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 1-7 of 7 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

A326386 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.

Original entry on oeis.org

8, 10, 14, 18, 22, 24, 27, 28, 32, 33, 34, 35, 38, 39, 44, 45, 46, 48, 50, 51, 52, 54, 55, 58, 60, 65, 66, 68, 69, 70, 74, 75, 76, 77, 78, 82, 84, 87, 88, 92, 94, 95, 96, 98, 99, 102, 104, 105, 106, 108, 112, 115, 116, 117, 118, 119, 120, 122, 123, 125, 126, 128, 130, 134, 135, 136
Offset: 1

Views

Author

Bernard Schott, Jul 12 2019

Keywords

Comments

As tau(m) = 2 * (1 + 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 A326379: non-oblong composites which have no Brazilian representation with three digits or more.

Examples

			tau(m) = 4 and beta(m)=1 for m = 8, 10, 14, 22, 27, 33, 34, 35, 38, ... 8 = 22_3,
tau(m) = 6 and beta(m)=2 for m = 18, 28, 32, 44, 45, 50, ... 18 = 33_5 = 22_8,
tau(m) = 8 and beta(m)=3 for m = 24, 54, 66, 70, ... 24 = 44_5 = 33_7 = 22_11,
tau(m) = 10 and beta(m) = 4: 48, 112, ... 48 = 66_7 = 44_11 = 33_15 = 22_23.

Links

Crossrefs

Cf. A000005 (tau), A220136 (beta).
Subsequence of A308874 and of A326379.
Cf. A326387 (non-oblongs with tau(m)/2), A326388 (non-oblongs with tau(m)/2 + 1), A326389 (non-oblongs with tau(m)/2 + 2).

Programs

  • PARI
    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 - 1); \\ Michel Marcus, Jul 15 2019

A308874 Composite numbers that are neither squares nor oblongs.

Original entry on oeis.org

8, 10, 14, 15, 18, 21, 22, 24, 26, 27, 28, 32, 33, 34, 35, 38, 39, 40, 44, 45, 46, 48, 50, 51, 52, 54, 55, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 74, 75, 76, 77, 78, 80, 82, 84, 85, 86, 87, 88, 91, 92, 93, 94, 95, 96, 98, 99, 102, 104, 105, 106, 108, 111, 112, 114
Offset: 1

Views

Author

Bernard Schott, Jul 12 2019

Keywords

Comments

A characterization: the terms of this sequence have Brazilian representations with repdigits of length = 2 and the number of these representations is beta'(n) = tau(n)/2 - 1.
Some examples (here tau(n) is the number of divisors of n):
tau(8) = 4 and 8 = 22_3, so: beta'(8) = tau(8)/2 - 1 = 1.
tau(15) = 4 and 15 = 1111_2 = 33_4, so beta'(15) = tau(15)/2 - 1 = 1.
tau(18) = 6 and 18 = 33_5 = 22_8, so beta'(18) = tau(18)/2 - 1 = 2.
tau(54) = 8 and 54 = 66_8 = 33_17 = 22_26, so beta'(54) = tau(54)/2 - 1 = 3.

Crossrefs

Cf. A002808 (composites), A000290 (squares), A000037 (nonsquares), A002378 (oblongs), A078358 (non-oblongs).
Subsequences: A326386, A326387, A326388, A326389.

Programs

  • PARI
    isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n;
isok(n) = !isprime(n) && !issquare(n) && !isoblong(n); \\ Michel Marcus, Jul 12 2019

A326388 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.

Original entry on oeis.org

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
Offset: 1

Views

Author

Bernard Schott, Jul 13 2019

Keywords

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, ...

Examples

			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.

Links

Crossrefs

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783, A308874 and A326381.
Cf. A326386 (non-oblongs with tau(m)/2 - 1), A326387 (non-oblongs with tau(m)/2), A326389 (non-oblongs with tau(m)/2 + 2).

Programs

  • PARI
    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 + 1); \\ Michel Marcus, Jul 15 2019

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

Views

Author

Bernard Schott, Jul 20 2019

Keywords

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), ...

Examples

			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.

Links

Crossrefs

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).

Programs

  • PARI
    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

Views

Author

Bernard Schott, Jul 21 2019

Keywords

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, ...

Examples

			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.

Links

Crossrefs

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).

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 1-7 of 7 results.

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