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A241946
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Numbers n equal to the sum of all the four-digit numbers formed without repetition from the digits of n.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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.
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.
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LINKS
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EXAMPLE
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37323 is in the sequence because 37323 = 2373 + 3233 + 3237 + 3273 + 3323 + 3373 + 3723 + 3732 + 3733 + 7323.
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MAPLE
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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:
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CROSSREFS
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KEYWORD
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nonn,base,fini,full
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AUTHOR
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STATUS
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approved
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