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A131639
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Numbers n such that the sum of all numbers formed by deleting one digit from n is equal to n.
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1
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1729404, 1800000, 13758846, 13800000, 14358846, 14400000, 15000000, 28758846, 28800000, 29358846, 29400000, 1107488889, 1107489042, 1111088889, 1111089042, 3277800000, 3281400000, 4388888889, 4388889042, 4392488889, 4392489042, 4500000000, 5607488889, 5607489042, 5611088889, 5611089042, 7777800000, 7781400000, 8888888889, 8888889042, 8892488889, 8892489042, 10000000000, 20000000000, 30000000000, 40000000000, 50000000000, 60000000000, 70000000000, 80000000000, 90000000000
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OFFSET
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1,1
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COMMENTS
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The sequence is complete. In general, a number x = x_1 x_2 ... x_n of n digits belongs to the sequence if its digits satisfy a certain Diophantine equation c_1*x_1 + c_2*x_2 + ... + c_n*x_n = 0, where the coefficients c_i depend on n. It is easy to verify that for n > 11 all the coefficient c_i are positive, so the equation does not admit a nonzero solution. - Giovanni Resta, Jul 20 2015
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LINKS
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FORMULA
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For a number with n digits there are n substrings generated by removing one digit from the original number. So for 12345, these are 2345, 1345, 1245, 1235, 1234. Sum(x) is defined as the sum of these substrings for a number x and the sequence above is those numbers such that sum(x) = x.
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EXAMPLE
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First term is 1729404 because sum(1729404) = 729404 + 129404 + 179404 + 172404 + 172904 + 172944 + 172940 = 1729404.
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PROG
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(PARI) isok(n)=d = digits(n); if (sumdigits(n)*(#d-2) % 9 , return (0)); s = 0; for (i=1, #d, nd = vector(#d-1, j, if (i > j, d[j], d[j+1])); s += subst(Pol(nd), x, 10); ); s == n; \\ Michel Marcus, Apr 24 2014
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CROSSREFS
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KEYWORD
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base,easy,nonn,full,fini
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AUTHOR
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Jon Ayres (jonathan.ayres(AT)ntlworld.com), Sep 05 2007
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EXTENSIONS
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STATUS
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approved
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