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.

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A305934 Powers of 3 that have exactly one digit '0' in base 10.

Original entry on oeis.org

59049, 14348907, 43046721, 129140163, 387420489, 3486784401, 847288609443, 68630377364883, 328256967394537077627, 26588814358957503287787, 717897987691852588770249, 6461081889226673298932241, 1144561273430837494885949696427, 22528399544939174411840147874772641, 67585198634817523235520443624317923
Offset: 1

Views

Author

M. F. Hasler (following a suggestion by Zak Seidov), Jun 14 2018

Keywords

Comments

Motivated by A030700: decimal expansion of 3^n contains no zeros (probably finite).
It appears that this sequence is finite. Is a(15) = 3^73 the last term?
There are no more terms through at least 3^(10^7) (which is a 4771213-digit number). It seems nearly certain that no power of 3 containing this many or more decimal digits could have fewer than two '0' digits. (Among numbers of the form 3^k with 73 < k <= 10^7, the only one having fewer than two '0' digits among its final 200 digits is 3^5028978.) - Jon E. Schoenfield, Jun 24 2018
The first 6 terms coincide with A305931: powers of 3 having at least one digit 0, with complement A238939 (within A000244: powers of 3) conjectured to be finite, too. Then, a(7..8) = A305931(9..10), etc.

Crossrefs

Cf. A030700: decimal expansion of 3^n contains no zeros (probably finite), A238939: powers of 3 with no digit '0' in their decimal expansion, A000244: powers of 3.
Subsequence of A305931: powers of 3 having at least one '0'.
Cf. A305933: row n = { k | 3^k has n digits '0' }.

Programs

  • Mathematica
    Select[3^Range[120], DigitCount[#, 10, 0] == 1 &] (* Michael De Vlieger, Jul 01 2018 *)
  • PARI
    for(n=1,99, #select(t->!t,digits(3^n))==1&& print1(3^n","))

Formula

a(n) = 3^A305933(1,n).

A305931 Powers of 3 having at least one digit '0' in their decimal representation.

Original entry on oeis.org

59049, 14348907, 43046721, 129140163, 387420489, 3486784401, 10460353203, 31381059609, 847288609443, 68630377364883, 205891132094649, 1853020188851841, 5559060566555523, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089, 4052555153018976267, 12157665459056928801
Offset: 1

Views

Author

M. F. Hasler, Jun 15 2018

Keywords

Comments

The analog of A298607 for 3^k instead of 2^k.
The complement A238939 is conjectured to have only 23 elements, the largest being 3^68. Thus, all larger powers of 3 are (conjectured to be) in this sequence. Each of the subsequences "powers of 3 with exactly n digits 0" is conjectured to be finite. Provided there is at least one such element for each n >= 0, this leads to a partition of the integers, given in A305933.

Crossrefs

Cf. A030700 = row 0 of A305933: decimal expansion of 3^n contains no zeros.
Complement (within A000244: powers of 3) of A238939: powers of 3 with no digit '0' in their decimal expansion.
Analog of A298607: powers of 2 with the digit '0' in their decimal expansion.
The first six terms coincide with the finite sequence A305934: powers of 3 having exactly one digit 0.

Programs

  • Mathematica
    Select[3^Range[0,40],DigitCount[#,10,0]>0&] (* Harvey P. Dale, May 30 2020 *)
  • PARI
    for(k=0,69, vecmin(digits(3^k))|| print1(3^k","))
    
  • PARI
    select( t->!vecmin(digits(t)), apply( k->3^k, [0..40]))

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]}

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","))

A306113 Largest k such that 3^k has exactly n digits 0 (in base 10), conjectured.

Original entry on oeis.org

68, 73, 136, 129, 205, 237, 317, 268, 251, 276, 343, 372, 389, 419, 565, 416, 494, 571, 637, 628, 713, 629, 638, 655, 735, 690, 862, 802, 750, 863, 826, 996, 976, 1008, 1085, 1026, 1130, 995, 962, 1082, 1136, 1064, 1176, 1084, 1215, 1354, 1298, 1275, 1226, 1468, 1353
Offset: 0

Views

Author

M. F. Hasler, Jun 22 2018

Keywords

Comments

a(0) is the largest term in A030700: exponents of powers of 3 without digit 0.
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. A063555: least k such that 3^k has n digits 0 in base 10.
Cf. A305943: number of k's such that 3^k has n digits 0.
Cf. A305933: row n lists exponents of 3^k with n digits 0.
Cf. A030700: { k | 3^k has no digit 0 } : row 0 of the above.
Cf. A238939: { 3^k having no digit 0 }.
Cf. A305930: number of 0's in 3^n.
Cf. A306112, ..., A306119: analog for 2^k, ..., 9^k.

Programs

  • PARI
    A306113_vec(nMax,M=99*nMax+199,x=3,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(x^k)),nMax)]=k);a[^-1]}

A358271 Product of the digits of 3^n.

Original entry on oeis.org

1, 3, 9, 14, 8, 24, 126, 112, 180, 1296, 0, 1372, 240, 3240, 217728, 0, 0, 0, 0, 24192, 0, 0, 0, 2709504, 6635520, 0, 66355200, 8534937600, 731566080, 0, 0, 10369949184, 0, 0, 399983754240, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6243870843076608000
Offset: 0

Views

Author

Joseph Caliendo, Nov 06 2022

Keywords

Comments

a(68) is likely the last nonzero term; see A030700 and A238939. - Michael S. Branicky, Nov 06 2022

Examples

			For a(0), 3^0 = 1 with product of digits 1;
for a(3), 3^3 = 27 with product of digits 2*7 = 14;
for a(10), 3^10 = 59049 with product of digits 5*9*0*4*9 = 0.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Times @@ IntegerDigits[3^n]; Array[a, 69, 0] (* Amiram Eldar, Nov 07 2022 *)
  • PARI
    a(n) = vecprod(digits(3^n)); \\ Michel Marcus, Nov 07 2022
  • Python
    from math import prod
    def a(n): return prod(map(int, str(3**n)))
    print([a(n) for n in range(69)]) # Michael S. Branicky, Nov 06 2022
    

Formula

a(n) = A007954(A000244(n)).

Extensions

More terms from Michael S. Branicky, Nov 06 2022
Previous Showing 11-16 of 16 results.