cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A326380 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

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
Offset: 1

Views

Author

Bernard Schott, Jul 03 2019

Keywords

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}.

Examples

			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.

Links

Crossrefs

Cf. A000005 (tau), A220136 (beta).
Cf. A085104 (Brazilian primes).
Subsequence of A167782.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).

Programs

  • PARI
    beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(n) = beta(n) == numdiv(n)/2; \\ Michel Marcus, Jul 03 2019

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.

Original entry on oeis.org

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

Views

Author

Bernard Schott, Jul 02 2019

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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).

Programs

  • PARI
    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

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

Views

Author

Michel Marcus, Jul 10 2019

Keywords

Comments

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)

Examples

			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

Links

Crossrefs

Intersection of A002378 and A290869.
Cf. A326378 (similar in no base), A326384 (similar in one base), A326385 (similar in 2 bases).

Programs

  • PARI
    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);

Extensions

a(11) from Chai Wah Wu, Jul 21 2019
a(12)-a(16) from Giovanni Resta, Jul 28 2019

A326384 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.

Original entry on oeis.org

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
Offset: 1

Views

Author

Bernard Schott, Jul 10 2019

Keywords

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.

Examples

			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.

Links

Crossrefs

Cf. A000005 (tau), A220136 (beta).
Subsequence of A002378 (oblong numbers) and of A167782.
Cf. A326378 (oblongs with tau(m)/2 - 2), A326385 (oblongs with tau(m)/2), A309062 (oblongs with tau(m)/2 + k, k >= 1).

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

Views

Author

Felix Fröhlich, Jul 16 2019

Keywords

Examples

			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)

Crossrefs

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).

Programs

  • PARI
    /* 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))))

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

Views

Author

Bernard Schott, Mar 29 2023

Keywords

Comments

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).

Examples

			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.

Links

Crossrefs

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

Programs

Showing 1-6 of 6 results.

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