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%I #27 Oct 31 2018 03:04:42
%S 1,10,18,21,72,100,101,111,134,231,246,505,682,1000,1010,1100,1122,
%T 2210,3103,4006,6008,10000,10001,10012,11101,15453,20101,29358,34698,
%U 56576,84304,100000,100010,100011,100100,100101,100110,100303,101000,101001,101010
%N Numbers k for which there exists some m such that k = Sum_{i=1..1+floor(log_10(k))} binomial(m, d_i), where d_i is the i-th digit of k.
%C Contains numbers of the form 10^k, k >= 0 so the sequence is infinite. - _David A. Corneth_, Oct 30 2018
%H Giovanni Resta, <a href="/A055481/b055481.txt">Table of n, a(n) for n = 1..500</a>
%e 3103 = C(22, 3) + C(22, 1) + C(22, 0) + C(22, 3).
%e C(k, 1) + C(k, 1) + C(k, 1) + C(k, 0) + C(k, 0) + C(k, 0) = 3k + 3 so all 6-digit numbers with 3 ones and 3 zeros are in the sequence. - _David A. Corneth_, Oct 30 2018
%t ok[n_] := Block[{d = IntegerDigits@n, k=1, v, x}, If[ Max@d <= 3, False =!= Reduce[ Total@ Binomial[x, d] == n && x>0, x, Integers], While[(v = Total@ Binomial[k, d]) < n, k++]; v == n]]; Select[ Range[10^5], ok] (* _Giovanni Resta_, Oct 30 2018 *)
%o (PARI) is(n) = my(d = digits(n)); for(i = 1, n, s = sum(j = 1, #d, binomial(i, d[j])); if(s >= n, return(s == n))) \\ _David A. Corneth_, Oct 30 2018
%Y Cf. A007088, A009994, A055482.
%K base,nonn
%O 1,2
%A _Erich Friedman_, Jun 27 2000
%E a(31)-a(41) from _Giovanni Resta_, Oct 30 2018