|
%I #14 Sep 27 2021 11:28:57
%S 9,27,31,32,35,39,45,57,81,105,123,127,128,131,135,141,145,146,149,
%T 150,154,157,158,161,163,164,167,171,175,176,179,183,189,195,199,200,
%U 203,207,213,219,223,224,227,231,237,249,267,271,272,275,279,285,297,321,345,369,387,391,392,395,399,405,417,441
%N Numbers n such that there are exactly three 1's in their factorial base representation (A007623).
%H Antti Karttunen, <a href="/A255343/b255343.txt">Table of n, a(n) for n = 1..6769</a>
%e The factorial base representation (A007623) of 9 is "111", which contains exactly three 1's, thus 3 is included in the sequence.
%e The f.b.r. of 27 is "1011", with exactly three 1's, thus 27 is included in the sequence.
%e The f.b.r. of 81 is "3111", with exactly three 1's, thus 81 is included in the sequence.
%t factBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > i!, i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= j!, m = m - j!; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; dList]; s = Table[FromDigits[factBaseIntDs[n]], {n, 480}]; Flatten@ Position[s, x_ /; DigitCount[x][[1]] == 3](* _Michael De Vlieger_, Apr 27 2015, after _Alonso del Arte_ at A007623 *)
%o (Scheme, with _Antti Karttunen_'s IntSeq-library)
%o (define A255343 (MATCHING-POS 1 0 (lambda (n) (= 3 (A257511 n)))))
%o (Python)
%o def fbr(n, p=2): # per _Indranil Ghosh_ in A007623
%o return n if n<p else fbr(n//p, p+1)*10 + n%p
%o def ok(n): return str(fbr(n)).count('1') == 3
%o print(list(filter(ok, range(442)))) # _Michael S. Branicky_, Sep 27 2021
%Y Cf. A007623, A257511, A255411, A255341, A255342.
%Y Subsequence of A256450.
%K nonn,base
%O 1,1
%A _Antti Karttunen_, Apr 27 2015
|
