A335066 Decimal numbers such that when they are written in all bases from 2 to 10 those numbers all share a common digit (the digit 0 or 1).

1, 81, 91, 109, 127, 360, 361, 417, 504, 540, 541, 631, 661, 720, 781, 841, 918, 981, 991, 1008, 1009, 1039, 1080, 1081, 1088, 1089, 1090, 1091, 1093, 1099, 1105, 1111, 1116, 1117, 1118, 1119, 1120, 1121, 1122, 1123, 1124, 1125, 1126, 1128, 1134, 1135, 1136, 1137, 1138, 1139

Offset: 1

Author

Scott R. Shannon and Zach J. Shannon, May 22 2020

Keywords

• nonn
• base

Comments

As base 2 is included the only possible common digit between all the bases is either a 0 or 1.

Examples

			1 is a term as 1 written in all bases is 1.
81 is a term as 81_2 = 1010001, 81_3 = 10000, 81_4 = 1101, 81_5 = 311, 81_6 = 213, 81_7 = 144, 81_8 121, 81_9 = 100, 81_10 = 81, all of which contain the digit 1.
360 is a term as 360_2 = 101101000, 360_3 = 111100, 360_4 = 11220, 360_5 = 2420, 360_6 = 1400, 360_7 = 1023, 360_8 = 550, 360_9 = 550, 360_10 = 360, all of which contain the digit 0.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A335051, A258107, A145100, A317725, A181929, A331565, A153114.

Programs

Python

def hasdigits01(n, b):
    has0, has1 = False, False
    while n >= b:
        n, r = divmod(n, b)
        if r == 0: has0 = True
        if r == 1: has1 = True
        if has0 and has1: return (True, True)
    return (has0, has1 or n==1)
def ok(n):
    all0, all1 = True, True
    for b in range(10, 1, -1):
        has0, has1 = hasdigits01(n, b)
        all0 &= has0; all1 &= has1
        if not all0 and not all1: return False
    return all0 or all1
print([k for k in range(1140) if ok(k)]) # Michael S. Branicky, May 23 2022