A065712 -id:A065712 - cp's OEIS frontend

A306112 Largest k such that 2^k has exactly n digits 0 (in base 10), conjectured.

86, 229, 231, 359, 283, 357, 475, 476, 649, 733, 648, 696, 824, 634, 732, 890, 895, 848, 823, 929, 1092, 1091, 1239, 1201, 1224, 1210, 1141, 1339, 1240, 1282, 1395, 1449, 1416, 1408, 1616, 1524, 1727, 1725, 1553, 1942, 1907, 1945, 1870, 1724, 1972, 1965, 2075, 1983, 2114, 2257, 2256

Offset: 0

M. F. Hasler, Jun 22 2018

a(0) is the largest term in A007377: exponents of powers of 2 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.

M. F. Hasler, Zeroless powers, OEIS Wiki, March 2014, updated 2018.
T. Khovanova, The 86-conjecture, Tanya Khovanova's Math Blog, Feb. 2011.
W. Schneider, No Zeros, 2000, updated 2003. (On web.archive.org--see A007496 for a cached copy.)

Cf. A031146: least k such that 2^k has n digits 0 in base 10.
Cf. A305942: number of k's such that 2^k has n digits 0.
Cf. A305932: row n lists exponents of 2^k with n digits 0.
Cf. A007377: { k | 2^k has no digit 0 } : row 0 of the above.
Cf. A238938: { 2^k having no digit 0 }.
Cf. A027870: number of 0's in 2^n (and A065712, A065710, A065714, A065715, A065716, A065717, A065718, A065719, A065744 for digits 1 .. 9).
Cf. A102483: 2^n contains no 0 in base 3.

A306112_vec(nMax,M=99*nMax+199,x=2,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(x^k)),nMax)]=k);a[^-1]}

A322849 Number of times 2^k (for k < 4) appears as a substring within 2^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 3, 3, 1, 3, 3, 2, 0, 3, 5, 5, 3, 3, 4, 4, 3, 3, 4, 6, 4, 3, 6, 7, 4, 4, 6, 3, 3, 5, 5, 6, 4, 5, 7, 5, 8, 8, 5, 7, 6, 7, 9, 9, 3, 5, 10, 5, 3, 11, 10, 7, 8, 6, 10, 7, 8, 11, 8, 9, 8, 7, 12, 15, 10, 8, 13, 7, 8, 15, 8, 9, 12, 14, 12, 6, 13

Offset: 0

Gaitz Soponski, Dec 28 2018

It appears that the only 0 in this sequence is a(16).

			n =  0, a(n) = 1, 2^n =     1 - solution is 1;
n =  1, a(n) = 1, 2^n =     2 - solution is 2;
n =  2, a(n) = 1, 2^n =     4 - solution is 4;
n =  3, a(n) = 1, 2^n =     8 - solution is 8;
n =  4, a(n) = 1, 2^n =    16 - solution is 1;
n =  5, a(n) = 1, 2^n =    32 - solution is 2;
n =  6, a(n) = 1, 2^n =    64 - solution is 4;
n =  7, a(n) = 3, 2^n =   128 - solutions are 1,2,8;
n = 14, a(n) = 3, 2^n = 16384 - solutions are 1,4,8;
n = 15, a(n) = 2, 2^n = 32768 - solutions are 2,8;
n = 16, a(n) = 0, 2^n = 65536 - no solutions.

Cf. A065712 (1), A065710 (2), A065715 (4), A065719 (8).

Cf. A322849.

Array[Total@ DigitCount[2^#, 10, {1, 2, 4, 8}] &, 85, 0] (* Michael De Vlieger, Dec 31 2018 *)

a(n) = #select(x->((x==1) || (x==2) || (x==4) || (x==8)), digits(2^n)); \\ Michel Marcus, Dec 30 2018

a(n) <= A322850(n), for n >= 4.

a(n) = A065712(n) + A065710(n) + A065715(n) + A065719(n). - Michel Marcus, Dec 30 2018

A158191 Attach the smallest prime to the end of the string a(n-1) so a(n) is also prime.

Original entry on oeis.org

2, 23, 233, 2333, 23333, 2333323, 23333237, 233332373, 23333237353, 2333323735319, 2333323735319149, 2333323735319149571, 23333237353191495713, 23333237353191495713131, 233332373531914957131313

Offset: 1

Sergio Pimentel, Mar 13 2009

a(279) has 1001 digits. - Michael S. Branicky, May 26 2023

			a(6) = 2333323 since a(5) = 23333 (prime) and 233333, 233335, 233337, 2333311, 2333313, 2333317 and 2333319 are all composite.

Michael S. Branicky, Table of n, a(n) for n = 1..278 (terms 1..200 from Harvey P. Dale)

Cf. A048549, A088603, A089703, A065712, A100893.

nxt[n_]:=Module[{k=3},While[CompositeQ[n*10^IntegerLength[k]+k],k = NextPrime[ k]];n*10^IntegerLength[k]+k]; NestList[nxt,2,20] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 13 2019 *)

from itertools import islice
from sympy import isprime, nextprime
def agen(): # generator of terms
    p, s = 2, "2"
    while True:
        yield p
        q = 2
        while not isprime(p:=int(s+str(q))):
            q = nextprime(q)
        s += str(q)
print(list(islice(agen(), 15))) # Michael S. Branicky, May 26 2023

More terms from Sean A. Irvine, Nov 29 2009

A322850 Number of times 2^k for k < n-1 appears as a substring within 2^n.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 3, 1, 2, 3, 3, 1, 3, 4, 3, 0, 3, 5, 5, 3, 3, 4, 4, 5, 4, 5, 6, 4, 3, 6, 8, 4, 4, 6, 3, 3, 5, 5, 6, 5, 6, 7, 5, 9, 9, 6, 8, 6, 7, 9, 9, 3, 5, 10, 5, 3, 11, 10, 8, 8, 6, 11, 7, 10, 13, 10, 10, 8, 7, 13, 16, 12, 9, 13, 8, 9, 16, 8, 9, 12, 15, 14, 7, 14, 9

Offset: 0

Gaitz Soponski, Dec 28 2018

			n =  0, a(n) = 0, 2^n =     1 - no solutions;
n =  1, a(n) = 0, 2^n =     2 - no solutions;
n =  2, a(n) = 0, 2^n =     4 - no solutions;
n =  3, a(n) = 0, 2^n =     8 - no solutions;
n =  4, a(n) = 1, 2^n =    16 - solution is 1;
n =  5, a(n) = 1, 2^n =    32 - solution is 2;
n =  6, a(n) = 1, 2^n =    64 - solution is 4;
n =  7, a(n) = 3, 2^n =   128 - solutions are 1,2,8;
n = 14, a(n) = 4, 2^n = 16384 - solutions are 1,4,8,16;
n = 15, a(n) = 3, 2^n = 32768 - solutions are 2,8,32;
n = 16, a(n) = 0, 2^n = 65536 - no solutions.

Cf. A065712 (1), A065710 (2), A065715 (4), A065719 (8).

Cf. A322849.

Array[If[# < 4, Total@ DigitCount[2^#, 10, 2^Range[0, Min[# - 1, 3]]], Total@ DigitCount[2^#, 10, {1, 2, 4, 8}]] &, 85, 0] (* Michael De Vlieger, Dec 31 2018 *)

isp2(n) = (n==1) || (n==2) || (ispower(n,,&k) && (k==2));
a(n) = {my(d=digits(2^n), nb = 0); for (i=1, #d-1, for (j=1, #d-i+1, my(nd = vector(i, k, d[j+k-1])); if (nd[1] != 0, nb += isp2(fromdigits(nd))););); nb;} \\ Michel Marcus, Dec 30 2018

a(n) >= A322849(n), for n >= 4.

cp's OEIS Frontend

A306112 Largest k such that 2^k has exactly n digits 0 (in base 10), conjectured.

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A322849 Number of times 2^k (for k < 4) appears as a substring within 2^n.

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A158191 Attach the smallest prime to the end of the string a(n-1) so a(n) is also prime.

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A322850 Number of times 2^k for k < n-1 appears as a substring within 2^n.

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