Displaying 1-10 of 14 results found.
Numbers that are repdigits with length > 2 in more than one base.
+10
13
31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 3906, 4095, 4557, 6560, 7566, 7812, 8191, 9114, 9331, 9841, 10507, 11349, 11718, 13671, 14043, 14763, 15132, 15624, 16383, 18291, 18662, 18915, 19608, 19682, 21845, 22351, 22698
COMMENTS
Definition requires "length > 2" because all numbers n > 2 are trivially represented as "11" in base n-1.
0 = 00 = 000 = 0000 = 00000 = 000000 = 0000000 = 00000000 = ... in any positional number representation (includes fixed base radix b > 1, mixed base radix with each b_i > 1, i >= 0, such as factorial and primorial based radix...)
The sequence definition should be read as:
Nonnegative integers that are repdigits with length > 2 in more than one fixed base radix b > 1.
Considering all fixed and mixed base radix would include many more nonnegative integers (but not the integers 1 to 6) which are repdigits with length > 2 in more than one radix. (End)
In this sequence data, the first number which is repdigit, with length > 2, in more than two bases is the twelfth Mersenne number 4095 with four Brazilian representations: M_12 = 4095 = 111111111111_2 = 333333_4 = 7777_8 = (15 15 15)_16.
The Mersenne number M_15 is the first number which is repdigit in exactly three bases with M_15 = 32767 = 111111111111111_2 = 77777_8 = (31 31 31)_32.
Only two numbers are repunits in more than one base: the Mersenne primes 31 and 8191 (Examples and A119598). Some numbers are once repunit and once multiple of a Brazilian prime such that Mersenne number M_9 = 511 = 7 * 73 = 111111111_2 = 7 * 111_8 = 777_8.
Some numbers are once repunit and once multiple of a composite repunit such that Mersenne number M_6 = 63 = 3 * 21 = 111111_2 = 3 * 111_4 = 333_4.
Some numbers are repdigits in two different bases: 546 = 666_9 = 222_16. (End)
EXAMPLE
31 is in the list because 31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90;
10507 = {19 19 19}_23 = 111_102.
MATHEMATICA
Select[Range[550], Function[n, 1 < Count[Range[2, n - 1], _?(And[Length@ DeleteCases[#, 0] == 1, Union[#][[2]] > 2] &@ DigitCount[n, #] &)]]] (* Michael De Vlieger, Aug 09 2017 *)
PROG
(PARI) isok(n)=my(nb = 0); for (b=2, n-1, d = digits(n, b); if ((#d > 2) && (#Set(d) == 1), nb++); if (nb > 1, return (1)); ); return (0); \\ Michel Marcus, Aug 08 2017
CROSSREFS
Cf. A167782 (numbers that are repdigits with length > 2 in some base). Cf. A053696 (numbers which are repunits in some base). Cf. A158235 (numbers n whose square is a repdigit in some base < n). Cf. A290869 (Numbers that are repdigits with length > 2 in more than two bases).
Numbers m such that beta(m) = tau(m)/2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
11
7, 13, 15, 21, 26, 40, 43, 57, 62, 73, 80, 85, 86, 91, 93, 111, 114, 124, 127, 129, 133, 146, 157, 170, 171, 172, 183, 211, 215, 219, 222, 228, 241, 242, 259, 266, 285, 292, 307, 312, 314, 333, 341, 343, 365, 366, 381, 399, 421, 422, 438, 444, 455, 463, 468, 471, 482, 507, 518, 532, 549, 553, 555, 585, 601, 614, 624
COMMENTS
As tau(m) = 2 * beta(m), the terms of this sequence are not squares. Indeed, there are 3 subsequences which realize a partition of this sequence (see examples):
1) Non-oblong composites which have only one Brazilian representation with three digits or more, they form A326387. 2) Oblong numbers that have exactly two Brazilian representations with three digits or more; these oblong integers are a subsequence of A167783 and form A326385. 3) Brazilian primes for which beta(p) = tau(p)/2 = 1, they are in A085104 \ {31, 8191}.
EXAMPLE
One example for each type:
15 = 1111_2 = 33_4 with tau(15) = 4 and beta(15) = 2.
3906 = 62 * 63 = 111111_5 = 666_25 = (42,42)_86 = (31,31)_125 = (21,21)_185 = (18,18)_216 = (14,14)_278 = 99_433 = 77_557 = 66_650 = 33_1301 = 22_1952, so tau(3906) = 24 with beta(3906) = 12.
43 = 111_6 is Brazilian prime, so tau(43) = 2 and beta(43) = 1.
PROG
(PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
Numbers m such that beta(m) = tau(m)/2 - 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
9
2, 3, 5, 8, 10, 11, 14, 17, 18, 19, 22, 23, 24, 27, 28, 29, 32, 33, 34, 35, 37, 38, 39, 41, 42, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 58, 59, 60, 61, 65, 66, 67, 68, 69, 70, 71, 74, 75, 76, 77, 78, 79, 82, 83, 84, 87, 88, 89, 92, 94, 95, 96, 97, 98, 99, 101, 102, 103, 104, 105, 106, 107, 108, 109, 112, 113, 115, 116
COMMENTS
As tau(m) = 2 * (1 + beta(m)), the terms of this sequence are not squares. Indeed, there are 3 subsequences which realize a partition of this sequence (see examples):
1) Non-oblong composites which have no Brazilian representation with three digits or more, they form A326386. 2) Oblong numbers that have only one Brazilian representation with three digits or more. These oblong integers are a subsequence of A167782 and form A326384. 3) Non Brazilian primes, then beta(p) = tau(p)/2 - 1 = 0.
EXAMPLE
One example for each type:
10 = 22_4 and tau(10) = 4 with beta(10) = 1.
42 = 6 * 7 = 222_4 = 33_13 = 22_20 and tau(42) = 8 with beta(42) = 3.
17 is no Brazilian prime with tau(17) = 2 and beta(17) = 0.
PROG
(PARI) beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136isok(n) = beta(n) == numdiv(n)/2 - 1; \\ Michel Marcus, Jul 03 2019
CROSSREFS
Cf. A220627 (subsequence of non Brazilian primes).
Numbers m such that beta(m) = tau(m)/2 + 2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
8
32767, 65535, 67053, 2097151, 4381419, 7174453, 9808617, 13938267, 14348906, 19617234, 21523360, 29425851, 39234468, 43046720, 48686547, 49043085, 58851702, 61035156, 68660319, 71270178, 78468936, 88277553, 98086170, 107894787, 115174101, 117703404, 134217727, 142540356, 175965517
COMMENTS
As tau(m) = 2 * (beta(m) - 2) , the terms of this sequence are not squares.
There are 2 subsequences which realize a partition of this sequence (see array in link and examples):
1) Non-oblong composites which have exactly three Brazilian representations with three digits or more, they are in A326389. 2) Oblong numbers that have exactly four Brazilian representations with three digits or more. These integers have been found through b-file of Rémy Sigrist in A290869. These oblong integers are a subsequence of A309062. There are no primes that satisfy this relation.
EXAMPLE
One example for each type:
1) The divisors of 32767 are {1, 7, 31, 151, 217, 1057, 4681, 32767} and tau(32767) = 8; also, 32767 = M_15 = R(15)_2 = 77777_8 = (31,31,31)_32 = (151,151)_216 = (31,31)_1056 = 77_4680 so beta(32767) = 6 with beta'(32767) = 3 and beta"(32767)= 3. The relation is beta(32767) = tau(32767)/2 + 2 = 6.
2) 61035156 = 7812 * 7813 is oblong with tau(61035156) = 144. The four Brazilian representations with three digits or more are 61035156 = R(12)_5 = 666666_25 = (31,31,31,31)_125 = (156,156,156)_625, so beta"(61035156) = 4 and beta(61035156) = tau(61035156)/2 + 2 = 74.
Non-oblong composites m such that beta(m) = tau(m)/2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
8
15, 21, 26, 40, 57, 62, 80, 85, 86, 91, 93, 111, 114, 124, 129, 133, 146, 170, 171, 172, 183, 215, 219, 222, 228, 242, 259, 266, 285, 292, 312, 314, 333, 341, 343, 365, 366, 381, 399, 422, 438, 444, 455, 468, 471, 482, 507, 518, 532, 549, 553
COMMENTS
As tau(m) = 2 * beta(m), the terms of this sequence are not squares.
The number of Brazilian representations of a non-oblong number m with repdigits of length = 2 is beta'(n) = tau(n)/2 - 1.
This sequence is the first subsequence of A326380: non-oblong composites which have only one Brazilian representation with three digits or more.
EXAMPLE
tau(m) = 4 and beta(m) = 2 for m = 15, 21, 26, 57, 62, 85, 86, ... with 15 = 1111_2 = 33_4.
tau(m) = 8 and beta(m) = 4 for m = 40 = 1111_3 = 55_7 = 44_9 = 22_19.
tau(m) = 10 and beta(m) = 5 for m = 80 = 2222_3 = 88_9 = 55_15 = 44_19 = 22_39.
PROG
(PARI) isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136isok(m) = !isprime(m) && !isoblong(m) && (beta(m) == numdiv(m)/2); \\ Michel Marcus, Jul 15 2019
CROSSREFS
Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326388 (non-oblongs with tau(m)/2 + 1), A326389 (non-oblongs with tau(m)/2 + 2).
Numbers m such that beta(m) = tau(m)/2 + 3 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
7
4095, 262143, 265720, 531440, 1048575, 5592405, 11184810, 122070312, 183105468, 193710244, 244140624, 268435455, 387420488
COMMENTS
As tau(m) = 2 * (beta(m) - 3), the terms of this sequence are not squares.
The current known terms are non-oblong composites that have exactly four Brazilian representations with three digits or more; but, maybe, there exist oblong integers that have exactly five Brazilian representations with three digits or more.
EXAMPLE
The 24 divisors of 4095 = M_12 are {1, 3, 5, 7, 9, 13, 15, 21, 35, 39, 45, 63, 65, 91, 105, 117, 195, 273, 315, 455, 585, 819, 1365, 4095} and tau(4095) = 24; also, 4095 = R(12)_2 = 333333_4 = 7777_8 = (15,15,15)_16, so, beta(4095) = 15 with beta'(4095)= 11 and beta''(4095) = 4. The relation is beta(4095) = tau(4095)/2 + 3 = 15 and 4095 is a term.
Non-oblong composites m such that beta(m) = tau(m)/2 + 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
7
63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 4557, 6560, 7566, 7812, 9114, 9331, 9841, 10507, 11349, 11718, 13671, 14043, 14763, 15132, 15624, 16383, 18291, 18662, 18915, 19608, 19682, 21845, 22351, 22698
COMMENTS
As tau(m) = 2 * (beta(m) - 1), the terms of this sequence are not squares.
The number of Brazilian representations of a non-oblong number m with repdigits of length = 2 is beta'(n) = tau(n)/2 - 1.
This sequence is the first subsequence of A326381: non-oblong composites which have exactly two Brazilian representations with three digits or more. Some Mersenne numbers belong to this sequence: M_6, M_8, M_9, M_10, M_14, ...
EXAMPLE
tau(m) = 4 and beta(m) = 3 for m = 511 with 511 = 111111111_2 = 777_8 = 77_72,
tau(m) = 6 and beta(m) = 4 for m = 63 with 63 = 111111_2 = 333_4 = 77_8 = 33_20,
tau(m) = 8 and beta(m) = 5 for m = 255 with 255 = 11111111_2 = 3333_4 = (15,15)_16 = 55_50 = 33_84,
tau(m) = 12 and beta(m) = 7 for m = 364 with 364 = 111111_3 = 4444_9 = (14,14)_25 = (13,13)_27 = 77_51 = 44_90 = 22_181.
PROG
(PARI) isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136isok(m) = !isprime(m) && !isoblong(m) && (beta(m) == numdiv(m)/2 + 1); \\ Michel Marcus, Jul 15 2019
CROSSREFS
Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326387 (non-oblongs with tau(m)/2), A326389 (non-oblongs with tau(m)/2 + 2).
Non-oblong numbers that are repdigits with length > 2 in exactly three bases.
+10
7
32767, 65535, 67053, 2097151, 4381419, 7174453, 9808617, 13938267, 14348906, 19617234, 21523360, 29425851, 39234468, 43046720, 48686547, 49043085, 58851702, 68660319, 71270178, 78468936, 88277553, 98086170, 107894787, 115174101, 117703404, 134217727, 142540356
COMMENTS
The number of Brazilian representations of a non-oblong number m with repdigits of length = 2 is beta'(m) = tau(m)/2 - 1. So, as here beta"(m) = 3, beta(m) = tau(m)/2 + 2 where beta(m) is the number of Brazilian representations of m. So, this sequence is the first subsequence of A326382. As tau(m) = 2 * (beta(m) - 2) is even, the terms of this sequence are not squares.
Some Mersenne numbers belong to this sequence: M_15 = a(1), M_16 = a(2), M_21 = a(4), M_27 = a(26), ...
EXAMPLE
tau(m) = 8 and beta(m) = 6 for m = 32767 with 32767 = R(15)_2 = 77777_8 = (31,31,31)_32.
tau(m) = 12 and beta(m) = 8 for m = 2097151 with 2097151 = R(21)_2 = 7777777_8 = (127,127,127)_128.
tau(m) = 16 and beta(m) = 10 with m = 67053 = (31,31,31)_46 = (21,21,21)_56 = 333_149.
PROG
(PARI) isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136isok(m) = !isprime(m) && !isoblong(m) && (beta(m) == numdiv(m)/2 + 2); \\ Jinyuan Wang, Aug 02 2019
CROSSREFS
Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326387 (non-oblongs with tau(m)/2), A326388 (non-oblongs with tau(m)/2 + 1), this sequence (non-oblongs with tau(m)/2 + 2), A326705 (non-oblongs with tau(m)/2 + k, k >=3).
Oblong composite numbers m such that beta(m) = tau(m)/2 - 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
+10
5
42, 156, 182, 342, 1406, 1640, 6162, 7140, 14280, 14762, 20880, 25440, 29412, 32942, 33306, 47742, 48620, 49952, 61256, 67860, 95172, 95790, 158802, 176820, 191406, 202950, 209306, 257556, 296480, 297570
COMMENTS
The number of Brazilian representations of an oblong number m with repdigits of length = 2 is beta'(n) = tau(n)/2 - 2.
This sequence is the second subsequence of A326379: oblong numbers that have only one Brazilian representation with three digits or more. Prime 2 is oblong and satisfies also beta(2) = tau(2)/2 - 1 = 0 but non-Brazilian primes are in A220627.
EXAMPLE
There are two types of such numbers:
1) m is repunit with 3 digits or more in only one base:
156 = 12 * 13 = 1111_5 = 66_25 = 44_38 = 33_51 = 22_77 with tau(156) = 12 and beta(156) = 5.
2) m is repdigit with 3 digits or more and digit >= 2 in only one base:
tau(m) = 8 and beta(m) = 3: 42 = 6*7 = 222_4 = 33_13 = 22_20,
tau(m) = 12 and beta(m)= 5: 342 = 18*19 = 666_7 = 99_37 = 66_56 = 33_113 = 22_170,
tau(m) = 16 and beta(m)= 7: 1640 = 40*41 = 2222_9 = (20,20)_81 = (10,10)_2 = 88_204 = 55_327 = 44_409 = 22_819.
CROSSREFS
Cf. A326378 (oblongs with tau(m)/2 - 2), A326385 (oblongs with tau(m)/2), A309062 (oblongs with tau(m)/2 + k, k >= 1).
Non-oblong numbers that are repdigits with length > 2 in more than three bases.
+10
3
4095, 262143, 265720, 531440, 1048575, 5592405, 11184810, 16777215, 122070312, 183105468, 193710244, 244140624, 268435455, 387420488, 435356467
COMMENTS
The number of Brazilian representations of a non-oblong number m with repdigits of length = 2 is beta'(m) = tau(m)/2 - 1. So, as here beta"(m) = r with r >= 4, beta(m) = tau(m)/2 + k with k >= 3 where beta(m) is the number of Brazilian representations of m.
As tau(m) = 2 * (beta(m) - k) is even, the terms of this sequence are not squares.
The terms which have exactly four Brazilian representations with three digits or more form the first subsequence of A326383. Indeed, for the given terms, the number of bases is 4, except for a(8) and a(15) where this number of bases is respectively 5 and 6 (see examples). Some Mersenne numbers belong to this sequence: M_12 = a(1), M_18 = a(2), M_20 = a(5), M_24 = a(8), M_28 = a(13), M_32, ...
EXAMPLE
If beta"(m)is the number of Brazilian representations with three digits or more of the integer m, then:
1) With beta"(m) = 4; tau(4095) = 24 and 4095 has exactly four Brazilian representations with three digits or more: [R(12)]_2 = 333333_4 = 7777_4 = (15,15,15)_16 and 11 representations with 2 digits, so beta(4095) = 15 and k = 3.
2) With beta"(m) = 5; tau(435356467) = 64 and 435356467 has exactly five Brazilian representations with three digits or more: R(12)_6 = 777777_36 = (43,43,43)_216 = (259,259,259)_1296 = (31,31,31)_3747 and has 31 representations with 2 digits, so beta(435356467) = 36 and k = 4.
3) With beta"(m)=6; tau(16777215)= 96 and 16777215 has exactly six Brazilian representations with three digits or more: [R(24)]_2 = 333333333333_4 = 7777777_8 = (15,15,15,15,15,15)_16 = (63,63,63,63)_64 = (255,255,255)_256 and 47 representations with 2 digits, so beta(16777215) = 53 and k = 5.
CROSSREFS
Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326387 (non-oblongs with tau(m)/2), A326388 (non-oblongs with tau(m)/2 + 1), A326389 (non-oblongs with tau(m)/2 + 2), this sequence (non-oblongs with tau(m/2) + k, k >= 3).
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