A326384 -id:A326384 - 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

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

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

A309193 Smallest oblong number that is a repdigit of length > 2 in exactly n bases.

Original entry on oeis.org

42, 3906, 641431602, 61035156

Offset: 1

Felix Fröhlich, Jul 16 2019

			From _Bernard Schott_, Jul 18 2019: (Start)
a(1) = 42 = 6*7 = 222_4.
a(2) = 3906 = 62*63 = 111111_5 = 666_25.
a(3) = 641431602 = 25326*25327 = 999999_37 = (342,342,342)_1469 = (54,54,54)_3446.
a(4) = 61035156 = 7812*7813 = 111111111111_5 = 666666_25 = (31,31,31)_125 = (156,156,156)_625. (End)

Cf. A002378, A326384 (oblongs repdigits of length > 2 in exactly 1 base), A326385 (oblongs repdigits of length > 2 in exactly 2 bases), A309062 (oblongs repdigits of length > 2 in more than 2 bases).

/* Functions isoblong, okrepu3 and dge3 after Michel Marcus in A309062 */
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; }
a(n) = for(k=1, oo, if(isoblong(k), if(dge3(k)==n, return(k))))

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

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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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A309193 Smallest oblong number that is a repdigit of length > 2 in exactly n bases.

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