cp's OEIS Frontend

A281724 Numbers k with the property that (sum of digits of k) times (number of digits of k) equals (sum of {each digit of k} raised to the power {number of digits in k}).

%I A281724 #31 Dec 26 2021 01:30:34
%S A281724 1,2,3,4,5,6,7,8,9,20,22,102,120,201,210
%N A281724 Numbers k with the property that (sum of digits of k) times (number of digits of k) equals (sum of {each digit of k} raised to the power {number of digits in k}).
%C A281724 Numbers with the property that q-p = 0, where p = (sum of digits of k) times (number of digits of k) (A110805) and q = (sum of {each digit of k} raised to the power {number of digits in k}) (A101337).
%C A281724 From _David Consiglio, Jr._, Jan 29 2017: (Start)
%C A281724 The sequence is finite and 210 is the last term. Proof:
%C A281724 1. Let Y = length(k).
%C A281724 2. The maximum value of digit_sum(k) is 9Y.
%C A281724 3. Since p = Y*digit_sum(k), p has a maximum value of 9Y^2.
%C A281724 4. 2^Y > 9Y^2 for all Y > 9. Therefore q > p for all numbers longer than 9 digits that contain any digits > 1.
%C A281724    a. Example: 1,000,000,002. q = 1^5 + 8*0^5 + 2^10 = 1025. The largest p value for a 10-digit number would be for 9,999,999,999 which has p = 10*(9*10) = 900. Since q > all possible p values at this size, any term of this sequence must either have fewer than 10 digits or contain only 0's and 1's as digits.
%C A281724 5. Now we can consider only numbers with 0 and 1 digits.
%C A281724 6. Let Z = number of 1 digits in a number k.
%C A281724 7. q = Z because q = Z*1^Y + (Y-Z)*0^Y = Z.
%C A281724 8. p = YZ because the sum of the digits is equal to the number of 1's.
%C A281724 9. Z = YZ only in the case of Y = 1. Thus, the only term of this sequence that contains only 0's and 1's has a length of only 1 digit. Thus, k = 1 is in this sequence.
%C A281724 10. Therefore a candidate number must have fewer than 10 digits if it contains a digit 2 or larger, and must have fewer than 2 digits if it does not. All numbers in this range have been checked, and no additional values of k with q = p have been found.
%C A281724 Thus the sequence is finite. (End)
%e A281724     9 is a term because             9^1 = 1*(9);
%e A281724    20 is a term because       2^2 + 0^2 = 2*(2 + 0);
%e A281724   210 is a term because 2^3 + 1^3 + 0^3 = 3*(2 + 1 + 0).
%t A281724 W = Array[Total[IntegerDigits[#]^IntegerLength[#]]&, 100]-Table[IntegerLength[n] * Total[IntegerDigits[n]], {n, 100}]; Flatten[Position[W,0]]
%Y A281724 Cf. A101337, A110805.
%K A281724 nonn,base,fini
%O A281724 1,2
%A A281724 _José de Jesús Camacho Medina_, Jan 28 2017