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 A131639 Numbers n such that the sum of all numbers formed by deleting one digit from n is equal to n. 1

%I

%S 1729404,1800000,13758846,13800000,14358846,14400000,15000000,

%T 28758846,28800000,29358846,29400000,1107488889,1107489042,1111088889,

%U 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

%N Numbers n such that the sum of all numbers formed by deleting one digit from n is equal to n.

%C 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

%F 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.

%e First term is 1729404 because sum(1729404) = 729404 + 129404 + 179404 + 172404 + 172904 + 172944 + 172940 = 1729404.

%o (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

%Y Cf. A093882.

%K base,easy,nonn,full,fini

%O 1,1

%A Jon Ayres (jonathan.ayres(AT)ntlworld.com), Sep 05 2007

%E a(12)-a(22) from _Donovan Johnson_, Jan 16 2011

%E a(23)-a(41) from _Anthony Sand_, Apr 24 2014

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Last modified August 19 18:51 EDT 2019. Contains 326133 sequences. (Running on oeis4.)