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.

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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

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).

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).

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

Views

Author

Bernard Schott, Aug 09 2019

Keywords

Comments

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).

Examples

			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.

Links

Crossrefs

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).

Programs

  • PARI
    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

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

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