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%I #11 May 03 2014 13:17:16
%S 167564622641,174977122641,175543159858,175543162247,183477122641,
%T 183518142444,191500000000,2779888721787,2784986175699,
%U 212148288981849,212148288982006,315131893491390,321400000000000,417586822240846,417586822241003,418112649991390
%N Numbers n such that the sum of all numbers formed by deleting two digits from n is equal to n.
%H Anthony Sand, <a href="/A241501/b241501.txt">Table of n, a(n) for n = 1..48</a>
%F For a number with n digits there are nC2 = n!/(n-2)!/2! substrings generated by removing two digits from the original number. So for 12345, these are 345, 245, 235, 234, 145, 135, 134, 125, 124, 123. 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 Sum(650000000000000) (15 digits) = 6000000000000 x 13 + 5000000000000 x 13 + 6500000000000 x (78 = 13C2) + 0.
%o (PARI) padbin(n, len) = {b = binary(n); while(length(b) < len, b = concat(0, b);); b;}
%o isok(n) = {d = digits(n); nb = #d; s = 0; for (j=1, 2^nb-1, if (hammingweight(j) == (nb-2), b = padbin(j, nb); nd = []; k = 1; for (i=1, nb, if (b[i], nd = concat(nd, d[k])); k++;); s += subst(Pol(nd), x, 10););); s == n;} \\ _Michel Marcus_, Apr 25 2014
%Y Cf. A131639 (n equal to sum of all numbers formed by deleting one digit from n).
%K nonn,base
%O 1,1
%A _Anthony Sand_, Apr 24 2014
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