A055481 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.

1, 10, 18, 21, 72, 100, 101, 111, 134, 231, 246, 505, 682, 1000, 1010, 1100, 1122, 2210, 3103, 4006, 6008, 10000, 10001, 10012, 11101, 15453, 20101, 29358, 34698, 56576, 84304, 100000, 100010, 100011, 100100, 100101, 100110, 100303, 101000, 101001, 101010

Offset: 1

Author

Erich Friedman, Jun 27 2000

Keywords

• base
• nonn

Comments

Contains numbers of the form 10^k, k >= 0 so the sequence is infinite. - David A. Corneth, Oct 30 2018

Examples

			3103 = C(22, 3) + C(22, 1) + C(22, 0) + C(22, 3).
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
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A007088, A009994, A055482.

Programs

Mathematica

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 *)

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

Extensions

a(31)-a(41) from Giovanni Resta, Oct 30 2018