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

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

Offset: 1

Author

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

Keywords

• base
• easy
• nonn
• full
• fini

Comments

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

Examples

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

Crossrefs

Cf. A093882.

Programs

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

Formula

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.

Extensions

a(12)-a(22) from Donovan Johnson, Jan 16 2011

a(23)-a(41) from Anthony Sand, Apr 24 2014