A241946 Numbers n equal to the sum of all the four-digit numbers formed without repetition from the digits of n.

1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, 2002, 2112, 2222, 2332, 2442, 2552, 2662, 2772, 2882, 2992, 3003, 3113, 3223, 3333, 3443, 3553, 3663, 3773, 3883, 3993, 4004, 4114, 4224, 4334, 4444, 4554, 4664, 4774, 4884, 4994, 5005, 5115, 5225

Offset: 1

Author

Michel Lagneau, May 03 2014

Keywords

• nonn
• base
• fini
• full

Comments

Let d(1)d(2)... d(q) denote the decimal expansion of a number n. Any decimal expansion of four-digits d(i)d(j)d(k)d(l) formed from the digits of n is such that ij>k>l.

This sequence is interesting because it contains more than just the only trivial palindromic values 1001, 1111, 1221,... The sequence is given by the union of subsets {palindromes with four digits from A056524} union {37323, 48015, 72468, 152658} and contains 94 elements. The last four elements are non-palindromic numbers.

But the generalization of this problem seems difficult, for example the case with the sum of all the three-digit numbers formed without repetition from the digits of n gives only 90 palindromic numbers 101, 111, 121,..., 989,999.

Examples

			37323 is in the sequence because 37323 =  2373 + 3233 + 3237 + 3273 + 3323 + 3373 + 3723 + 3732 + 3733 + 7323.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..93

Crossrefs

Cf. A241899.

Programs

Maple

with(numtheory):
for n from 1000 to 10000 do:
     lst:={}:k:=0:x:=convert(n,base,10):n1:=nops(x):
        for i from 1 to n1 do:
          for j from i+1 to n1 do:
            for m from j+1 to n1 do:
              for q from m+1 to n1 do:
            lst:=lst union {x[i]+10*x[j]+100*x[m]+1000*x[q]}:
            od:
          od:
        od:
        od:
           for a from n1 by -1 to 1 do:
             for b from a-1 by -1 to 1 do:
               for c from b-1 by -1 to 1 do:
                 for d from c-1 by -1 to 1 do:
               lst:=lst union
               {x[a]+10*x[b]+100*x[c]+1000*x[d]}:
               od:
             od:
            od:
            od:
           n2:=nops(lst):s:=sum('lst[i]', 'i'=1..n2):
           if s=n
             then
             printf(`%d, `,n):
             else
           fi:
  od: