login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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.
0
1, 2, 54, 13122, 15258789062500
OFFSET
2,2
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.
EXAMPLE
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)
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Scott R. Shannon, Dec 18 2022
STATUS
approved