A261433 - OEIS

%I #41 Jul 10 2018 10:15:50

%S 5,378,91882,3762938,46478818,564426414

%N k-digit integers equal to the sum of the k-th powers of the tens' complements of their digits.

%C The terms of the sequence could be called "Shy n n-digit numbers" as suggested by Geoffrey Campbell, Long Term Visitor (Visiting Fellow), Mathematical Sciences Institute, Australian National University, cf Links.

%C In base 10, x is a "Shy k n-digit number" if it is an n-digit (d_i) number such that x = Sum_{i=1..n}{(10-d_i)^k}. For instance, 2240 is a "Shy 3 4-digit number": (10 - 2)^3 + (10 - 2)^3 + (10 - 4)^3 + (10 - 0)^3 = 512 + 512 + 216 + 1000 = 2240. Again, 2149042 is a "Shy 6 7-digit number": (10 - 2)^6 + (10 - 1)^6 + (10 - 4)^6 + (10 - 9)^6 + (10 - 0)^6 + (10 - 4)^6 + (10 - 2)^6 =  262144 + 531441 + 46656 + 1 + 1000000 + 46656 + 262144 = 2149042.

%C It is not known if the sequence is finite. At least there are no other terms up to 18-digit numbers (as tested by Marco Cecchi at LinkedIn link).

%C If there are further terms, they are greater than 10^33. - _Giovanni Resta_, Aug 20 2015

%C Subsequence of A052382. Sequence is finite and complete as verified by exhaustive search since all terms have 60 or fewer digits. Since all terms are zeroless, they are less than k*9^k which would be less than 10^(k-1) (i.e., have fewer than k digits) if k > 60. - _Chai Wah Wu_, Apr 07 2018

%H Geoffrey Campbell, <a href="https://www.linkedin.com/grp/post/4510047-6038963758880010243"> Related to Narcissistic numbers: the Shy numbers</a>, Number Theory group on LinkedIn.com

%H Marco Cecchi, <a href="/A261433/a261433.txt">Python program based on partitions</a>.

%e (10 - 5)^1 = 5,

%e (10 - 3)^3 + (10 - 7)^3 + (10 - 8)^3 = 343 + 27 + 8 = 378,

%e (10 - 9)^5 + (10 - 1)^5 + (10 - 8)^5 + (10 - 8)^5 + (10 - 2)^5 = 1 + 59049 + 32 + 32 + 32768 = 91882, etc.

%p with(numtheory): P:=proc(q) local a,b,c,k,n;

%p for n from 1 to q do a:=ilog10(n)+1; b:=0; c:=n;

%p for k from 1 to a do b:=b+(10-(c mod 10))^a; c:=trunc(c/10); od;

%p if b=n then print(n); fi; od; end: P(10^9);

%t Select[Range[10^5], # == Total[(10 - IntegerDigits@ #)^ IntegerLength[#]] &] (* _Giovanni Resta_, Aug 20 2015 *)

%o (PARI) isok(n) = (d = digits(n)) && (sum(k=1, #d, (10-d[k])^#d) == n); \\ _Michel Marcus_, Aug 24 2015

%o (Python)

%o from itertools import combinations_with_replacement

%o A261433_list = []

%o for k in range(1,10):

%o     a, k10 = tuple([i**k for i in range(10,0,-1)]), 10**k

%o     for b in combinations_with_replacement(range(1,10),k):

%o         x = sum(list(map(lambda y:a[y],b)))

%o         if x < k10 and tuple(int(d) for d in sorted(str(x))) == b:

%o             A261433_list.append(x)

%o A261433_list = sorted(A261433_list) # _Chai Wah Wu_, Aug 25 2015, updated Apr 06, 2018

%Y Cf. A005188, A052382.

%K nonn,base,fini,full

%O 1,1

%A _Paolo P. Lava_, Aug 20 2015

%E a(4)-a(6) found by Aleksander Zujev