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 1-10 of 12 results. Next

A167782 Numbers that are repdigits with length > 2 in some base.

Original entry on oeis.org

0, 7, 13, 15, 21, 26, 31, 40, 42, 43, 57, 62, 63, 73, 80, 85, 86, 91, 93, 111, 114, 121, 124, 127, 129, 133, 146, 156, 157, 170, 171, 172, 182, 183, 211, 215, 219, 222, 228, 241, 242, 255, 259, 266, 273, 285, 292, 307, 312, 314, 333, 341, 342, 343, 364, 365, 366
Offset: 1

Views

Author

Andrew Weimholt, Nov 12 2009

Keywords

Comments

Definition requires "length > 2" because all numbers n > 2 are trivially represented as "11" in base n-1.
0 included at the suggestion of Franklin T. Adams-Watters (and others) as 0 = 000 in any base.

Examples

			26 is a term because 26_10 = 222_3.
		

Crossrefs

Cf. A010785 (Repdigits (base 10)).
Cf. A167783 (Numbers that are repdigits with length > 2 in more than one base).
Cf. A053696 (Numbers which are repunits in some base).
Cf. A158235 (Numbers n whose square can be represented as a repdigit number in some base < n).

Programs

  • PARI
    /* In PARI versions < 2.6, define: digits(n,b) = if(n=b^2+b+1,d=digits(n,b);if(is_repdigit(d),print(n," = ",d," base ",b));b++)) \\ Michael B. Porter

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.
		

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

A290869 Numbers that are repdigits with length > 2 in more than two bases.

Original entry on oeis.org

4095, 32767, 65535, 67053, 262143, 265720, 531440, 1048575, 2097151, 4381419, 5592405, 7174453, 9808617, 11184810, 13938267, 14348906, 16777215, 19617234, 21523360, 29425851, 39234468, 43046720, 48686547, 49043085, 58851702, 61035156, 68660319, 71270178
Offset: 1

Views

Author

Michel Marcus, Aug 13 2017

Keywords

Comments

Most of the terms of A167783 are repdigits with length > 2 in only two bases.
For any b > 1 and k > 2, b^(4*k)-1 is a repdigit with length > 2 in bases b, b^2 and b^4; hence this sequence is infinite. - Rémy Sigrist, Aug 19 2017

Examples

			67053 is a repdigit with more than 2 digits in three bases: [31,31,31]_46, [21,21,21]_56, [3,3,3]_149.
		

Crossrefs

Subsequence of A167783.

Programs

  • PARI
    isok(n)=my(nb = 0); for (b=2, n-1, d = digits(n, b); if ((#d > 2) && (#Set(d) == 1), nb++); if (nb > 2, return (1)); ); return (0);
    (C#) See Links section.
    (C++) See Links section.

Extensions

More terms from Giovanni Resta, Aug 13 2017

A326382 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

32767, 65535, 67053, 2097151, 4381419, 7174453, 9808617, 13938267, 14348906, 19617234, 21523360, 29425851, 39234468, 43046720, 48686547, 49043085, 58851702, 61035156, 68660319, 71270178, 78468936, 88277553, 98086170, 107894787, 115174101, 117703404, 134217727, 142540356, 175965517
Offset: 1

Views

Author

Bernard Schott, Jul 08 2019

Keywords

Comments

As tau(m) = 2 * (beta(m) - 2) , the terms of this sequence are not squares.
There are 2 subsequences which realize a partition of this sequence (see array in link and examples):
1) Non-oblong composites which have exactly three Brazilian representations with three digits or more, they are in A326389.
2) Oblong numbers that have exactly four Brazilian representations with three digits or more. These integers have been found through b-file of Rémy Sigrist in A290869. These oblong integers are a subsequence of A309062.
There are no primes that satisfy this relation.

Examples

			One example for each type:
1) The divisors of 32767 are {1, 7, 31, 151, 217, 1057, 4681, 32767} and tau(32767) = 8; also, 32767 = M_15 = R(15)_2 = 77777_8 = (31,31,31)_32 = (151,151)_216 = (31,31)_1056 = 77_4680 so beta(32767) = 6 with beta'(32767) = 3 and beta"(32767)= 3. The relation is beta(32767) = tau(32767)/2 + 2 = 6.
2) 61035156 = 7812 * 7813 is oblong with tau(61035156) = 144. The four Brazilian representations with three digits or more are 61035156 = R(12)_5 = 666666_25 = (31,31,31,31)_125 = (156,156,156)_625, so beta"(61035156) = 4 and beta(61035156) = tau(61035156)/2 + 2 = 74.
		

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), this sequence (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).

Extensions

Missing a(18) inserted by Bernard Schott, Jul 20 2019

A326381 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

31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 4557, 6560, 7566, 7812, 8191, 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 07 2019

Keywords

Comments

As tau(m) = 2 * (beta(m) - 1), the terms of this sequence are not squares.
There are 3 subsequences which realize a partition of this sequence (see examples):
1) Non-oblong composites which have exactly two Brazilian representations with three digits or more, they form A326388.
2) Oblong numbers that have exactly three Brazilian representations with three digits or more; thanks to Michel Marcus, who found the smallest, 641431602. These oblong integers are a subsequence of A290869 and A309062.
3) The two Brazilian primes 31 and 8191 of the Goormaghtigh conjecture (A119598) for which beta(p) = tau(p)/2 + 1 = 2.

Examples

			One example for each type:
1) 63 = 111111_2 = 333_4 = 77_8 = 33_20 with tau(63) = 6 and beta(63) = 4.
2) 641431602 = 25326 * 25327 is oblong with tau(641431602) = 256. The three Brazilian representations with three digits or more of 641431602 are 999999_37 = (342,342,342)_1369 = (54,54,54)_3446, so beta"(641431602) = 3 and beta(641431602) = tau(641431602)/2 + 1 = 129.
3) 31 = 11111_2 = 111_5 and 8191 = 1111111111111_2 = 11_90 with beta(p) = tau(p)/2 + 1 = 2.
		

Crossrefs

Cf. A000005 (tau), A220136 (beta).
Cf. A119598 (Goormaghtigh conjecture).
Subsequence of A167783.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), 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 08 2019

A326383 Numbers m such that beta(m) = tau(m)/2 + 3 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

4095, 262143, 265720, 531440, 1048575, 5592405, 11184810, 122070312, 183105468, 193710244, 244140624, 268435455, 387420488
Offset: 1

Views

Author

Bernard Schott, Jul 08 2019

Keywords

Comments

As tau(m) = 2 * (beta(m) - 3), the terms of this sequence are not squares.
The current known terms are non-oblong composites that have exactly four Brazilian representations with three digits or more; but, maybe, there exist oblong integers that have exactly five Brazilian representations with three digits or more.

Examples

			The 24 divisors of 4095 = M_12 are {1, 3, 5, 7, 9, 13, 15, 21, 35, 39, 45, 63, 65, 91, 105, 117, 195, 273, 315, 455, 585, 819, 1365, 4095} and tau(4095) = 24; also, 4095 = R(12)_2 = 333333_4 = 7777_8 = (15,15,15)_16, so, beta(4095) = 15 with beta'(4095)= 11 and beta''(4095) = 4. The relation is beta(4095) = tau(4095)/2 + 3 = 15 and 4095 is a term.
		

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).

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

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 A326380: oblong numbers that have exactly two Brazilian representations with three digits or more.

Examples

			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.
		

Crossrefs

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).

Extensions

a(6)-a(25) from Giovanni Resta, Jul 11 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.
		

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.
		

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.
		

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).
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