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%I #12 May 17 2014 04:03:51
%S 1001,1111,1221,1331,1441,1551,1661,1771,1881,1991,2002,2112,2222,
%T 2332,2442,2552,2662,2772,2882,2992,3003,3113,3223,3333,3443,3553,
%U 3663,3773,3883,3993,4004,4114,4224,4334,4444,4554,4664,4774,4884,4994,5005,5115,5225
%N Numbers n equal to the sum of all the four-digit numbers formed without repetition from the digits of n.
%C 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 i<j<k<l or i>j>k>l.
%C 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.
%C 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.
%H Michel Lagneau, <a href="/A241946/b241946.txt">Table of n, a(n) for n = 1..93</a>
%e 37323 is in the sequence because 37323 = 2373 + 3233 + 3237 + 3273 + 3323 + 3373 + 3723 + 3732 + 3733 + 7323.
%p with(numtheory):
%p for n from 1000 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 for m from j+1 to n1 do:
%p for q from m+1 to n1 do:
%p lst:=lst union {x[i]+10*x[j]+100*x[m]+1000*x[q]}:
%p od:
%p od:
%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 for c from b-1 by -1 to 1 do:
%p for d from c-1 by -1 to 1 do:
%p lst:=lst union
%p {x[a]+10*x[b]+100*x[c]+1000*x[d]}:
%p od:
%p od:
%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:
%Y Cf. A241899.
%K nonn,base,fini,full
%O 1,1
%A _Michel Lagneau_, May 03 2014
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