A290869 -id:A290869 - cp's OEIS frontend

A167783 Numbers that are repdigits with length > 2 in more than one base.

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

Offset: 1

Andrew Weimholt, Nov 12 2009

Definition requires "length > 2" because all numbers n > 2 are trivially represented as "11" in base n-1.
From Daniel Forgues, Nov 13 2009: (Start)
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)
From Bernard Schott, Aug 08 2017: (Start)
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)

			31 is in the list because 31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90;
10507 = {19 19 19}_23 = 111_102.

David Trimas, Table of n, a(n) for n = 1..8424 (first 158 terms from Michel Marcus)
David Trimas, Wolfram Cloud Implementation of A167783
Wolfram Demonstrations Project, Mixed Radix Number Representations

Cf. A167782 (numbers that are repdigits with length > 2 in some base).
Cf. A010785 (repdigits (base 10)).
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).

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 *)

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

a(41)-a(44) from Bernard Schott, Aug 08 2017

A326382 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.

Original entry on oeis.org

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

Offset: 1

Bernard Schott, Jul 08 2019

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.

			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.

Bernard Schott, Array of relations beta = f(tau)
Index entries for sequences related to Brazilian numbers

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783 and A290869.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), this sequence (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).
Cf. A309062, A326389.

Missing a(18) inserted by Bernard Schott, Jul 20 2019

A309062 Oblong numbers that are repdigits with length > 2 in more than two bases.

Original entry on oeis.org

61035156, 641431602, 38146972656, 70607384120, 953674316406, 5824521280620, 23841857910156, 51472783023662, 145655559307440, 463255047212960, 1838956877846660, 14901161193847656, 37523658824249780, 88453695801367260, 166354152295794960, 416972378738246240

Offset: 1

Michel Marcus, Jul 10 2019

All initial terms come from the b-file in A290869.
For the given terms, the number of bases are respectively 4, 3, 3, 4, 4, 4, 4, 3, 4, 3 and 4.
A003463(64), A003463(24) (confirmed) and A003463(36) are candidates for 5, 6 and 7 bases representations.
From Bernard Schott, Jul 24 2019: (Start)
The terms of this sequence are necessarily of the form (b^(2*q) - 1)/4 with q > 2 and b = 4*m+1 with m > 0, but when b = c^2 is an odd square (A016754), then some terms can also have the form (b^(2*q+1) - 1)/4 as a(8) and a(23). If these terms have representations in u bases, the values of (b, 2*q or 2*q+1, u) for the first eleven terms are respectively (5, 12, 4), (37, 6, 3), (5, 16, 3), (9, 12, 4), (5, 18, 4), (13, 12, 4), (5, 20, 4), (9, 15, 3), (17, 12, 4), (9, 16, 3) and (21, 12, 4).
For any b = 4*m+1 with m > 0 and r > 2, (b^(4*r) - 1)/4 is an oblong repdigit with length > 2 in at least bases b, b^2 and b^4; hence this sequence is infinite.
(End)
From Chai Wah Wu, Jul 24 2019: (Start)
Other values of (b, q, u) for which (b^(2*q) - 1)/4 is a term with representations in u bases:
(5, 12, 6), (5, 14, 4), (5, 15, 6), (9, 9, 4), (9, 10, 4), (13, 8, 3), (13, 9, 4), (17, 8, 3), (29, 6, 4), (33, 6, 4), (37, 6, 4), (41, 6, 4), (45, 6, 4).
(End)
From Bernard Schott, Jul 24 2019: (Start)
Theorem: if tau(2*q) = r > 4, (b^(2*q) - 1)/4 is a term that has exactly r-2 representations as repdigits with length > 2 in bases that are powers of b.
There exist cases where a term also has representation in another base that is not power of b. For instance a(2), see example, where base 3446 is not a perfect power of 37.
Conclusion: if m = (b^(2*q) - 1)/4 is a term and if beta"(m) is the number of representations of this term as repdigits with length > 2, then, beta"(m) >= tau(2*q) - 2. (End)

			From _Bernard Schott_, Jul 18 2019: (Start)
a(1) = 61035156 = 7812*7813 = 111111111111_5 = 666666_25 = (31,31,31)_125 = (156,156,156)_625.
a(2) = 641431602 = 25326*25327 = 999999_37 = (342,342,342)_1469 = (54,54,54)_3446.
(End)
a(11) = 1838956877846660 = 42883060*42883061 = 555555555555_21 = (110, 110, 110, 110, 110, 110)_441 = (2315, 2315, 2315, 2315)_9261 = (48620, 48620, 48620)_194481. - _Chai Wah Wu_, Jul 24 2019

Giovanni Resta, Table of n, a(n) for n = 1..26

Intersection of A002378 and A290869.

Cf. A326378 (similar in no base), A326384 (similar in one base), A326385 (similar in 2 bases).

isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378
okrepu3(b, target, lim) = {my(k = 3, nb = 0, x); while ((x=(b^k-1)/(b-1)) <= target, if (x==target, nb++); k++); nb;}
dge3(n) = {my(d=divisors(n), nb=0, ndi, limi); for (i=1, #d, ndi = n/d[i]; limi = sqrtint(ndi); for (k=d[i]+1, limi, nb += okrepu3(k, ndi, limi););); nb;}
isok(n) = isoblong(n) && (dge3(n) >= 3);

a(11) from Chai Wah Wu, Jul 21 2019

a(12)-a(16) from Giovanni Resta, Jul 28 2019

A326381 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.

Original entry on oeis.org

31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 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

Offset: 1

Bernard Schott, Jul 07 2019

As tau(m) = 2 * (beta(m) - 1), the terms of this sequence are not squares.
There are 3 subsequences which realize a partition of this sequence (see examples):
1) Non-oblong composites which have exactly two Brazilian representations with three digits or more, they form A326388.
2) Oblong numbers that have exactly three Brazilian representations with three digits or more; thanks to Michel Marcus, who found the smallest, 641431602. These oblong integers are a subsequence of A290869 and A309062.
3) The two Brazilian primes 31 and 8191 of the Goormaghtigh conjecture (A119598) for which beta(p) = tau(p)/2 + 1 = 2.

			One example for each type:
1) 63 = 111111_2 = 333_4 = 77_8 = 33_20 with tau(63) = 6 and beta(63) = 4.
2) 641431602 = 25326 * 25327 is oblong with tau(641431602) = 256. The three Brazilian representations with three digits or more of 641431602 are 999999_37 = (342,342,342)_1369 = (54,54,54)_3446, so beta"(641431602) = 3 and beta(641431602) = tau(641431602)/2 + 1 = 129.
3) 31 = 11111_2 = 111_5 and 8191 = 1111111111111_2 = 11_90 with beta(p) = tau(p)/2 + 1 = 2.

Wikipedia, Goormaghtigh conjecture.
Index entries for sequences related to Brazilian numbers.

Cf. A000005 (tau), A220136 (beta).
Cf. A119598 (Goormaghtigh conjecture).
Subsequence of A167783.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).

beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(n) = beta(n) == numdiv(n)/2 + 1; \\ Michel Marcus, Jul 08 2019

A326383 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.

Original entry on oeis.org

4095, 262143, 265720, 531440, 1048575, 5592405, 11184810, 122070312, 183105468, 193710244, 244140624, 268435455, 387420488

Offset: 1

Bernard Schott, Jul 08 2019

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.

			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.

Index entries for sequences related to Brazilian numbers

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783 and A290869.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2).

A326389 Non-oblong numbers that are repdigits with length > 2 in exactly three bases.

Original entry on oeis.org

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

Offset: 1

Bernard Schott, Jul 20 2019

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), ...

			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.

Bernard Schott, Array of relations beta = f(tau)
Index entries for sequences related to Brazilian numbers

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783, A290869, A308874 and A326382.
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).

isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378
beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(m) = !isprime(m) && !isoblong(m) && (beta(m) == numdiv(m)/2 + 2); \\ Jinyuan Wang, Aug 02 2019

A326705 Non-oblong numbers that are repdigits with length > 2 in more than three bases.

Original entry on oeis.org

4095, 262143, 265720, 531440, 1048575, 5592405, 11184810, 16777215, 122070312, 183105468, 193710244, 244140624, 268435455, 387420488, 435356467

Offset: 1

Bernard Schott, Jul 21 2019

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, ...

			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.

Bernard Schott, Array for the relations beta = f(tau)
Index entries for sequences related to Brazilian numbers

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783, A290869 and A308874.
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).

A290969 The least positive integer that is a repdigit with length > 2 in exactly n bases.

Original entry on oeis.org

1, 7, 31, 32767, 4095, 435356467, 16777215, 68719476735, 281474976710655

Offset: 0

Bernard Schott, Aug 16 2017

a(10) <= 1152921504606846975 = 2^60 - 1.
a(7) and following terms > 3*10^9. - Giovanni Resta, Aug 16 2017
a(7) <= 2^36-1 and a(8) <= 2^48-1. - Michel Marcus, Aug 17 2017
In fact, we have equality in both cases. - Rémy Sigrist, Aug 21 2017
Except for a(5) = (6^12 - 1) / 5, all the numbers in the data through a(8) are Mersenne numbers A000225. - Bernard Schott, Aug 27 2017

			a(1) = 7 = 111_2.
a(2) = 31 = 11111_2 = 111_5.
a(3) = 32767 = (R_15)_2 = 77777_8 = (31,31,31)_32.

Cf. A000225, A053696, A125134, A167782, A167783, A290869.

a(7) from Rémy Sigrist, Aug 19 2017

a(8) from Rémy Sigrist, Aug 21 2017

A326706 Numbers m such that beta(m) = tau(m)/2 + k for some k >= 4, where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

Original entry on oeis.org

16777215, 435356467, 1073741823, 68719476735, 1099511627775, 4398046511103, 35184372088831, 281474976710655, 14901161193847656, 18014398509481983

Offset: 1

Bernard Schott, Aug 09 2019

As tau(m) = 2 * (beta(m) - k) is even, the terms of this sequence are not squares.
There are two classes of terms (see array in link and examples):
1) Non-oblong composites which have five or more Brazilian representations with three digits or more, they form a subsequence of A326705. The smallest example is a(1) = 16777215 = M_24.
2) Oblong numbers that have six or more Brazilian representations with three digits or more, they form a subsequence of A309062. The smallest example is a(9) (see 2nd example).
For a(1) to a(10), the numbers k are respectively 5, 4, 5, 6, 5, 5, 4, 7, 4 and 5.
Some Mersenne numbers are terms: M_24 = a(1), M_30 = a(3), M_36 = a(4), M_40 = a(5), M_42 = a(6), M_45 = a(7), M_48 = a(8), M_54 = a(10).

			One example of each type:
1) Non-oblong with beta"(m) = 5; tau(435356467) = 64 and 435356467 = (6^12 - 1)/5 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.
2) Oblong with beta"(m) = 6; tau(14901161193847656) = 768 and 14901161193847656 = (5^24 - 1)/4 = 122070312*122070313 is oblong. The six Brazilian representations with three digits or more of this term are R(24)_5 = 666666666666_25 = (31,31,31,31,31,31,31,31)_125 = (156,156,156,156,156)_625, =(3906,3906,3906,3906)_15625 = (97656,97656,97656)_390625 so beta"(14901161193847656) = 6 and beta(61035156) = (tau(61035156)/2 - 2) + 6 = 388 and k = 4.

Bernard Schott, Array of relations beta = f(tau)
Index entries for sequences related to Brazilian numbers

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783 and A290869.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3), this sequence (tau(m)/2 + k, k >= 4).
Cf. A291592 (Mersenne numbers).

okrepu3(b, target, lim) = {my(k = 3, nb = 0, x); while ((x=(b^k-1)/(b-1)) <= target, if (x==target, nb++); k++); nb; }
dge3(n, d) = {my(nb=0, ndi, limi); for (i=1, #d, ndi = n/d[i]; limi = sqrtint(ndi); for (k=d[i]+1, limi, nb += okrepu3(k, ndi, limi); ); ); nb; }
deq2(n, d) = {my(nb=0, nk); for (k=1, #d\2, nk = (n - d[k])/d[k]; if (nk > d[k], nb++); ); nb; }
beta(n) = {if (n<3, return (0)); my(d=divisors(n)); deq2(n, d) + dge3(n, d) - 1; }
isok(n) = beta(n) - numdiv(n)/2 > = 4; \\ Michel Marcus, Aug 10 2019

cp's OEIS Frontend

A167783 Numbers that are repdigits with length > 2 in more than one base.

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A326382 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.

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A309062 Oblong numbers that are repdigits with length > 2 in more than two bases.

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A326381 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.

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A326383 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.

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A326389 Non-oblong numbers that are repdigits with length > 2 in exactly three bases.

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A326705 Non-oblong numbers that are repdigits with length > 2 in more than three bases.

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A290969 The least positive integer that is a repdigit with length > 2 in exactly n bases.

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A326706 Numbers m such that beta(m) = tau(m)/2 + k for some k >= 4, where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

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