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A052382
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Numbers without 0 as a digit, a.k.a. zeroless numbers.
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131
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1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 111, 112, 113, 114
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OFFSET
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1,2
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COMMENTS
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The entries 1 to 79 match the corresponding subsequence of A043095, but then 81, 91-98, 100, 102, etc. are only in one of the two sequences. - R. J. Mathar, Oct 13 2008
Complement of A011540; A168046(a(n)) = 1; A054054(a(n)) > 0; A007602, A038186, A038618, A052041, A052043, and A052045 are subsequences. - Reinhard Zumkeller, Apr 25 2012, Apr 07 2011, Dec 01 2009
a(n) = n written in base 9 where zeros are not allowed but nines are. The nine distinct digits used are 1, 2, 3, ..., 9 instead of 0, 1, 2, ..., 8. To obtain this sequence from the "canonical" base 9 sequence with zeros allowed, just replace any 0 with a 9 and then subtract one from the group of digits situated on the left. For example, 9^3 = 729 (10) (in base 10) = 1000 (9) (in base 9) = 889 (9-{0}) (in base 9 without zeros) because 100 (9) = [9-1]9 = 89 (9-{0}) and thus 1000 (9) = [89-1]9 = 889 (9-{0}). - Robin Garcia, Jan 15 2014
From Hieronymus Fischer, May 28 2014: (Start)
Inversion: Given a term m, the index n such that a(n) = m can be calculated by A052382_inverse(m) = m - sum_{1<=j<=k} floor(m/10^j)*9^(j-1), where k := floor(log_10(m)) [see Prog section for an implementation in Smalltalk].
Example 1: A052382_inverse(137) = 137 - (floor(137/10) + floor(137/100)*9) = 137 - (13*1 + 1*9) = 137 - 22 = 115.
Example 2: A052382_inverse(4321) = 4321 - (floor(4321/10) + floor(4321/100)*9 + floor(4321/1000)*81) = 4321 - (432*1 + 43*9 + 4*81) = 4321 - (432 + 387 + 324) = 3178. (End)
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LINKS
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Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
K. Mahler, On the generating function of the integers with a missing digit, J. Indian Math. Soc. 15A (1951), 34-40.
Eric Weisstein's World of Mathematics, Zerofree
Index entries for 10-automatic sequences.
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FORMULA
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a(n+1) = f(a(n)) with f(x) = 1 + if x mod 10 < 9 then x else 10*f([x/10]). - Reinhard Zumkeller, Nov 15 2009
From Hieronymus Fischer, Apr 30, May 30 and Jun 08 2012: (Start)
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 9)*10^j, where m = floor(log_9(8*n + 1)), b(j) = floor((8*n + 1 - 9^m)/(8*9^j)).
Also: a(n) = Sum_{j=0..m-1} (1 + A010878(b(j)))*10^j.
Special values:
a(k*(9^n - 1)/8) = k*(10^n - 1)/9, k=1..9.
a((17*9^n - 9)/8) = 2*10^n - 1.
a((9^n - 1)/8 - 1) = 10^(n-1) - 1, n > 1.
Inequalities:
a(n) <= (10^log_9(8*n + 1)-1)/9, equality holds for n=(9^k-1)/8, k>0.
a(n) > (9/10)*(10^log_9(8*n + 1)-1)/9, n > 0.
Lower and upper limits:
lim inf a(n)/10^log_9(8*n) = 1/10, for n --> infinity.
lim sup a(n)/10^log_9(8*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/8)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(9/8)*(1 - 10z(j)^9 + 9z(j)^10)/((1-z(j))(1-z(j)^9)), where z(j) = x^9^j.
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1 - 10(x^9^j)^9 + 9(x^9^j)^10)*x^9^j*f_j(x)/(1-x^9^j), where f_j(x) = 10^j*x^((9^j-1)/8)/(1-(x^9^j)^9). Here, the f_j obey the recurrence f_0(x) = 1/(1-x^9), f_(j+1)(x) = 10x*f_j(x^9).
Also: g(x) = (1/(1-x))*((Sum{k=0..8} h_(9,k)(x)) - 9*h_(9,9)(x)), where h_(9,k)(x) = Sum_{j>=0} 10^j*x^((9^(j+1)-1)/8)*x^(k*9^j)/(1-x^9^(j+1)).
Generic formulas for analogous sequences with numbers expressed in base p and only using the digits 1, 2, 3, ... d, where 1 < d < p:
a(n) = Sum_{j=0..m-1} (1 + b(j) mod d)*p^j, where m = floor(log_d((d-1)*n+1)), b(j) = floor(((d-1)*n+1-d^m)/((d-1)*d^j)).
Special values:
a(k*(d^n-1)/(d-1)) = k*(10^n-1)/9, k=1..d.
a(d*((2d-1)*d^(n-1)-1)/(d-1)) = ((d+9)*10^n-d)/9 = 10^n + d*(10^n-1)/9.
a((d^n-1)/(d-1)-1) = d*(10^(n-1)-1)/9, n > 1.
Inequalities:
a(n) <= (10^log_d((d-1)*n+1)-1)/9, equality holds for n = (d^k-1)/(d-1), k > 0.
a(n) > (d/10)*(10^log_d((d-1)*n+1)-1)/9, n > 0.
Lower and upper limits:
lim inf a(n)/10^log_d((d-1)*n) = d/90, for n --> infinity.
lim sup a(n)/10^log_d((d-1)*n) = 1/9, for n --> infinity.
G.f.: g(x) = (1/(1-x)) Sum_{j>=0} (1 - (d+1)(x^d^j)^d + d(x^d^j)^(d+1))*x^d^j*f_j(x)/(1-x^d^j), where f_j(x) = p^j*x^((d^j-1)/(d-1))/(1-(x^d^j)^d). Here, the f_j obey the recursion f_0(x) = 1/(1-x^d), f_(j+1)(x) = px*f_j(x^d).
(End)
A052382 = { n | A054054(n) > 0 }. - M. F. Hasler, Jan 23 2013
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EXAMPLE
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From Hieronymus Fischer, May 30 2012, Jun 06 2012, and May 28 2014: (Start)
a(10^3) = 1331.
a(10^4) = 14641.
a(10^5) = 162151.
a(10^6) = 1783661.
a(10^7) = 19731371.
a(10^8) = 228145181.
a(10^9) = 2519596991.
a(10^10)= 27726678111.
a(10^20)= 819218515128277959321.
a(10^50)= 2.3492563334554748559828922...*10^53
a(10^100)= 5.657155154549942913548713...*10^104.
a(10^1000)= 7.974498368372833883261615...*10^1047.
(End)
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MAPLE
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a:= proc(n) local d, l, m; m:= n; l:= NULL;
while m>0 do d:= irem(m, 9, 'm');
if d=0 then d:=9; m:= m-1 fi;
l:= d, l
od; parse(cat(l))
end:
seq(a(n), n=1..100); # Alois P. Heinz, Jan 11 2015
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MATHEMATICA
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A052382 = Select[Range[100], DigitCount[#, 10, 0] == 0 &] (* Alonso del Arte, Mar 10 2011 *)
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PROG
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(Haskell)
a052382 n = a052382_list !! (n-1)
a052382_list = iterate f 1 where
f x = 1 + if r < 9 then x else 10 * f x' where (x', r) = divMod x 10
-- Reinhard Zumkeller, Mar 08 2015, Apr 07 2011
(MAGMA) [ n: n in [1..114] | not 0 in Intseq(n) ]; // Bruno Berselli, May 28 2011
(sh) seq 0 1000 | grep -v 0; # Joerg Arndt, May 29 2011
(PARI) is_A052382(n)= n=divrem(n, 10); until(!n=divrem(n[1], 10), n[2]||return); 1 \\ Code corrected for PARI vers. > 2.5. But the code below is much faster. - M. F. Hasler, Sep 25 2011, edited Jan 23 2013
(PARI) is_A052382(n)=n&&vecmin(digits(n)) \\ i.e., truth value equivalent to A054054. - M. F. Hasler, Jan 23 2013
(PARI) a(n) = for (w=0, oo, if (n >= 9^w, n -= 9^w, return ((10^w-1)/9 + fromdigits(digits(n, 9))))) \\ Rémy Sigrist, Jul 26 2017
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(Smalltalk)
A052382
"Answers the n-th term of A052382, where n is the receiver."
^self zerofree: 10
[by Hieronymus Fischer, May 28 2014]
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(Smalltalk)
A052382_inverse
"Answers that index n which satisfy A052382(n) = m, where m is the receiver.”
^self zerofree_inverse: 10
[by Hieronymus Fischer, May 28 2014]
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(Smalltalk)
zerofree: base
"Answers the n-th zerofree number in base base, where n is the receiver. Valid for base > 2.
Usage: n zerofree: b [b = 10 for this sequence]
Answer: a(n)"
| n m s c bi ci d |
n := self.
c := base - 1.
m := (base - 2) * n + 1 integerFloorLog: c.
d := n - (((c raisedToInteger: m) - 1)//(base - 2)).
bi := 1.
ci := 1.
s := 0.
1 to: m
do:
[:i |
s := (d // ci \\ c + 1) * bi + s.
bi := base * bi.
ci := c * ci].
^s
[by Hieronymus Fischer, May 28 2014]
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(Smalltalk)
zerofree_inverse: base
"Answers the index n such that the n-th zerofree number in base base is = m, where m is the receiver. Valid for base > 2.
Usage: m zerofree_inverse: b [b = 10 for this sequence]
Answer: n"
| m p q s |
m := self.
s := 0.
p := base.
q := 1.
[p < m] whileTrue:
[s := m // p * q + s.
p := base * p.
q := (base - 1) * q].
^m - s
[by Hieronymus Fischer, May 28 2014]
(Python)
A052382 = [n for n in range(1, 10**5) if not str(n).count('0')]
# Chai Wah Wu, Aug 26 2014
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CROSSREFS
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Cf. A004719, A052040, different from A067251.
Cf. A055640, A046034, A007931, A007932, A084544, A084545.
Column k=9 of A214676.
Cf. A066484 (subsequence).
Cf. A011540, A043489.
Cf. A052383 (without 1), A052404 (without 2), A052405 (without 3), A052406 (without 4), A052413 (without 5), A052414 (without 6), A052419 (without 7), A052421 (without 8), A007095 (without 9).
Cf. A099542 (subsequence).
Zeroless numbers in some other bases <= 10: A000042 (base 2), A032924 (base 3), A023705 (base 4), A248910 (base 6), A255805 (base 8), A255808 (base 9).
Cf. A059405 (subsequence).
Sequence in context: A023804 A209931 A067251 * A055572 A052040 A069570
Adjacent sequences: A052379 A052380 A052381 * A052383 A052384 A052385
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KEYWORD
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base,easy,nonn
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AUTHOR
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Henry Bottomley, Mar 13 2000
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EXTENSIONS
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Typos in formula section corrected by Hieronymus Fischer, May 30 2012
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STATUS
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approved
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