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.

A359183 a(n) is the smallest number such that when written in all bases from base 2 to base n its leading digit equals the base - 1.

Original entry on oeis.org

1, 2, 54, 13122, 15258789062500
Offset: 2

Views

Author

Scott R. Shannon, Dec 18 2022

Keywords

Comments

Each term can be represented in some base < n as a number < n multiplied by the base to some power. The terms given in the data section are a(2) = 1, a(3) = 2, a(4) = 54 = 2*3^3, a(5) = 13122 = 2*3^8, a(6) = 15258789062500 = 4*5^18, a(7) = 8158...4608 (186 digits) = 3*4^308. The other known terms (too large to write in the data section) are a(8) = 9532...8658 (3448 digits) = 2*3^7226, a(9) = a(10) = 9123...2500 (10344 digits) = 4*5^14798.
Assuming a(11) exists, it is greater than 10^22500.

Examples

			a(2) = 1 as 1 = 1_2, which has 1 = 2 - 1 as its leading digit.
a(3) = 2 as 2 = 10_2 = 2_3, which have 1 = 2 - 1 and 2 = 3 - 1 as their leading digits.
a(4) = 54 as 54 = 110110_2 = 2000_3 = 312_4, which have 1 = 2 - 1, 2 = 3 - 1 and 3 = 4 - 1 as their leading digits.
a(5) = 13122 as 13122 = 11001101000010_2 = 200000000_3 = 3031002_4 = 404442_5, which have 1 = 2 - 1, 2 = 3 - 1, 3 = 4 - 1 and 4 = 5 - 1 as their leading digits.
a(6) = 15258789062500 as 15258789062500 = 110000010110110101100111010011101100100_2 = 2000000201121020121212112011_3 = 3132002312230322131210_4 = 4000000000000000000_5 = 52241442501204004_6, which have 1 = 2 - 1, 2 = 3 - 1, 3 = 4 - 1, 4 = 5 - 1 and 5 = 6 - 1 as their leading digits.
a(7) = 81582795696655426358720748526459181157825502882872103403434619627581986794626\
  90448473536034793921827874140100908746255557234586263455831973302268738547817\
  2585724832003163984432734404608 (Too large to include in the DATA section)
		

Crossrefs

Programs

  • Python
    from math import floor, log
    def a(n):
        arr = []
        p = 0
        while True:
            for m in range(1, n):
                for b in range(2, max(3, n)):
                        k = m*b**p
                        if k in arr:
                            continue
                        arr.append(k)
                        q = 1
                        for b in range(3, n+1):
                            if floor(k/b**floor(log(k)/log(b))) != b-1:
                                q = 0
                                break
                        if q:
                            return k
            p += 1
    # Christoph B. Kassir, Feb 10 2023