, 371, 520, 584, 645, 1189, 1456, 1457, 1547, 1611, 2240, 2241, 2458, 2729, 2755, 3240, 3689, 3744, 3745, 47314, 79225, 177922, 177954, 368764, 369788, 786656, 786657, 787680, 787681, 811239, 812263, 819424, 819425, 820448, 820449, 909360
Offset: 1
Examples
645 is in the sequence because 645 is 285 in hexadecimal and 2^3 + 8^3 + 5^3 = 645. (The exponent 3 is the number of hexadecimal digits.)
Links
- Joseph Myers, Table of n, a(n) for n=1..293 (the full list of terms, from Winter)
- Henk Koppelaar and Peyman Nasehpour, On Hardy's Apology Numbers, arXiv:2008.08187 [math.NT], 2020.
- Eric Weisstein's World of Mathematics, Narcissistic Number
- D. T. Winter, Table of Armstrong Numbers
Crossrefs
Programs
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Mathematica
Select[Range[10^7], # == Total[IntegerDigits[#, 16]^IntegerLength[#, 16]] &] (* Michael De Vlieger, Nov 04 2020 *)
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PARI
isok(n) = {my(b=16, d=digits(n, b), e=#d); sum(k=1, #d, d[k]^e) == n;} \\ Michel Marcus, Feb 25 2019
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PARI
select( is_A161953(n)={n==vecsum([d^#n|d<-n=digits(n,16)])}, [1..10^5]) \\ M. F. Hasler, Nov 22 2019
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Python
from itertools import islice, combinations_with_replacement def A161953_gen(): # generator of terms for k in range(1,74): a = tuple(i**k for i in range(16)) yield from (x[0] for x in sorted(filter(lambda x:x[0] > 0 and tuple(int(d,16) for d in sorted(hex(x[0])[2:])) == x[1], \ ((sum(map(lambda y:a[y],b)),b) for b in combinations_with_replacement(range(16),k))))) A161953_list = list(islice(A161953_gen(),30)) # Chai Wah Wu, Apr 21 2022
Extensions
Terms sorted in increasing order by Pontus von Brömssen, Mar 03 2019
Comments