cp's OEIS Frontend

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

%I A131639 #25 Nov 16 2017 15:52:23
%S A131639 1729404,1800000,13758846,13800000,14358846,14400000,15000000,
%T A131639 28758846,28800000,29358846,29400000,1107488889,1107489042,1111088889,
%U A131639 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 A131639 Numbers n such that the sum of all numbers formed by deleting one digit from n is equal to n.
%C A131639 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 A131639 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 A131639 First term is 1729404 because sum(1729404) = 729404 + 129404 + 179404 + 172404 + 172904 + 172944 + 172940 = 1729404.
%o A131639 (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 A131639 Cf. A093882.
%K A131639 base,easy,nonn,full,fini
%O A131639 1,1
%A A131639 Jon Ayres (jonathan.ayres(AT)ntlworld.com), Sep 05 2007
%E A131639 a(12)-a(22) from _Donovan Johnson_, Jan 16 2011
%E A131639 a(23)-a(41) from _Anthony Sand_, Apr 24 2014