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.

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

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Author

Bernard Schott, Jul 03 2019

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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).

Programs

  • PARI
    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

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

Views

Author

Bernard Schott, Jul 14 2019

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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).

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); \\ 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

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

Views

Author

Bernard Schott, Aug 04 2019

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf. A002182, A279930, A309039, A329383.

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).
Showing 1-7 of 7 results.

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