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 1-8 of 8 results.

A095862 Numbers whose number of decimal digits and number of divisors are equal.

Original entry on oeis.org

1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 121, 169, 289, 361, 529, 841, 961, 1003, 1006, 1007, 1011, 1018, 1027, 1037, 1041, 1042, 1043, 1046, 1047, 1055, 1057, 1059, 1067, 1073, 1077, 1079, 1081, 1082, 1094, 1099
Offset: 1

Views

Author

Ray Chandler, Jun 28 2004

Keywords

Comments

An element of this sequence is prime iff it has 2 digits, which is the case for a(2)=11 through a(22); sequence A096489 lists exactly these and the leading term a(1)=1 (the only noncomposite number which is not prime). - M. F. Hasler, Nov 29 2007

Crossrefs

Cf. A096489 (= a(1)..a(22) = noncomposite elements of this sequence).
Cf. A135772-A135779 (analog for bases 2...9).

Programs

  • Maple
    q:= n-> is(length(n)=numtheory[tau](n)):
    select(q, [$1..1100])[];  # Alois P. Heinz, Jun 18 2021
  • Mathematica
    Select[Range[1100],IntegerLength[#]==DivisorSigma[0,#]&] (* Harvey P. Dale, Oct 19 2015 *)

Extensions

Edited by M. F. Hasler, Nov 29 2007

A135772 Numbers having equal number of divisors and binary digits.

Original entry on oeis.org

1, 2, 3, 4, 8, 10, 14, 15, 16, 32, 44, 45, 50, 52, 63, 64, 128, 130, 135, 136, 138, 152, 154, 165, 170, 174, 182, 184, 186, 189, 190, 195, 222, 230, 231, 232, 238, 246, 248, 250, 255, 256, 441, 484, 512, 567, 592, 656, 688, 752, 848, 891, 944, 976
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Examples

			a(1) = 1 since 1 has 1 divisor and 1 binary digit.
a(2), a(3) = 2, 3 since 2 = 10_2 and 3 = 11_2 have 2 divisors and 2 binary digits.
a(4) = 4 = 100_2 is the only number with 3 binary digits having 3 divisors.
8, 10, 14, 15 have 4 binary digits and 4 divisors.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[500], DivisorSigma[0, #] == IntegerLength[#, 2] &] (* G. C. Greubel, Nov 08 2016 *)
  • PARI
    for(d=1,10,for(n=2^(d-1),2^d-1,d==numdiv(n)&print1(n", ")))
    
  • Python
    from sympy import divisor_count
    def ok(n): return divisor_count(n) == n.bit_length()
    print(list(filter(ok, range(1, 977)))) # Michael S. Branicky, Jul 29 2021

A135778 Numbers having number of divisors equal to number of digits in base 8.

Original entry on oeis.org

1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 121, 169, 289, 361, 514, 515, 517, 519, 526, 527, 533, 535, 537, 538, 542, 543, 545, 551, 553, 554, 559, 562, 565, 566, 573, 579, 581, 583, 586, 589, 591, 597, 611, 614, 622, 623, 626, 629, 633, 634
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Comments

Since 8 is not a prime, no element > 1 of the sequence A001018(k)=8^k (having k+1 digits in base 8, but much more divisors) can be member of this sequence. Also, no power of a prime less than 8 can be in the sequence, since it will always have fewer divisors than digits in base 8. However all powers of 11 up to 11^6 are in this sequence, having the same number of digits (in base 8) than the same power of 8 (since 6 = floor(log(11/8)/log(8))) and also that number of divisors (since 11 is prime).

Examples

			a(1) = 1 since 1 has 1 divisor and 1 digit (in base 8 as in any other base).
They are followed by the primes (having 2 divisors {1,p}) between 8 and 8^2 - 1 (to have 2 digits in base 8).
Then come the squares of primes (3 divisors) between 8^2 = 100_8 and 8^3 - 1 = 777_8.
These are followed by all semiprimes and cubes of primes (4 divisors) between 8^3 and 8^4 - 1.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1000],IntegerLength[#,8]==DivisorSigma[0,#]&] (* Harvey P. Dale, Mar 04 2016 *)
  • PARI
    for(d=1,4,for(n=8^(d-1),8^d-1,d==numdiv(n)&print1(n", ")))

Extensions

More terms from Harvey P. Dale, Mar 04 2016

A135773 Numbers having number of divisors equal to number of digits in base 3.

Original entry on oeis.org

1, 3, 5, 7, 9, 25, 27, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 81, 243, 244, 245, 261, 268, 275, 279, 284, 292, 316, 325, 332, 333, 338, 356, 363, 369, 387, 388, 404, 412, 423, 425, 428, 436, 452, 475, 477, 507, 508, 524, 531, 539, 548, 549
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Comments

Since 3 is a prime, any power 3^k has k+1 divisors { 3^i ; i=0..k } and the same number of digits in base 3; thus the sequence A000244(k) = 3^k is a subsequence of this one. Note that no number in between 3^4 and 3^5, neither in between 3^6 and 3^7, is in this sequence.

Examples

			a(1) = 1 since 1 has 1 divisor and 1 digit (in base 3).
2 has 2 divisors but only 1 digit in base 3, so it is not member of the sequence.
a(2)..a(4) = 3, 5, 7 all have 2 divisors and 2 digits in base 3.
81 = 3^4 = 10000_3 is the only number with 5 divisors and 5 digits in base 3, so it is followed by 243 = 3^5 = 100000_3.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[500], DivisorSigma[0, #] == IntegerLength[#, 3] &] (* G. C. Greubel, Nov 08 2016 *)
  • PARI
    for(d=1,6,for(n=3^(d-1),3^d-1,d==numdiv(n)&print1(n", ")))

A135774 Numbers having number of divisors equal to number of digits in base 4.

Original entry on oeis.org

1, 5, 7, 11, 13, 25, 49, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 115, 118, 119, 122, 123, 125, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 166, 177, 178, 183, 185, 187, 194, 201, 202, 203, 205, 206, 209, 213, 214, 215, 217, 218
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Comments

Since 4 is not a prime, no element > 1 of the sequence A000302(k)=4^k (having k+1 digits in base 4 but 2k+1 divisors) can be member of this sequence. However all powers of 5 up to 5^6 are in this sequence, having the same number of digits (in base 4) than the same power of 4 (since (5/4)^6 < 4 < (5/4)^7) and also that number of divisors.

Examples

			a(1) = 1 since 1 has 1 divisor and 1 digit (in base 4 as in any other base).
a(2)..a(5) = 5, 7, 11, 13 are the primes (to have 2 divisors {1,p}) between 4 and 4^2 - 1 (to have 2 digits in base 4).
a(6), a(7) = 25, 49 are the squares of primes (3 divisors) between 4^2 = 100[4] and 4^3 - 1 = 333_4.
They are followed by all semiprimes and cubes of primes (4 divisors) between 4^3 and 4^4 - 1.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[500], DivisorSigma[0, #] == IntegerLength[#, 4] &] (* G. C. Greubel, Nov 08 2016 *)
  • PARI
    for(d=1,4,for(n=4^(d-1),4^d-1,d==numdiv(n)&print1(n", ")))

A135775 Numbers having number of divisors equal to number of digits in base 5.

Original entry on oeis.org

1, 5, 7, 11, 13, 17, 19, 23, 25, 49, 121, 125, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 166, 177, 178, 183, 185, 187, 194, 201, 202, 203, 205, 206, 209, 213, 214, 215, 217, 218, 219, 221, 226, 235, 237, 247, 249, 253, 254, 259, 262, 265, 267
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Comments

Since 5 is a prime, any power 5^k has k+1 divisors { 5^i ; i=0..k } and the same number of digits in base 5; thus the sequence A000351(k)=5^k is a subsequence of this one. It also includes the powers of 7 up to 7^4, since (7/5)^4 < 5 < (7/5)^5.

Examples

			a(1) = 1 since 1 has 1 divisor and 1 digit (in base 5).
2,3,4 have 2 resp. 3 divisors but only 1 digit in base 5, so they are not members of the sequence.
a(2) = 5 = 10_5 has 2 divisors { 1, 5 } and 2 digits in base 5, so it is (the second term) in this sequence.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[300],DivisorSigma[0,#]==IntegerLength[#,5]&] (* Harvey P. Dale, Mar 14 2013 *)
  • PARI
    for(d=1,4,for(n=5^(d-1),5^d-1,d==numdiv(n)&print1(n", ")))

A135776 Numbers having number of divisors equal to number of digits in base 6.

Original entry on oeis.org

1, 7, 11, 13, 17, 19, 23, 29, 31, 49, 121, 169, 217, 218, 219, 221, 226, 235, 237, 247, 249, 253, 254, 259, 262, 265, 267, 274, 278, 287, 291, 295, 298, 299, 301, 302, 303, 305, 309, 314, 319, 321, 323, 326, 327, 329, 334, 335, 339, 341, 343, 346, 355, 358
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Comments

Since 6 is not a prime, no element > 1 of the sequence A000400(k)=6^k (having k+1 digits in base 6, but much more divisors) can be a member of this sequence. However, all powers of 7 up to 7^11 are in this sequence, having the same number of digits (in base 6) as the same power of 6 (since 11 = floor(log(7/6)/log(6))) and also that number of divisors (since 7 is prime).

Examples

			a(1) = 1 since 1 has 1 divisor and 1 digit (in base 6 as in any other base).
They are followed by the primes (having 2 divisors {1,p}) between 6 and 6^2 - 1 (to have 2 digits in base 6).
Then come the squares of primes (3 divisors) between 6^2 = 100_6 and 6^3 - 1 = 555_6.
These are followed by all semiprimes and cubes of primes (4 divisors) between 6^3 and 6^4 - 1.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[500], DivisorSigma[0, #] == IntegerLength[#, 6] &] (* G. C. Greubel, Nov 08 2016 *)
  • PARI
    for(d=1,4,for(n=6^(d-1),6^d-1,d==numdiv(n)&print1(n", ")))

A135777 Numbers having number of divisors equal to number of digits in base 7.

Original entry on oeis.org

1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 121, 169, 289, 343, 346, 355, 358, 362, 365, 371, 377, 381, 382, 386, 391, 393, 394, 395, 398, 403, 407, 411, 413, 415, 417, 422, 427, 437, 445, 446, 447, 451, 453, 454, 458, 466, 469, 471, 473, 478, 481
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2007

Keywords

Comments

Since 7 is a prime, any power 7^k has k+1 divisors { 7^i ; i=0..k } and the same number of digits in base 7; thus the sequence A000420(k)=7^k is a subsequence of this one.

Examples

			a(1) = 1 since 1 has 1 divisor and 1 digit (in base 7 as in any other base).
All other numbers have at least 2 divisors so there are no other members of the sequence below a(2) = 7 = 10_7 having 2 divisors { 1, 7 } and 2 digits in base 7.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[500],DivisorSigma[0,#]==IntegerLength[#,7]&] (* Harvey P. Dale, Feb 14 2015 *)
  • PARI
    for(d=1,4,for(n=7^(d-1),7^d-1,d==numdiv(n)&print1(n", ")))
Showing 1-8 of 8 results.