A241899 -id:A241899 - cp's OEIS frontend

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

Michel Lagneau, May 03 2014

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.

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

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

Cf. A241899.

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:

A319274 Osiris or Digit re-assembly numbers: numbers that are equal to the sum of permutations of subsamples of their own digits.

Original entry on oeis.org

132, 264, 396, 8991, 10545, 35964, 255530, 1559844, 9299907, 47755078, 89599104, 167264994, 283797162, 473995260, 3929996070, 6379993620, 10009998999, 11111111110, 22222222220, 33333333330, 44444444440, 55555555550, 66666666660, 77777777770, 88888888880, 99999999990

Offset: 1

Pieter Post, Sep 16 2018

This sequence differs from A241754 because this sequence uses permutations only once.
Permutations are of the same length k, leading zeros are allowed.
The k's in the sequence are: 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 7, 10, 10, 10, 10, 10, 10, 10, 10, 10, 6, 7, 7, 8, 7, 9, 9.

			10545 = 014 + 015 + 041 + 045 + 051 + 054 + 055 + 104 + 105 + 140 + 145 + 150 + 154 + 155 + 401 + 405 + 410 + 415 + 450 + 451 + 455 + 501 + 504 + 505 + 510 + 514 + 515 + 540 + 541 + 545 + 550 + 551 + 554.

Giovanni Resta, Table of n, a(n) for n = 1..33 (terms < 10^16)

Cf. A047726, A179239, A241754, A241899.

import itertools
def getData(a, b):
    dig = (itertools.permutations(str(a), b))
    for d in dig:
        yield d
for w in range(2, 6):
    kk=int(w*'1')
    for i in range (kk, 10**(w+3), kk):
        m=[]
        get = getData(i, w)
        while True:
            try:
                n = next(get)
                ee=int("".join((n)))
                if ee not in m:
                    m.append(ee)
            except StopIteration:
                if sum (m)==i and len(m)>1:
                    m.sort()
                    print (sum(m), len(m), m, i)
                break

a(12)-a(26) from Giovanni Resta, Sep 16 2018

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

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