A101337 Sum of (each digit of n raised to the power (number of digits in n)).
1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 9, 10, 13, 18, 25, 34, 45, 58, 73, 90, 16, 17, 20, 25, 32, 41, 52, 65, 80, 97, 25, 26, 29, 34, 41, 50, 61, 74, 89, 106, 36, 37, 40, 45, 52, 61, 72, 85, 100, 117, 49, 50, 53, 58, 65
Offset: 1
Examples
a(75) = 7^2 + 5^2 = 74 and a(705) = 7^3 + 0^3 + 5^3 = 468. a(1.02e59 - 1) = 102429587095122578993551250282047487264694110769657513064859 ~ 1.024e59 is an example of n close to the limit beyond which a(n) < n for all n. - _M. F. Hasler_, Nov 17 2019
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Wikipedia, Narcissistic number, as of Nov 18 2019.
Programs
-
Magma
f:=func
; [f(n):n in [1..75]]; // Marius A. Burtea, Nov 18 2019 -
Mathematica
Array[Total[IntegerDigits[#]^IntegerLength[#]]&,80] (* Harvey P. Dale, Aug 27 2011 *)
-
PARI
a(n)=my(d=digits(n)); sum(i=1,#d, d[i]^#d) \\ Charles R Greathouse IV, Aug 10 2017
-
PARI
apply( A101337(n)=vecsum([d^#n|d<-n=digits(n)]), [0..99]) \\ M. F. Hasler, Nov 17 2019
-
Python
def A101337(n): s = str(n) l = len(s) return sum(int(d)**l for d in s) # Chai Wah Wu, Feb 26 2019
Formula
a(n) <= A055642(n)*9^A055642(n) with equality for all n = 10^k - 1. Write n = 10^x to get a(n) < n when 1+log_10(x+1) < (x+1)(1-log_10(9)) <=> x > 59.85. It appears that a(n) < n already for all n > 1.02*10^59. - M. F. Hasler, Nov 17 2019
Extensions
Name changed by Axel Harvey, Dec 26 2011
Edited by M. F. Hasler, Nov 17 2019
Comments