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 1-3 of 3 results.

A370612 The smallest number whose prime factor concatenation, when written in base n, does not contain 0 and contains all digits 1,...,(n-1) at least once.

Original entry on oeis.org

3, 5, 14, 133, 706, 2490, 24258, 217230, 2992890, 24674730, 647850030, 4208072190, 82417704810
Offset: 2

Views

Author

Chai Wah Wu, Apr 30 2024

Keywords

Comments

All terms are squarefree. Many thanks to Michael Branicky for pointing out errors in the terms in the original submission.

Examples

			a(2) = 3 = 3 whose prime factor in base 2 is: 11.
a(3) = 5 = 5 whose prime factor in base 3 is: 12.
a(4) = 14 = 2*7 whose prime factors in base 4 are: 2, 13.
a(5) = 133 = 7*19 whose prime factors in base 5 are: 12, 34.
a(6) = 706 = 2*353 whose prime factors in base 6 are: 2, 1345.
a(7) = 2490 = 2*3*5*83 whose prime factors in base 7 are: 2, 3, 5, 146.
a(8) = 24258 = 2*3*13*311 whose prime factors in base 8 are: 2, 3, 15, 467.
a(9) = 217230 = 2*3*5*13*557 whose prime factors in base 9 are: 2, 3, 5, 14, 678.
a(10) = 2992890 = 2*3*5*67*1489.
a(11) = 24674730 = 2*3*5*19*73*593 whose prime factors in base 11 are: 2, 3, 5, 18, 67, 49a.
a(12) = 647850030 = 2*3*5*19*1136579 whose prime factors in base 12 are: 2, 3, 5, 17, 4698ab.
a(13) = 4208072190 = 2*3*5*7*61*89*3691 whose prime factors in base 13 are: 2, 3, 5, 7, 49, 6b, 18ac.
a(14) = 82417704810 = 2*3*5*7*23*937*18211 whose prime factors in base 14 are: 2, 3, 5, 7, 19, 4ad, 68cb.
		

Crossrefs

Programs

  • Python
    from math import factorial
    from itertools import count
    from sympy import primefactors
    from sympy.ntheory import digits
    def A370612(n): return next(k for k in count(max(factorial(n-1),2)) if 0 not in (s:=set.union(*(set(digits(p,n)[1:]) for p in primefactors(k)))) and len(s) == n-1)

Formula

(n-1)! <= a(n) <= A371194(n).

Extensions

a(13)-(14) from Dominic McCarty, Jan 07 2025

A371511 a(n) is the smallest prime such that its representation in base n contains each of the digits 0,1,...,n-2 at least once and does not contain the digit n-1.

Original entry on oeis.org

3, 73, 683, 8521, 123323, 2140069, 43720693, 1012356487, 26411157737, 749149003087, 23459877380431, 798411310382011, 29471615863458281, 1158045600182881261, 48851274656431280857, 2193475267557861578041, 104737172422274885174411, 5257403213296398892278377
Offset: 3

Views

Author

Chai Wah Wu, Apr 10 2024

Keywords

Comments

Conjecture: for n>3, a(n) has digit sum 2+(n-2)(n-1)/2 if n is of the form 4k+3 and has digit sum 1+(n-2)(n-1)/2 otherwise.

Examples

			The corresponding base-n representations are:
n   a(n) in base n
------------------------
3   10
4   1021
5   10213
6   103241
7   1022354
8   10123645
9   101236457
10  1012356487
11  10223456798
12  10123459a867
13  1012345678a9b
14  1012345678c9ab
15  1022345678a9cdb
16  10123456789acbed
		

Crossrefs

Programs

  • Python
    from math import gcd
    from sympy import nextprime
    from sympy.ntheory import digits
    def A371511(n):
        m, j = n, 0
        if n > 3:
            for j in range(1,n-1):
                if gcd((n*(n-1)>>1)+j,n-1) == 1:
                     break
        if j == 0:
            for i in range(2,n-1):
                m = n*m+i
        elif j == 1:
            for i in range(1,n-1):
                m = n*m+i
        else:
            for i in range(2,1+j):
                m = n*m+i
            for i in range(j,n-1):
                m = n*m+i
        m -= 1
        while True:
            s = digits(m:=nextprime(m), n)[1:]
            if n-1 not in s and len(set(s))==n-1:
                return m

Formula

For n>3, a(n) >= (n^(n-1)-n)/(n-1)^2 + n^(n-1). If n = 4k+3 for k>0, then a(n) >= (n^(n-1)-n)/(n-1)^2 + n^(n-1) + n^(n-3) .

A371512 a(n) is the smallest prime such that its representation in base n contains each of the digits 1,...,n-2 at least once and does not contain the digit 0 nor the digit n-1.

Original entry on oeis.org

13, 37, 163, 1861, 22481, 304949, 5455573, 112345687, 2831681057, 68057976031, 1953952652167, 61390449569437, 2224884906436873, 77181689614101181, 3052505832274232281, 129003238915759600789, 6090208982148446231753, 276667213296398892309917, 13944042713948404997174231
Offset: 3

Views

Author

Chai Wah Wu, Apr 10 2024

Keywords

Comments

Conjecture: for n>3, a(n) has digit sum 2+(n-2)(n-1)/2 if n is of the form 4k+3 and has digit sum 1+(n-2)(n-1)/2 otherwise.

Examples

			The corresponding base-n representations are:
n   a(n) in base n
------------------------
3   111
4   211
5   1123
6   12341
7   122354
8   1123465
9   11234567
10  112345687
11  1223456987
12  1123458a967
13  112345678ba9
14  11234567a8bc9
15  122345678acb9d
16  1123456789ceabd
		

Crossrefs

Programs

  • Python
    from math import gcd
    from sympy import nextprime
    from sympy.ntheory import digits
    def A371512(n):
        m, j = 1, 0
        if n > 3:
            for j in range(1,n-1):
                if gcd((n*(n-1)>>1)+j,n-1) == 1:
                     break
        if j == 0:
            for i in range(2,n-1):
                m = n*m+i
        elif j == 1:
            for i in range(1,n-1):
                m = n*m+i
        else:
            for i in range(2,1+j):
                m = n*m+i
            for i in range(j,n-1):
                m = n*m+i
        m -= 1
        while True:
            s = digits(m:=nextprime(m), n)[1:]
            if (not (0 in s or n-1 in s)) and len(set(s))==n-2:
                return m

Formula

For n>=3, a(n) >= (n^(n-1)-n)/(n-1)^2 + n^(n-2). If n = 4k+3 for k>0, then a(n) >= (n^(n-1)-n)/(n-1)^2 + n^(n-2) + n^(n-3) .
Showing 1-3 of 3 results.