cp's OEIS Frontend

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.

%I A359183 #43 Feb 26 2023 19:40:44
%S A359183 1,2,54,13122,15258789062500
%N 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.
%C A359183 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.
%C A359183 Assuming a(11) exists, it is greater than 10^22500.
%e A359183 a(2) = 1 as 1 = 1_2, which has 1 = 2 - 1 as its leading digit.
%e A359183 a(3) = 2 as 2 = 10_2 = 2_3, which have 1 = 2 - 1 and 2 = 3 - 1 as their leading digits.
%e A359183 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.
%e A359183 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.
%e A359183 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.
%e A359183 a(7) = 81582795696655426358720748526459181157825502882872103403434619627581986794626\
%e A359183   90448473536034793921827874140100908746255557234586263455831973302268738547817\
%e A359183   2585724832003163984432734404608 (Too large to include in the DATA section)
%o A359183 (Python)
%o A359183 from math import floor, log
%o A359183 def a(n):
%o A359183     arr = []
%o A359183     p = 0
%o A359183     while True:
%o A359183         for m in range(1, n):
%o A359183             for b in range(2, max(3, n)):
%o A359183                     k = m*b**p
%o A359183                     if k in arr:
%o A359183                         continue
%o A359183                     arr.append(k)
%o A359183                     q = 1
%o A359183                     for b in range(3, n+1):
%o A359183                         if floor(k/b**floor(log(k)/log(b))) != b-1:
%o A359183                             q = 0
%o A359183                             break
%o A359183                     if q:
%o A359183                         return k
%o A359183         p += 1
%o A359183 # _Christoph B. Kassir_, Feb 10 2023
%Y A359183 Cf. A347053, A004053, A258107, A181929, A307254, A307255.
%K A359183 nonn,base
%O A359183 2,2
%A A359183 _Scott R. Shannon_, Dec 18 2022