A326380 -id:A326380 - cp's OEIS frontend

A326378 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.

6, 12, 20, 30, 56, 72, 90, 110, 132, 210, 240, 272, 306, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1482, 1560, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450, 2550, 2652, 2756, 2862, 2970, 3080, 3192, 3306, 3422, 3540, 3660, 3782

Offset: 1

Bernard Schott, Jul 02 2019

As tau(m) = 2 * (2 + beta(m)), the terms of this sequence are not squares. Indeed, there exists only one family that satisfies this relation and these integers are exactly the oblong numbers that have no Brazilian representation with three digits or more.

There are no integers such as beta(m) = tau(m)/2 - q with q >= 3.

			1) tau(m) = 4 and beta(m) = 0: m = 6 which is not Brazilian.
2) tau(m) = 6 and beta(m) = 1: m = 12, 20.
   12 = 3 * 4 = 22_5, 20 = 4 * 5 = 22_9.
3) tau(m) = 8 and beta(m) = 2: m = 30, 56, 110, 506, 2162, 3422, ...
   30 = 5 * 6 = 33_9 = 22_14, 56 = 7 * 8 = 44_13 = 22_27.
4) tau(m) = 10 and beta(m) = 3: m = 272, ...
   272 = 16 * 17 = 88_32 = 44_67 = 22_135.
5) tau(m) = 12 and beta(m) = 4: m = 72, 90, 132, 306, 380, 650, 812, 992, ...
   72 = 8 * 9 = 66_11 = 44_17 = 33_23 = 22_35.

Amiram Eldar, Table of n, a(n) for n = 1..800
Bernard Schott, Relation beta = f(tau).
Index entries for sequences related to Brazilian numbers.

Cf. A000005 (tau), A220136 (beta).
Subsequence of A002378 (oblong numbers).
Cf. A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).
Cf. A326384 (oblongs with tau(m)/2 - 1), A326385 (oblongs with tau(m)/2).

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

A326379 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

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

Offset: 1

Bernard Schott, Jul 03 2019

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.

			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.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for sequences related to Brazilian numbers.

Cf. A000005 (tau), A220136 (beta).
Cf. A220627 (subsequence of non Brazilian primes).
Cf. A326378 (tau(m)/2 - 2), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), 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 03 2019

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

A326387 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.

Original entry on oeis.org

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

Offset: 1

Bernard Schott, Jul 14 2019

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.

			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.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for sequences related to Brazilian numbers.

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A308874 and A326380.
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).

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); \\ Michel Marcus, Jul 15 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).

A326385 Oblong 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.

Original entry on oeis.org

3906, 37830, 97656, 132860, 1206702, 2441406, 6034392, 10761680, 21441530, 96855122, 148705830, 203932680, 322866992, 747612306, 871696100, 1187526060, 1525878906, 1743939360, 2075941406, 3460321800, 5541090282, 8574111812, 9455714840, 12880093590, 18854722656

Offset: 1

Bernard Schott, Jul 10 2019

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 A326380: oblong numbers that have exactly two Brazilian representations with three digits or more.

			3906 = 62 * 63 is oblong, tau(3906) = 24, beta(3906) = 12 with beta'(3906) = 10 and beta"(3906) = 2: 3906 = 111111_5 = 666_25 = (42,42)_92 = (31,31)_125 = (21,21)_185 = (18,18)_216 = (14,14)_278 = 99_433 = 77_557 = 66_650 = 33_130 = 22_1952.

Cf. A000005 (tau), A220136 (beta).
Subsequence of A002378 (oblong numbers) and of A167783.
Cf. A326378 (oblongs with tau(m)/2 - 2), A326384 (oblongs with tau(m)/2 - 1), A309062 (oblongs with tau(m)/2 + k, k >= 1).

a(6)-a(25) from Giovanni Resta, Jul 11 2019

A309493 Highly Brazilian numbers (A329383) that are not highly composite numbers (A002182).

Original entry on oeis.org

7, 15, 40, 336, 1440, 5405400

Offset: 1

Bernard Schott, Aug 04 2019

Is this sequence finite or infinite?
Indeed, from 6486480 to 321253732800, that is, during 41 successive terms (maybe more?), highly Brazilian numbers are the same as highly composite numbers.
The data for this sequence comes from the new terms in the b-file of A066044 found by Giovanni Resta.
Why are these six numbers HB (highly Brazilian) and not HC (highly composite)? (See link Why HB and not HC? for more details)
1) For 7, 15 and 40, it is because they have a Brazilian representation with 3 or 4 digits and belong to A326380 (see examples).
2) For 336, 1440 and 5405400, it is because each of these three terms HB r is non-oblong, belong to A326386 and the greatest HC m less than r is oblong with the same number of divisors.
a(7) > A329383(91) = 321253732800.

			a(1) = 7 because 7 is the smallest Brazilian number with 7 = 111_2 so beta(7) = 1, as tau(7) = tau(2) = 2, 7 is highly Brazilian but cannot be highly composite.
a(2) = 15 because 15 is the smallest integer 2-Brazilian with 15 = 1111_2 = 33_4 and beta(15) = 2, as tau(15) = tau(6) = 4, 15 is highly Brazilian but not highly composite.
a(3) = 40 because 40 is the smallest integer 4-Brazilian with 40 = 1111_3 = 55_7 = 44_9 = 22_19 so beta(40) = 4, as tau(40) = tau(24) = 8, 40 is highly Brazilian but not highly composite.
a(4) = 336 because beta(336) = 9 and tau(336) = tau(240) = 20.
a(5) = 1440 because beta(1440) = 17 and tau(1440) = tau(1260) = 36.
a(6) = 5405400 because beta(5405400) = 191 and tau(5405400) = tau(4324320) = 384.

Bernard Schott, Why HB and not HC?

Cf. A002182, A279930, A309039, A329383.

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

A361914 Primes that are repunits with three or more digits for exactly one base b >= 2.

Original entry on oeis.org

7, 13, 43, 73, 127, 157, 211, 241, 307, 421, 463, 601, 757, 1093, 1123, 1483, 1723, 2551, 2801, 2971, 3307, 3541, 3907, 4423, 4831, 5113, 5701, 6007, 6163, 6481, 8011, 9901, 10303, 11131, 12211, 12433, 13807, 14281, 17293, 19183, 19531, 20023, 20593, 21757, 22621, 22651, 23563

Offset: 1

Bernard Schott, Mar 29 2023

Brazilian primes that have exactly one Brazilian representation as a repunit.
As these primes p satisfy beta(p) = tau(p) / 2 (= 1), where beta = A220136 and tau = A000005, this sequence is a subsequence of A326380.
Equals A085104 \ {31, 8191}, since according to the Goormaghtigh conjecture (link), 31 and 8191 which are both Mersenne numbers, are the only primes which are Brazilian in two different bases.
The three following sequences realize a partition of the set of primes: A220627 (primes not Brazilian), this sequence (primes 1-Brazilian) and {31,8191} (primes 2-Brazilian).

			7 = 111_2 is a term.
13 = 111_3 is a term.
19 = 11_18 is not a term.
31 = 11111_5 = 111_5 is not a term.
127 = 1111111_2 is a term.
8191 = 1111111111111_2 = 111_90 is not a term.

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

Cf. A000005, A085104, A220136, A220627, A326380.
Equals A326380 \ {A326385 Union A326387}.
Subsequence of A288783.

q[n_] := Count[Range[2, n - 2], ?(Length[d = IntegerDigits[n, #]] > 2 && Length[Union[d]] == 1 &)] == 1;; Select[Prime[Range[3000]], q] (* _Amiram Eldar, Mar 30 2023 *)

cp's OEIS Frontend

A326378 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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A326379 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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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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A326387 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.

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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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A326385 Oblong 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.

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A309493 Highly Brazilian numbers (A329383) that are not highly composite numbers (A002182).

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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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