cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-30 of 30 results.

A238940 Powers of 4 without the digit '0' in their decimal expansion.

Original entry on oeis.org

1, 4, 16, 64, 256, 16384, 65536, 262144, 16777216, 268435456, 4294967296, 17179869184, 68719476736, 4722366482869645213696, 75557863725914323419136, 77371252455336267181195264
Offset: 1

Views

Author

M. F. Hasler, Mar 07 2014

Keywords

Comments

Conjectured to be finite and complete. See the OEIS wiki page for further information, references and links.

Crossrefs

For the zeroless numbers (powers x^n), see A238938, A238939, A238940, A195948, A238936, A195908, A195946, A195945, A195942, A195943.
For the corresponding exponents, see A007377, A008839, A030700, A030701, A030702, A030703, A030704, A030705, A030706, A195944.
For other related sequences, see A052382, A027870, A102483.

Programs

  • Mathematica
    Select[4^Range[0,50],DigitCount[#,10,0]==0&] (* Harvey P. Dale, Aug 31 2021 *)
  • PARI
    for(n=0,99,vecmin(digits(4^n))&& print1(4^n","))

Formula

a(n)=4^A030701(n).

Extensions

Keyword:fini removed by Jianing Song, Jan 28 2023 as finiteness is only conjectured.

A071531 Smallest exponent r such that n^r contains at least one zero digit (in base 10).

Original entry on oeis.org

10, 10, 5, 8, 9, 4, 4, 5, 1, 5, 4, 6, 7, 4, 3, 7, 4, 4, 1, 5, 3, 6, 6, 4, 6, 5, 5, 4, 1, 6, 2, 2, 3, 4, 5, 3, 4, 5, 1, 5, 3, 3, 4, 2, 5, 2, 2, 2, 1, 2, 2, 2, 4, 2, 5, 4, 6, 3, 1, 5, 6, 3, 2, 4, 6, 3, 9, 3, 1, 2, 6, 3, 3, 4, 8, 4, 2, 3, 1, 4, 5, 5, 2, 4, 3, 3, 6, 3, 1, 5, 5, 3, 3, 2, 7, 2, 2, 2, 1, 1, 1, 1, 1, 1
Offset: 2

Views

Author

Paul Stoeber (paul.stoeber(AT)stud.tu-ilmenau.de), Jun 02 2002

Keywords

Comments

For all n, a(n) is at most 40000, as shown below. Is 10 an upper bound?
If n has d digits, the numbers n, n^2, ..., n^k have a total of about N = k*(k+1)*d/2, and if these were chosen randomly the probability of having no zeros would be (9/10)^N. The expected number of d-digit numbers n with f(n)>k would be 9*10^(d-1)*(9/10)^N. If k >= 7, (9/10)^(k*(k+1)/2)*10 < 1 so we would expect heuristically that there should be only finitely many n with f(n) > 7. - Robert Israel, Jan 15 2015
The similar definition using "...exactly one digit 0..." would be ill-defined for all multiples of 100 and others (1001, ...). - M. F. Hasler, Jun 25 2018
When r=40000, one of the last five digits of n^r is always 0. Working modulo 10^5, we have 2^r=9736 and 5^r=90625, and both of these are idempotent; also, if gcd(n,10)=1, then n^r=1, and if 10|n, then n^r=0. Therefore the last five digits of n^r are always either 00000, 00001, 09736, or 90625. In particular, a(n) <= 40000. - Mikhail Lavrov, Nov 18 2021

Examples

			a(4)=5 because 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024 (has zero digit).
		

Crossrefs

Cf. A305941 for the actual powers n^k.
Cf. A007377, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, A195944: decimal expansion of k^n contains no zeros, k = 2, 3, 4, ...
Cf. A305932, A305933, A305924, ..., A305929: row n = {k: x^k has n 0's}, x = 2, 3, ..., 9.
Cf. A305942, ..., A305947, A305938, A305939: #{k: x^k has n 0's}, x = 2, 3, ..., 9.
Cf. A306112, ..., A306119: largest k: x^k has n 0's; x = 2, 3, ..., 9.

Programs

  • Maple
    f:= proc(n) local j;
    for j from 1 do if has(convert(n^j,base,10),0) then return j fi od:
    end proc:
    seq(f(n),n=2..100); # Robert Israel, Jan 15 2015
  • Mathematica
    zd[n_]:=Module[{r=1},While[DigitCount[n^r,10,0]==0,r++];r]; Array[zd,110,2] (* Harvey P. Dale, Apr 15 2012 *)
  • PARI
    A071531(n)=for(k=1, oo, vecmin(digits(n^k))||return(k)) \\ M. F. Hasler, Jun 23 2018
  • Python
    def a(n):
        r, p = 1, n
        while 1:
            if "0" in str(p):
                return r
            r += 1
            p *= n
    [a(n) for n in range(2, 100)] # Tim Peters, May 19 2005
    

Formula

a(n) >= 1 with equality iff n is in A011540 \ {0} = {10, 20, ..., 100, 101, ...}. - M. F. Hasler, Jun 23 2018

A239010 Exponents m such that the decimal expansion of 5^m exhibits its first zero from the right later than any previous exponent.

Original entry on oeis.org

0, 2, 3, 5, 6, 9, 11, 15, 17, 18, 25, 26, 30, 33, 57, 58, 153, 1839, 3290, 4081, 16431, 577839, 2190974, 15167023, 23155442, 24477994, 36290003, 53687441, 62497567, 181850218, 790111167, 872257561, 4531889178, 26964400609, 32626158305, 268600630073
Offset: 1

Views

Author

Keywords

Comments

Assume that a zero precedes all decimal expansions. This will take care of those cases in A008839.
Inspired by the seqfan list discussion Re: "possible sequence", beginning with David Wilson 7:57 PM Mar 06 2014 and continued by M. F. Hasler, Allan C. Wechsler and Franklin T. Adams-Watters.
Highest position known is 232th digit from the right for a(33). - Bert Dobbelaere, Jan 21 2019

Crossrefs

Programs

  • Mathematica
    f[n_] := Position[ Reverse@ Join[{0}, IntegerDigits[ PowerMod[5, n, 10^500]]], 0, 1, 1][[1, 1]]; k = mx = 0; lst = {}; While[k < 100000001, c = f[k]; If[c > mx, mx = c; AppendTo[ lst, k]; Print@ k]; k++]; lst

Extensions

a(30)-a(33) from Bert Dobbelaere, Jan 21 2019
a(34)-a(36) from Chai Wah Wu, Jan 18 2020

A305925 Irregular table read by rows in which row n >= 0 lists all k >= 0 such that the decimal representation of 5^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58, 8, 12, 14, 15, 16, 19, 20, 21, 22, 25, 26, 31, 41, 42, 43, 85, 13, 23, 24, 27, 28, 29, 32, 36, 37, 56, 57, 107, 34, 35, 38, 39, 50, 54, 59, 74, 75, 84, 112, 40, 44, 46, 47, 49, 51, 60, 73, 78, 79, 82, 83, 86, 88, 89, 95, 96, 97, 106, 113, 127
Offset: 0

Views

Author

M. F. Hasler, Jun 19 2018

Keywords

Comments

The set of (nonempty) rows is a partition of the nonnegative integers.
Read as a flattened sequence, a permutation of the nonnegative integers.
In the same way, another choice of (basis, digit, base) = (m, d, b) different from (5, 0, 10) will yield a similar partition of the nonnegative integers, trivial if m is a multiple of b.
It remains an open problem to provide a proof that the rows are complete, in the same way as each of the terms of A020665 is unproved.
We can also decide that the rows are to be truncated as soon as no term is found within a sufficiently large search limit. (For all of the displayed rows, there is no additional term up to many orders of magnitude beyond the last term.) That way the rows are well-defined, but we are no more guaranteed to get a partition of the integers.
The author finds this sequence "nice", i.e., appealing (as well as, e.g., the variant A305933 for basis 3) in view of the idea of partitioning the integers in such an elementary yet highly nontrivial way, and the remarkable fact that the rows are just roughly one line long. Will this property remain for large n, or else, how will the row lengths evolve?

Examples

			The table reads:
n \ k's
0 : 0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58 (cf. A008839)
1 : 8, 12, 14, 15, 16, 19, 20, 21, 22, 25, 26, 31, 41, 42, 43, 85
2 : 13, 23, 24, 27, 28, 29, 32, 36, 37, 56, 57, 107
3 : 34, 35, 38, 39, 50, 54, 59, 74, 75, 84, 112
4 : 40, 44, 46, 47, 49, 51, 60, 73, 78, 79, 82, 83, 86, 88, 89, 95, 96, 97, 106, 113, 127
5 : 48, 55, 61, 67, 77, 91, 102, 110, 111, 126, 148, 157
...
The first column is A063585: least k such that 5^k has n digits '0' in base 10.
Row lengths are 16, 16, 12, 11, 21, 12, 17, 14, 16, 17, 14, 13, 16, 18, 13, 14, 10, 10, 21, 7,... (A305945).
Last terms of the rows are (58, 85, 107, 112, 127, 157, 155, 194, 198, 238, 323, 237, 218, 301, 303, 324, 339, 476, 321, 284, ...), A306115.
The inverse permutation is (0, 1, 2, 3, 4, 5, 6, 7, 16, 8, 9, 10, 17, 32, 18, 19, 20, 11, 12, 21, 22, 23, 24, 33, 34, 25, 26, 35, 36, 37, 13, 27,...), not in OEIS.
		

Crossrefs

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305924 (analog for 4^k), ..., A305929 (analog for 9^k).

Programs

  • Mathematica
    mx = 1000; g[n_] := g[n] = DigitCount[5^n, 10, 0]; f[n_] := Select[Range@mx, g@# == n &]; Table[f@n, {n, 0, 4}] // Flatten (* Robert G. Wilson v, Jun 20 2018 *)
  • PARI
    apply( A305925_row(n,M=60*(n+1))=select(k->#select(d->!d,digits(5^k))==n,[0..M]), [0..19])

A305945 Number of powers of 5 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

16, 16, 12, 11, 21, 12, 17, 14, 16, 17, 14, 13, 16, 18, 13, 14, 10, 10, 21, 7, 19, 13, 15, 13, 10, 15, 12, 15, 11, 11, 15, 10, 9, 15, 17, 16, 13, 12, 12, 11, 14, 9, 14, 15, 16, 14, 13, 14, 15, 24, 14
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) = 16 is the number of terms in A008839 and in A195948, which includes the power 5^0 = 1.
These are the row lengths of A305925. It remains an open problem to provide a proof that these rows are complete (as are all terms of A020665), but the search has been pushed to many orders of magnitude beyond the largest known term, and the probability of finding an additional term is vanishing, cf. Khovanova link.

Crossrefs

Cf. A030701 (= row 0 of A305925): k such that 5^k has no 0's; A195948: these powers 4^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063585 (= column 1 of A305925): least k such that 5^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

Programs

  • PARI
    A305945(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(5^k))==n)
    
  • PARI
    A305945_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(5^k)),nMax)]++);a[^-1]}

A195985 Least prime such that p^2 is a zeroless n-digit number.

Original entry on oeis.org

2, 5, 11, 37, 107, 337, 1061, 3343, 10559, 33343, 105517, 333337, 1054133, 3333373, 10540931, 33333359, 105409309, 333333361, 1054092869, 3333333413, 10540925639, 33333333343, 105409255363, 333333333367, 1054092553583, 3333333333383, 10540925534207
Offset: 1

Views

Author

M. F. Hasler, Sep 26 2011

Keywords

Examples

			a(1)^2=4, a(2)^2=25, a(3)^2=121, a(4)^2=1369 are the least squares of primes with 1, 2, 3 resp. 4 digits, and these digits are all nonzero.
a(5)=107 since 101^2=10201 and 103^2=10609 both contain a zero digit, but 107^2=11449 does not.
a(1000)=[10^500/3]+10210 (500 digits), since primes below sqrt(10^999) = 10^499*sqrt(10) ~ 3.162e499 have squares of less than 1000 digits, between sqrt(10^999) and 10^500/3 = sqrt(10^1000/9) ~ 3.333...e499 they have at least one zero digit. Finally, the 7 primes between 10^500/3 and a(1000) also happen to have a "0" digit in their square, but not so
  a(1000)^2 = 11111...11111791755555...55555659792849
  = [10^500/9]*(10^500+5) + 6806*10^500+104237294.
		

Crossrefs

Programs

  • PARI
    a(n)={ my(p=sqrtint(10^n\9)-1); until( is_A052382(p^2), p=nextprime(p+2));p}

A050726 Decimal expansion of 5^n contains no pair of consecutive equal digits (probably finite).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 14, 15, 16, 17, 18, 19, 20, 21, 22, 27, 28, 29, 32
Offset: 0

Views

Author

Patrick De Geest, Sep 15 1999

Keywords

Comments

No further terms up to 3000. - Harvey P. Dale, May 11 2011
No further terms up to 80000. - Ivan N. Ianakiev, Aug 31 2019

Examples

			5^32 = 23283064365386962890625.
		

Crossrefs

Programs

  • Maple
    q:= n-> (s-> andmap(i-> s[i]<>s[i+1], [$1..length(s)-1]))(""||(5^n)):
    select(q, [$0..200])[];  # Alois P. Heinz, Mar 07 2024
  • Mathematica
    Select[Range[0,3000],And@@(First[#]!=Last[#]&/@Partition[ IntegerDigits[ 5^#],2,1])&] (* Harvey P. Dale, May 11 2011 *)
    Select[Range[0,40],SequenceCount[IntegerDigits[5^#],{x_,x_}]==0&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 12 2020 *)

A252482 Exponents n such that the decimal expansion of the power 12^n contains no zeros.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 8, 10, 14, 20, 26
Offset: 1

Views

Author

M. F. Hasler, Dec 17 2014

Keywords

Comments

Conjectured to be finite.
See A245853 for the actual powers 12^a(n).

Crossrefs

For zeroless powers x^n, see A238938 (x=2), A238939, A238940, A195948, A238936, A195908, A245852, A240945 (k=9), A195946 (x=11), A245853, A195945; A195942, A195943, A103662.
For the corresponding exponents, see A007377, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, this sequence A252482, A195944.
For other related sequences, see A052382, A027870, A102483, A103663.

Programs

  • Mathematica
    Select[Range[0,30],DigitCount[12^#,10,0]==0&] (* Harvey P. Dale, Apr 06 2019 *)
  • PARI
    for(n=0,9e9,vecmin(digits(12^n))&&print1(n","))

A253638 Number of zeros in the decimal expansion of 5^n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 0, 1, 2, 0, 3, 3, 2, 2, 3, 3, 4, 1, 1, 1, 4, 7, 4, 4, 5, 4, 3, 4, 6, 6, 3, 5, 2, 2, 0, 3, 4, 5, 6, 7, 8, 6, 6, 5, 7, 8, 8, 6, 8, 4, 3, 3, 6, 5, 4, 4, 8, 7, 4, 4, 3, 1, 4, 6, 4, 4, 6, 5, 6, 7, 6, 4, 4, 4, 6, 9, 12, 8, 5, 9, 7, 6, 4, 2, 9, 8, 5, 5, 3, 4, 6, 6, 9, 14, 12, 12, 12, 12, 13
Offset: 0

Views

Author

Zak Seidov, Jan 07 2015

Keywords

Comments

Probably a(58) is the last 0 term.

Examples

			5^57 = 6938893903907228377647697925567626953125, 2 zeros hence a(57) = 2,
5^58 = 34694469519536141888238489627838134765625, no zeros hence a(58) = 0,
5^59 = 173472347597680709441192448139190673828125, 3 zeros hence a(59) = 3.
		

Crossrefs

Cf. A008839 (no zeros in 5^n), A055641 (number of zeros for n), A000351 (5^n).

Programs

  • Mathematica
    Table[Count[IntegerDigits[5^n],0],{n,0,200}]
  • PARI
    a(n) = my(d = digits(5^n)); sum(i=1, #d, d[i] == 0); \\ Michel Marcus, Jan 15 2015

Formula

a(n) = A055641(A000351(n)). - Michel Marcus, Jan 15 2015

A306115 Largest k such that 5^k has exactly n digits 0 (in base 10), conjectured.

Original entry on oeis.org

58, 85, 107, 112, 127, 157, 155, 194, 198, 238, 323, 237, 218, 301, 303, 324, 339, 476, 321, 284, 496, 421, 475, 415, 537, 447, 494, 538, 531, 439, 473, 546, 587, 588, 642, 690, 769, 689, 687, 686, 757, 732, 683, 826, 733, 825, 833, 810, 827, 888, 966
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) is the largest term in A008839: exponents of powers of 5 without digit 0 in base 10.
There is no proof for any of the terms, just as for any term of A020665 and many similar / related sequences. However, the search has been pushed to many magnitudes beyond the largest known term, and the probability of any of the terms being wrong is extremely small, cf., e.g., the Khovanova link.

Crossrefs

Cf. A063585: least k such that 5^k has n digits 0 in base 10.
Cf. A305945: number of k's such that 5^k has n digits 0.
Cf. A305925: row n lists exponents of 5^k with n digits 0.
Cf. A008839: { k | 5^k has no digit 0 } : row 0 of the above.
Cf. A195948: { 5^k having no digit 0 }.
Cf. A020665: largest k such that n^k has no digit 0 in base 10.
Cf. A071531: least k such that n^k contains a digit 0 in base 10.
Cf. A103663: least x such that x^n has no digit 0 in base 10.
Cf. A306112, ..., A306119: analog for 2^k, ..., 9^k.

Programs

  • PARI
    A306115_vec(nMax,M=99*nMax+199,x=5,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(x^k)),nMax)]=k);a[^-1]}
Previous Showing 21-30 of 30 results.