[go: up one dir, main page]

login
Search: a052041 -id:a052041
     Sort: relevance | references | number | modified | created      Format: long | short | data
Numbers without 0 as a digit, a.k.a. zeroless numbers.
+10
212
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
OFFSET
1,2
COMMENTS
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)
The sum of the reciprocals of these numbers from a(1)=1 to infinity, called the Kempner series, is convergent towards a limit: 23.103447... whose decimal expansion is in A082839. - Bernard Schott, Feb 23 2019
Integer n > 0 is encoded using bijective base-9 numeration, see Wikipedia link below. - Alois P. Heinz, Feb 16 2020
REFERENCES
Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, p. 258.
LINKS
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, Kempner Series.
Eric Weisstein's World of Mathematics, Zerofree
FORMULA
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, Jun 08 2012, Feb 17 2019: (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.
a(9*n + k) = 10*a(n) + k, k=1..9.
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) <= (1/9)*((8*n+1)^(1/log_10(9)) - 1), equality holds for n=(9^k-1)/8, k>0.
a(n) > (1/10)*(8*n+1)^(1/log_10(9)) - 1), n > 0.
Lower and upper limits:
lim inf a(n)/10^log_9(8*n) = 1/10, for n -> infinity.
lim inf a(n)/n^(1/log_10(9)) = 8^(1/log_10(9))/10, for n -> infinity.
lim sup a(n)/10^log_9(8*n) = 1/9, for n -> infinity.
lim sup a(n)/n^(1/log_10(9)) = 8^(1/log_10(9))/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
From Hieronymus Fischer, Feb 20 2019: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 0.696899720...
Sum_{n>=1} 1/a(n)^2 = 1.6269683705819...
Sum_{n>=1} 1/a(n) = 23.1034479... = A082839. This so-called Kempner series converges very slowly. For the calculation of the sum, it is helpful to use the following fraction of partial sums, which converges rapidly:
lim_{n->infinity} (Sum_{k=p(n)..p(n+1)-1} 1/a(k)) / (Sum_{k=p(n-1)..p(n)-1} 1/a(k)) = 9/10, where p(n) = (9^n-1)/8, n > 1.
(End)
EXAMPLE
For k >= 0, a(10^k) = (1, 11, 121, 1331, 14641, 162151, 1783661, 19731371, ...) = A325203(k). - Hieronymus Fischer, May 30 2012 and Jun 06 2012; edited by M. F. Hasler, Jan 13 2020
MAPLE
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
MATHEMATICA
A052382 = Select[Range[100], DigitCount[#, 10, 0] == 0 &] (* Alonso del Arte, Mar 10 2011 *)
PROG
(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) select( {is_A052382(n)=n&&vecmin(digits(n))}, [0..111]) \\ actually: is_A052382 = (bool) A054054. - M. F. Hasler, Jan 23 2013, edited Jan 13 2020
(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
(PARI)
apply( {A052382(n, L=logint(n, 9))=fromdigits(digits(n-9^L>>3, 9))+10^L\9}, [1..100])
next_A052382(n, d=digits(n+=1))={for(i=1, #d, d[i]|| return(n-n%(d=10^(#d-i+1))+d\9)); n} \\ least a(k) > n. Used in A038618.
( {A052382_vec(n, M=1)=M--; vector(n, i, M=next_A052382(M))} )(99) \\ n terms >= M
\\ See OEIS Wiki page (cf. LINKS) for more programs. - M. F. Hasler, Jan 11 2020
(Smalltalk)
"Answers the n-th term of A052382, where n is the receiver."
^self zerofree: 10
A052382_inverse
"Answers that index n which satisfy A052382(n) = m, where m is the receiver.”
^self zerofree_inverse: 10
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
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
CROSSREFS
Cf. A004719, A052040, different from A067251.
Column k=9 of A214676.
Cf. A011540 (complement), A043489, A054054, A168046.
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).
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. A082839 (sum of reciprocals).
Cf. A038618 (subset of primes)
KEYWORD
base,easy,nonn
AUTHOR
Henry Bottomley, Mar 13 2000
EXTENSIONS
Typos in formula section corrected by Hieronymus Fischer, May 30 2012
STATUS
approved
Squares of A002280 or numbers (666...6)^2.
+10
27
0, 36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556, 444444444443555555555556, 44444444444435555555555556, 4444444444444355555555555556
OFFSET
0,2
COMMENTS
A transformation of the Wonderful Demlo numbers (A002477).
FORMULA
a(n) = A002280(n)^2 = (6 * A002275(n))^2 = 36 * (A002275(n))^2.
a(n) = (6*(10^n-1)/9)^2 = (4/9)*(10^(2*n) - 2*10^n + 1), which is n-1 4's, followed by a 3, n-1 5's and a 6. - Ignacio Larrosa Cañestro, Feb 26 2005
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = ((A002278(n-1)*10+3)*10^(n-1)+A002279(n-1))*10+6 for n>0.
a(n) = A002283(n)*A002278(n). (End)
G.f.: 36*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Arkadiusz Wesolowski, Dec 26 2011
EXAMPLE
a(2) = 66^2 = 4356.
From Reinhard Zumkeller, May 31 2010: (Start)
n=1: ..................... 36 = 9 * 4;
n=2: ................... 4356 = 99 * 44;
n=3: ................. 443556 = 999 * 444;
n=4: ............... 44435556 = 9999 * 4444;
n=5: ............. 4444355556 = 99999 * 44444;
n=6: ........... 444443555556 = 999999 * 444444;
n=7: ......... 44444435555556 = 9999999 * 4444444;
n=8: ....... 4444444355555556 = 99999999 * 44444444;
n=9: ..... 444444443555555556 = 999999999 * 444444444. (End)
MATHEMATICA
Table[FromDigits[PadRight[{}, n, 6]]^2, {n, 0, 20}] (* or *) LinearRecurrence[ {111, -1110, 1000}, {0, 36, 4356}, 20] (* Harvey P. Dale, May 20 2021 *)
KEYWORD
nonn,easy
AUTHOR
Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002
EXTENSIONS
Edited by Alois P. Heinz, Aug 21 2019 (merged with A102794, submitted by Richard C. Schroeppel, Feb 26 2005)
STATUS
approved
Numbers whose square is zeroless.
+10
20
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, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 46, 54, 56, 57, 58, 59, 61, 62, 63, 65, 66, 67, 68, 69, 72, 73, 74, 75, 76, 77, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 96
OFFSET
1,2
COMMENTS
This sequence is infinite, since 33...334^2 = 11...11155...556, for example. This answers an open problem stated in HAKMEM. - Karl W. Heuer, Aug 19 2015
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Michael Beeler, R. William Gosper, and Rich Schroeppel, "HAKMEM" Item 36, Memo 239, Artificial Intelligence Laboratory, Massachusetts Institute of Technology, Cambridge, Mass., 1972.
EXAMPLE
From Jon E. Schoenfield, Aug 16 2021: (Start)
31 is a term: 31^2 = 961 has no 0's among its digits.
32 is not a term, because 32^2 = 1024. (End)
MATHEMATICA
Select[Range[0, 100], DigitCount[#^2, 10, 0]==0 &] (* Vincenzo Librandi, Feb 22 2015 *)
PROG
(PARI) is(n)=vecmin(digits(n^2))>0 \\ Charles R Greathouse IV, Oct 11 2013
(Magma) [n: n in [1..100] | not 0 in Intseq(n^2)]; // Vincenzo Librandi, Feb 22 2015
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Dec 15 1999
STATUS
approved
Squares of primes lacking the digit zero in their decimal expansion.
+10
13
4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 3481, 3721, 4489, 5329, 6241, 6889, 7921, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 24649, 26569, 27889, 29929, 32761, 36481, 37249, 44521, 49729, 51529, 52441, 54289
OFFSET
1,1
COMMENTS
Intersection of A052382 and A001248; A168046(a(n))*A064911(a(n))*A010052(a(n)) = 1. - Reinhard Zumkeller, Dec 01 2009
LINKS
Eric Weisstein's World of Mathematics, Zerofree - Reinhard Zumkeller, Dec 01 2009
MATHEMATICA
Select[Prime[Range[100]]^2, DigitCount[#, 10, 0]==0&] (* Harvey P. Dale, Mar 18 2012 *)
PROG
(PARI) is(n)=my(d=digits(n)); vecsort(d)[1]&&issquare(n, &n)&&isprime(n) \\ Charles R Greathouse IV, Jun 05 2013
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Dec 15 1999
STATUS
approved
a(n) is the square of one plus the number consisting of n 3's.
+10
13
1, 16, 1156, 111556, 11115556, 1111155556, 111111555556, 11111115555556, 1111111155555556, 111111111555555556, 11111111115555555556, 1111111111155555555556, 111111111111555555555556, 11111111111115555555555556, 1111111111111155555555555556, 111111111111111555555555555556
OFFSET
0,2
COMMENTS
Old name was: The number (333...334)^2.
An infinite sequence of squares with no zeros in base 10.
a(n) = A104265(2n) for n > 0. - Chai Wah Wu, Mar 24 2020
REFERENCES
Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 31 at p. 61.
Italo Ghersi, Matematica dilettevole e curiosa, pp. 111-112, Hoepli, Milano, 1967. [Vincenzo Librandi, Dec 31 2008]
LINKS
Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
FORMULA
From R. J. Mathar, Jan 06 2009: (Start)
a(n) = (100^n + 4*10^n + 4)/9.
G.f.: (1 - 95*x + 490*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)
E.g.f.: exp(x)*(4 + 4*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024
MAPLE
a:= n-> (1+parse(cat(0, 3$n)))^2:
seq(a(n), n=0..20); # Alois P. Heinz, Sep 03 2018
MATHEMATICA
Table[(10^n + 2)^2/9, {n, 0, 20}] (* Paolo Xausa, Jun 26 2024 *)
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Ron Knott, Feb 27 2005
EXTENSIONS
New name from Alois P. Heinz, Sep 03 2018
STATUS
approved
Primes that lack the digit zero in the decimal expansion of their squares.
+10
12
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 59, 61, 67, 73, 79, 83, 89, 107, 109, 113, 127, 131, 137, 139, 157, 163, 167, 173, 181, 191, 193, 211, 223, 227, 229, 233, 239, 263, 269, 271, 277, 281, 293, 307, 311, 313, 337, 359, 367, 373, 379, 383, 389, 409, 419, 421, 431
OFFSET
1,1
FORMULA
a(n) = sqrt(A052043(n)). - Zak Seidov, Dec 27 2014
EXAMPLE
The primes 47, 53 and 71 are not in the sequence because 47^2=2209, 53^2=2809 and 71^2=5041 contain zeros in their decimal representation.
MATHEMATICA
fQ[n_] := DigitCount[n^2][[-1]] == 0; Select[Prime@ Range@ 80, fQ] (* Robert G. Wilson v, Aug 22 2012 *)
PROG
(PARI) {p=2; for(k=1, 10^2, if(vecmin(digits(p^2))>0,
print1(p", ")); p=nextprime(1+p))}\\ Zak Seidov, Dec 24 2014
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Dec 15 1999
STATUS
approved
Number of n-digit squares with no zero digits.
+10
11
3, 6, 19, 44, 136, 376, 1061, 2985, 8431, 24009, 67983, 193359, 549697, 1563545, 4446173, 12650545, 35999714, 102439796, 291532841, 829634988, 2360947327, 6719171580, 19122499510, 54423038535, 154888366195
OFFSET
1,1
COMMENTS
Comments from David W. Wilson, Feb 26 2005: (Start)
"There are approximately s(d) = (10^d)^(1/2) - (10^(d-1))^(1/2) d-digit squares. A random d-digit number has the probability p(d) = (9/10)^(d-1) of being zeroless (exponent d-1 as opposed to d because the first digit is not zero). So we expect p(d)s(d) zeroless d-digit squares.
"For d = 1 through 12, we get (truncating): 1, 5, 15, 44, 127, 363, 1034, 2943, 8377, 23841, 67854, 193117, ... The elements grow approximately geometrically with limit ratio (9/10)*10^(1/2) = 2.846+.
"The same naive estimate can easily be generalize to k-th powers, giving the estimate s(d) = (10^d)^(1/k) - (10^(d-1))^(1/k) for d-digit k-th powers. p(d) remains the same. The resulting estimates have ratio (9/10)*10^(1/k).
"We should expect an infinite number of zeroless k-th powers when this ratio is >= 1, which it is for k <= 21. For k >= 22, the ratio is < 1 and we should expect a finite number of zeroless k-th powers." (End)
EXAMPLE
a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.
PROG
(Python)
def aupton(terms):
c, k, kk = [0 for i in range(terms)], 1, 1
while kk < 10**terms:
s = str(kk)
c[len(s)-1], k, kk = c[len(s)-1] + (s.count('0')==0), k+1, kk + 2*k + 1
return c
print(aupton(14)) # Michael S. Branicky, Mar 06 2021
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Reinhard Zumkeller and Ron Knott, Feb 26 2005
EXTENSIONS
a(14)-a(18) from Donovan Johnson, Nov 05 2009
a(19)-a(21) from Donovan Johnson, Mar 23 2011
a(22)-a(25) from Donovan Johnson, Jan 29 2013
STATUS
approved
Primes whose cubes lack zeros.
+10
10
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 53, 61, 71, 83, 97, 113, 137, 139, 151, 157, 167, 173, 179, 181, 191, 197, 211, 233, 239, 241, 251, 257, 263, 277, 283, 293, 307, 331, 337, 347, 353, 359, 373, 379, 383, 389, 409, 421, 433, 457, 461, 463, 499, 503
OFFSET
1,1
LINKS
MATHEMATICA
Select[Prime[Range[100]], FreeQ[IntegerDigits[#^3], 0]&] (* Harvey P. Dale, Jan 06 2017 *)
PROG
(PARI) isok(n) = isprime(n) && vecmin(digits(n^3)); \\ Michel Marcus, Jan 06 2014
CROSSREFS
KEYWORD
base,nonn
AUTHOR
G. L. Honaker, Jr., Dec 07 1999
EXTENSIONS
More terms from Michel Marcus, Jan 06 2014
STATUS
approved
Numbers k such that (2*10^(5*k) - 10^(4*k) + 2*10^(3*k) + 10^(2*k) + 10^k + 1)/3 is prime.
+10
10
OFFSET
1,3
COMMENTS
The Baxter-Hickerson function provides a number whose cube lacks zeros.
The next term is > 4400. - Jason Earls, Sep 10 2005
The next term is > 20000 (found using pfgw64). - Patrick De Geest, Jul 22 2012
LINKS
Ed Pegg, Jr., Fun with numbers.
Eric Weisstein's World of Mathematics, Baxter-Hickerson Function.
MAPLE
f := n->(2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3;
PROG
(PARI) is(n)=isprime((2*10^(5*n)-10^(4*n)+2*10^(3*n)+10^(2*n)+10^n+1)/3) \\ Charles R Greathouse IV, Feb 17 2017
CROSSREFS
KEYWORD
hard,nonn
AUTHOR
G. L. Honaker, Jr., Dec 11 1999
STATUS
approved
Primes of form (2*10^(5n) - 10^(4n) + 2*10^(3n) + 10^(2n) + 10^n + 1)/3.
+10
10
2, 64037, 66666663333334000000033333336666667
OFFSET
1,1
COMMENTS
The Baxter-Hickerson function provides a number whose cube lacks zeros.
LINKS
Ed Pegg, Jr., Fun with numbers.
Eric Weisstein's World of Mathematics, Baxter-Hickerson Function.
CROSSREFS
KEYWORD
hard,nonn,bref
AUTHOR
G. L. Honaker, Jr., Dec 11 1999
EXTENSIONS
Next term has 665 digits.
STATUS
approved

Search completed in 0.017 seconds