%I #20 Nov 16 2017 15:52:25
%S 11,22,33,44,55,66,77,88,99,132,264,396
%N Numbers n equal to the sum of all the two-digit numbers formed without repetition from the digits of n.
%C The sequence is complete because the maximum number of partitions of an m-digit number n into sets of 2-digit numbers is 2*binomial(m,2) = m*(m-1). (See the Oblong numbers A002378.) The maximum value of the sum of all the two-digit numbers is M = 99*m*(m-1) where m = floor(log_10(n) + 1). But M < n for n > 1188 because the solution of the equation 99*m*(m-1) = n with m = floor(log_10(n) + 1) is n = 1188.
%e 132 is in the sequence because 132 = 12 + 13 + 21 + 23 + 31 + 32.
%p with(numtheory):
%p for n from 10 to 10000 do:
%p lst:={}:k:=0:x:=convert(n,base,10):n1:=nops(x):
%p for i from 1 to n1 do:
%p for j from i+1 to n1 do:
%p lst:=lst union {x[i]+10*x[j]}:
%p od:
%p od:
%p for a from n1 by -1 to 1 do:
%p for b from a-1 by -1 to 1 do:
%p lst:=lst union
%p {x[a]+10*x[b]}:
%p od:
%p od:
%p n2:=nops(lst):s:=sum('lst[i]', 'i'=1..n2):
%p if s=n
%p then
%p printf(`%d, `,n):
%p else
%p fi:
%p od:
%t Select[Range[10,400],Total[FromDigits/@Permutations[IntegerDigits[#],{2}]]==#&] (* _Ivan N. Ianakiev_, Oct 24 2014 *)
%Y Cf. A002378.
%K nonn,base,fini,full
%O 10,1
%A _Michel Lagneau_, May 01 2014