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.

Showing 1-9 of 9 results.

A305929 Irregular table: row n >= 0 lists all k >= 0 such that the decimal representation of 9^k has n digits '0' (conjectured).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 12, 13, 14, 17, 34, 5, 8, 9, 10, 25, 26, 36, 11, 15, 16, 18, 19, 20, 21, 22, 23, 24, 28, 29, 30, 31, 32, 48, 54, 68, 41, 45, 56, 33, 35, 37, 44, 49, 53, 58, 64, 65, 38, 39, 40, 43, 46, 51, 52, 59, 61, 67, 82, 83, 106, 42, 47, 62, 66, 69, 72, 73, 76, 84, 89, 144, 27, 50
Offset: 0

Views

Author

M. F. Hasler, Jun 19 2018

Keywords

Comments

The set of (nonempty) rows forms 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 (9, 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 it is no longer guaranteed to have 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, 6, 7, 12, 13, 14, 17, 34 (= A030705)
1 : 5, 8, 9, 10, 25, 26, 36
2 : 11, 15, 16, 18, 19, 20, 21, 22, 23, 24, 28, 29, 30, 31, 32, 48, 54, 68
3 : 41, 45, 56
4 : 33, 35, 37, 44, 49, 53, 58, 64, 65
5 : 38, 39, 40, 43, 46, 51, 52, 59, 61, 67, 82, 83, 106
...
Column 0 is A063626: least k such that 9^k has n digits '0' in base 10.
Row lengths are 12, 7, 18, 3, 9, 13, 11, 11, 6, 9, 17, 15, 12, 9, 11, 6, 9, 9, ... (A305939).
Last element of the rows (largest exponent such that 9^k has exactly n digits 0) are (34, 36, 68, 56, 65, 106, 144, 134, 119, 138, 154, ...), A306119.
Inverse permutation is (0, 1, 2, 3, 4, 12, 5, 6, 13, 14, 15, 19, 7, 8, 9, 20, 21, 10, 22, 23, 24, 25, 26, 27, 28, 16, 17, 73, 29, 30, 31, 32, ...), not in OEIS.
		

Crossrefs

Cf. A305932 (analog for 2^k), A305933 (analog for 3^k), A305924 (analog for 4^k), ..., A305928 (analog for 8^k).

Programs

  • Mathematica
    mx = 1000; g[n_] := g[n] = DigitCount[9^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( A305929_row(n,M=50*(n+1))=select(k->#select(d->!d,digits(9^k))==n,[0..M]), [0..10])
    print(apply(t->#t,%)"\n"apply(vecmax,%)"\n"apply(t->t-1,Vec(vecsort(concat(%),,1)[1..99]))) \\ to show row lengths, last terms and the inverse permutation

Formula

Row n consists of the integers in (row n of A305933 divided by 2).

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

A305947 Number of powers of 7 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

10, 11, 12, 13, 9, 10, 9, 7, 10, 14, 21, 10, 18, 7, 11, 11, 12, 15, 17, 10, 11, 6, 10, 16, 13, 9, 7, 9, 11, 12, 10, 16, 7, 16, 9, 14, 13, 13, 9, 17, 14, 12, 11, 9, 13, 9, 12, 12, 9, 12, 14
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) = 10 is the number of terms in A030703 and in A195908, which includes the power 7^0 = 1.
These are the row lengths of A305927. It remains an open problem to provide a proof that these rows are complete (as for 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. A030703 (= row 0 of A305927): k such that 7^k has no 0's; A195908: these powers 7^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063606 (= column 1 of A305927): least k such that 7^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305946, A305938, A305939 (analog for 9^k).

Programs

  • PARI
    A305947(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(7^k))==n)
    
  • PARI
    A305947_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(7^k)),nMax)]++);a[^-1]}

A305938 Number of powers of 8 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

14, 11, 15, 11, 6, 12, 10, 7, 14, 21, 9, 9, 15, 8, 6, 10, 8, 13, 11, 13, 7, 10, 12, 8, 16, 10, 10, 10, 9, 14, 18, 11, 15, 12, 9, 9, 10, 17, 8, 12, 8, 12, 9, 8, 8, 12, 10, 17, 12, 6, 16
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) = 14 is the number of terms in A030704 and in A195946, which includes the power 7^0 = 1.
These are the row lengths of A305928. 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. A030704 (= row 0 of A305928): k such that 8^k has no 0's; A195946: these powers 8^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063616 (= column 1 of A305928): least k such that 8^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305939 (analog for 9^k).

Programs

  • PARI
    A305947(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(8^k))==n)
    
  • PARI
    A305947_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(8^k)),nMax)]++);a[^-1]}

A306119 Largest k such that 9^k has exactly n digits 0 (in base 10), conjectured.

Original entry on oeis.org

34, 36, 68, 56, 65, 106, 144, 134, 119, 138, 154, 186, 194, 191, 219, 208, 247, 267, 199, 314, 292, 263, 319, 303, 307, 345, 431, 401, 375, 388, 413, 498, 488, 504, 465, 513, 565, 464, 481, 541, 568, 532, 588, 542, 600, 677, 649, 633, 613, 734, 627
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) is the largest term in A030705: exponents of powers of 9 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. A063626: least k such that 9^k has n digits 0 in base 10.
Cf. A305939: number of k's such that 9^k has n digits 0.
Cf. A305929: row n lists exponents of 9^k with n digits 0.
Cf. A030705: { k | 9^k has no digit 0 } : row 0 of the above.
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, ..., A306118: analog for 2^k, ..., 8^k.

Programs

  • PARI
    A306119_vec(nMax,M=99*nMax+199,x=9,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(x^k)),nMax)]=k);a[^-1]}

Extensions

Data corrected thanks to a remark by R. J. Mathar, by M. F. Hasler, Feb 11 2023

A305946 Number of powers of 6 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

14, 10, 17, 16, 11, 14, 10, 8, 12, 19, 9, 16, 13, 11, 10, 10, 11, 10, 10, 17, 7, 15, 14, 16, 13, 22, 12, 17, 15, 17, 7, 6, 14, 22, 13, 19, 14, 12, 15, 7, 11, 14, 6, 12, 9, 12, 9, 14, 13, 15, 21
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) = 14 is the number of terms in A030702 and in A195948, which includes the power 6^0 = 1.
These are the row lengths of A305926. It remains an open problem to provide a proof that these rows are complete (as for 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. A030702 = row 0 of A305926: k such that 6^k has no 0's; A238936: these powers 6^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063596 = column 1 of A305926: least k such that 6^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

Programs

  • PARI
    A305946(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(6^k))==n)
    
  • PARI
    A305946_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(6^k)),nMax)]++);a[^-1]}

A305943 Number of powers of 3 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

23, 15, 31, 13, 18, 23, 23, 25, 16, 17, 28, 25, 22, 20, 18, 21, 19, 19, 18, 24, 33, 17, 17, 18, 17, 14, 21, 26, 25, 23, 24, 29, 17, 22, 18, 21, 27, 26, 20, 21, 13, 27, 24, 12, 18, 24, 16, 17, 15, 30, 24, 32, 24, 12, 16, 16, 23, 23, 20, 23, 19, 23, 10, 21, 20, 21, 23, 20, 19, 23, 23, 22, 16, 18, 20, 20, 13, 15, 25, 24, 28, 24, 21, 16, 14, 23, 21, 19, 23, 19, 27, 26, 22, 18, 27, 16, 31, 21, 18, 25, 24
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) = 23 is the number of terms in A030700 and in A238939, which include the power 3^0 = 1.
These are the row lengths of A305933. It remains an open problem to provide a proof that these rows are complete (as for 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 vanishingly small, cf. Khovanova link.

Crossrefs

Cf. A030700 = row 0 of A305933: k s.th. 3^k has no '0'; A238939: these powers 3^k.
Cf. A305931, A305934: powers of 3 with at least / exactly one '0'.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063555 = column 1 of A305933: least k such that 3^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

Programs

  • PARI
    A305943(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(3^k))==n)
    
  • PARI
    A305943_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(3^k)),nMax)]++);a[^-1]}

A305944 Number of powers of 4 having exactly n digits '0' (in base 10), conjectured.

Original entry on oeis.org

16, 22, 17, 14, 11, 19, 15, 15, 21, 20, 17, 22, 12, 13, 17, 24, 16, 19, 8, 17, 11, 15, 17, 15, 20, 17, 18, 20, 17, 21, 16, 19, 16, 14, 15, 19, 20, 24, 7, 16, 13, 14, 13, 14, 22, 22, 15, 18, 16, 16, 25
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) = 16 is the number of terms in A030701 and in A238940, which includes the power 4^0 = 1.
These are the row lengths of A305924. 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 A305924: k such that 4^k has no 0's; A238940: these powers 4^k.
Cf. A020665: largest k such that n^k has no '0's.
Cf. A063575 = column 1 of A305924: least k such that 4^k has n digits '0' in base 10.
Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

Programs

  • PARI
    A305944(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(4^k))==n)
    
  • PARI
    A305944_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(4^k)),nMax)]++);a[^-1]}

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]}
Showing 1-9 of 9 results.