A331561
The base-10 numbers with a digit product > 0 and which when written in bases 3,4,5,6,7,8,9 have three or more other base representations with the same digit product.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 12563124891, 115233863842, 123858133813, 363254652118, 1324658354423, 1511864334458, 1825693128524, 2321856215149, 2632654133853, 3146626254542, 3521445321466, 12462982162122, 12496523158865, 13129883155583, 13443165514365, 14213435966581
Offset: 1
6 is a term as 6_10 = 6_7 = 6_8 = 6_9, so it has three other base representations where the digit product also equals 6.
12563124891 is a term as 12563124891_10 = 5434343211123_6 = 623216541162_7 = 135464411233_8, so it has three other base representations where the digit product also equals 103680.
115233863842 is a term as 115233863842_10 = 124534313342214_6 = 11216413452466_7 = 1532436234242_8, so it has three other base representations where the digit product also equals 829440.
A335051
a(n) is the smallest decimal number > 1 such that when it is written in all bases from base 2 to base n those numbers all contain both 0 and 1.
Original entry on oeis.org
2, 9, 19, 28, 145, 384, 1128, 2601, 2601, 101256, 103824, 382010, 572101, 971400, 1773017, 1773017, 22873201, 64041048, 64041048, 1193875201, 2496140640, 10729882801, 21660922801, 120068616277, 333679563001, 427313653201, 427313653201, 10436523921264, 10868368953601
Offset: 2
a(3) = 9 as 9_2 = 1001 and 9_3 = 100, both of which contain a 0 and 1.
a(6) = 145 as 145_2 = 10010001, 145_3 = 12101, 145_4 = 2101, 145_5 = 1040, 145_6 = 401, all of which contain a 0 and 1.
a(9) = 2601 as 2601_2 = 101000101001, 2601_3 = 10120100, 2601_4 = 220221, 2601_5 = 40401, 2602_6 = 20013, 2601_7 = 10404, 2601_8 = 5051, 2601_9 = 3510, all of which contain a 0 and 1. Note that, as 2601 also contains a 0 and 1, a(10) = 2601.
a(16) = 1773017 as 1773017_2 = 110110000110111011001, 1773017_3 = 10100002010022, 1773017_4 = 12300313121, 1773017_5 = 423214032, 1773017_6 = 102000225, 1773017_7 = 21033101, 1773017_8 = 6606731, 1773017_9 = 3302108, 1773017_10 = 1773017, 1773017_11 = 1001104, 1773017_12 = 716075, 1773017_13 = 4A102C, 1773017_14 = 342201, 1773017_15 = 250512, 1773017_16 = 1B0DD9, all of which contain a 0 and 1.
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a[n_] := Block[{k=2}, While[ AnyTrue[ Range[n, 2, -1], ! SubsetQ[ IntegerDigits[k, #], {0, 1}] &], k++]; k]; a /@ Range[2, 13] (* Giovanni Resta, May 24 2020 *)
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from numba import njit
@njit
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
return has0 and (has1 or n==1)
@njit
def a(n, start=2):
k = start
while True:
for b in range(n, 1, -1):
if not hasdigits01(k, b): break
else: return k
k += 1
anm1 = 2
for n in range(2, 21):
an = a(n, start=anm1)
print(an, end=", ")
anm1 = an # Michael S. Branicky, Feb 09 2021
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).
Original entry on oeis.org
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
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.
-
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
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