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 1-6 of 6 results.

A092849 Number of primes <= A092800(n).

Original entry on oeis.org

2, 13, 87, 631, 4884, 39876, 336756, 2914140, 25683614, 229595323, 2075911061, 18944394204, 174219183788, 1612626081244, 15010195375589, 140387627837380, 1318553324421089, 12430152478650578, 117566893367826528, 1115249538463126642, 10607400237283187171, 101131327003506262770, 966289700136857769743
Offset: 1

Views

Author

Enoch Haga, Mar 07 2004

Keywords

Examples

			Below 10^1 there are 4 primes: 2 + 3 + 5 + 7 = 17. The rounded mean is 17/4 =~ 4. There are 2 primes less than 4: 2 and 3, so a(1) = 2.
		

Crossrefs

Formula

a(n) = PrimePi(A092800(n)) = PrimePi(A046731(n)/A006880(n)). - Robert G. Wilson v, Jan 19 2007

Extensions

a(9)-a(13) from Robert G. Wilson v, Jan 19 2007
a(14)-a(23) from Amiram Eldar, Jun 14 2024

A092850 Number of primes between A092800(n) and 10^n.

Original entry on oeis.org

2, 12, 81, 598, 4708, 38622, 327823, 2847315, 25163920, 225457188, 2042143752, 18663517814, 171846353051, 1592315669558, 14834375047080, 138850713196545, 1305003833233144, 12309801809090282, 116490773908518079, 1105570064097792198, 10519869248735544757, 100335959685809643520
Offset: 1

Views

Author

Enoch Haga, Mar 07 2004

Keywords

Examples

			Below 10^1 there are 4 primes: 2 + 3 + 5 + 7 = 17. The rounded mean is 17/4 =~ 4. There are 2 primes > 4: 5 and 7, so a(1) = 2.
		

Crossrefs

Formula

a(n) = PrimePi(10^n) - PrimePi(A092800(n)) = PrimePi(10^n) - PrimePi(A046731(n)/A006880(n)). - Robert G. Wilson v, Jan 19 2007
a(n) = A006880(n) - A092849(n). - Amiram Eldar, Jun 14 2024

Extensions

a(9)-a(13) from Robert G. Wilson v, Jan 19 2007
a(14)-a(22) from Amiram Eldar, Jun 14 2024

A092802 Mean (rounded) of composite numbers below 10^n.

Original entry on oeis.org

7, 53, 510, 5047, 50279, 501844, 5012809, 50094049, 500720912, 5005703554, 50046261712, 500382762350, 5003219541311, 50027458290511, 500236958622730, 5002065769284422, 50018168989895296, 500161047150051653, 5001437349858735047, 50012907314623137795, 500116546694296773760
Offset: 1

Views

Author

Enoch Haga, Mar 06 2004

Keywords

Comments

As with the primes in A092800, each succeeding term seems a close multiple of that preceding.

Examples

			a(3) = 510 because the mean of composites below 10^3 is 510.087... .
		

Crossrefs

Formula

Add composites and divide total by the count of composites below 10^n.
a(n) = floor((10^n(10^n + 1)/2 - A046731(n) - 1)/(10^n - A006880(n) - 1)). - Charles R Greathouse IV, Sep 25 2008

Extensions

More terms from Charles R Greathouse IV, Sep 25 2008

A091716 Standard deviation (rounded) of primes below 10^n.

Original entry on oeis.org

2, 29, 298, 2962, 29412, 292821, 2921863, 29170821, 291324189, 2910238255, 29078387910, 290589147156, 2904276036695
Offset: 1

Views

Author

Enoch Haga, Mar 05 2004

Keywords

Comments

It appears that a good estimate for the standard deviation of primes below 10^(n+1) is about 10 times the term for 10^n.
Heuristically, if we use a model where each positive integer x has probability approximately 1/log(x) of being prime, we should expect the standard deviation of the primes below N to be approximately N/sqrt(12). - Robert Israel, Sep 23 2014

Examples

			a(6) = 292821 (rounded from 292820.634) because this is the computed and rounded sample standard deviation of the 78498 primes below 10^6.
		

References

  • John E. Freund, Modern elementary statistics, 5th ed. (Prentice-Hall, 1979), pp. 42-47

Crossrefs

Programs

  • Maple
    seq(round(Statistics:-StandardDeviation(select(isprime, [$2 .. 10^n-1]))),n=1..7); # Robert Israel, Sep 23 2014

Extensions

a(9)-a(13) from Hiroaki Yamanouchi, Sep 23 2014

A092851 Difference in count of primes <= mean and > mean below 10^n in A092849 and A092850.

Original entry on oeis.org

0, 1, 6, 33, 176, 1254, 8933, 66825, 519694, 4138135, 33767309, 280876390, 2372830737, 20310411686, 175820328509, 1536914640835, 13549491187945, 120350669560296, 1076119459308449, 9679474365334444, 87530988547642414, 795367317696619250
Offset: 1

Views

Author

Enoch Haga, Mar 07 2004

Keywords

Examples

			a(3) = 6 because the count at 10^3 in A092849 is 87 and in A092850 it is 81. 87 - 81 = 6.
		

Crossrefs

Formula

a(n) = A092849(n)-A092850(n).

Extensions

a(14)-a(22) from Amiram Eldar, Jun 14 2024

A377571 a(n) is a n-digit number; for k = 1..n, its k-th digit is the most frequent k-th digit among n-digit prime numbers; in case of a tie, preference is given to the least digit.

Original entry on oeis.org

2, 13, 157, 1223, 12127, 104993, 1000597, 10289067, 100080553, 1000447633, 10015225131
Offset: 1

Views

Author

Rémy Sigrist, Nov 01 2024

Keywords

Comments

Although each digit taken independently is the most likely to be in that position for a prime number, overall, a term is not necessarily a prime number; for example, a(5) = 67 * 181 is composite.
For n>1, a(n)'s last digit is either 1, 3, 7 or 9. The prime number theorem says that the number of primes <= n is not linear with n, but grows sparser at a rate of n/log(n) so we expect the first digit of a(n) to be equal to 1 and the k-th digit of a(n) to be 0 for k>1 fixed and as n -> oo. - Chai Wah Wu, Nov 06 2024

Examples

			For n = 4: the frequency of digits among 4-digit prime numbers, and the corresponding most frequent digits, are:
  Digit    0     1     2     3    4    5    6    7    8    9  Most frequent
  -----  ---  ----  ----  ----  ---  ---  ---  ---  ---  ---  -------------
  1st      0   135*  127   120  119  114  117  107  110  112              1
  2nd    112    95   116*  104  104  107  115  104  106   98              2
  3rd    105   107   116*  110  103  106  104  101  105  104              2
  4th      0   266     0   268*   0    0    0  262    0  265              3
- so a(4) = 1223.
		

Crossrefs

Programs

  • PARI
    a(n, base = 10) = { my (f = vector(n, k, vector(base))); forprime (p = base^(n-1), base^n-1, my (d = digits(p, base)); for (k = 1, n, f[k][1+d[k]]++;);); my (b = vector(n), i); for (k = 1, n, vecmax(f[k], &i); b[k] = i-1;); fromdigits(b, base); }
    
  • Python
    from sympy import primerange
    def A377571(n):
        c = [[0]*10 for i in range(n)]
        for p in primerange(10**(n-1),10**n):
            for i, j in enumerate(str(p)):
                c[i][int(j)]+=1
        return int(''.join(str(c[i].index(max(c[i]))) for i in range(n))) # Chai Wah Wu, Nov 06 2024
Showing 1-6 of 6 results.