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 1-4 of 4 results.

A068601 a(n) = n^3 - 1.

Original entry on oeis.org

0, 7, 26, 63, 124, 215, 342, 511, 728, 999, 1330, 1727, 2196, 2743, 3374, 4095, 4912, 5831, 6858, 7999, 9260, 10647, 12166, 13823, 15624, 17575, 19682, 21951, 24388, 26999, 29790, 32767, 35936, 39303, 42874, 46655, 50652, 54871, 59318, 63999, 68920
Offset: 1

Views

Author

Naohiro Nomoto, Mar 28 2002

Keywords

Comments

a(n) is the least positive integer k such that k can only contain 'n-1' in exactly 2 different bases B, where 1 < B <= k.
Apart from the first term, the same as A135300. - R. J. Mathar, Apr 29 2008
A058895(n)^3 + a(n)^3 + A033562(n)^3 = A185065(n)^3. - Vincenzo Librandi, Mar 13 2012
Numbers k such that for every nonnegative integer m, k^(3*m+1) + k^(3*m) is a cube. - Arkadiusz Wesolowski, Aug 10 2013

Examples

			For n=6; 215 written in bases 6 and 42 is 555, 55 and (555, 55) are exactly 2 different bases.
		

Crossrefs

Programs

Formula

Partial sums of A003215, hex (or centered hexagonal) numbers: 3*n(n+1)+1. - Jonathan Vos Post, Mar 16 2006
G.f.: x^2*(7-2*x+x^2)/(1-x)^4. - Colin Barker, Feb 12 2012
4*a(m^2-2*m+2) = (m^2-m+1)^3 + (m^2-m-1)^3 + (m^2-3*m+3)^3 + (m^2-3*m+1)^3. - Bruno Berselli, Jun 23 2014
a(n) = Sum_{i=1..n-1} (i+1)^3 - i^3. - Wesley Ivan Hurt, Jul 23 2014
Sum_{n>=2} 1/a(n) = Sum_{n>=1} (zeta(3*n) - 1) = A339604. - Amiram Eldar, Nov 06 2020
Product_{n>=2} (1 + 1/a(n)) = 3*Pi*sech(sqrt(3)*Pi/2). - Amiram Eldar, Jan 20 2021
E.g.f.: 1 + exp(x)*(x^3 + 3*x^2 + x - 1). - Stefano Spezia, Jul 06 2021

A129296 Number of divisors of n^2 - 1 that are not greater than n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 5, 3, 5, 3, 8, 2, 8, 4, 6, 4, 9, 2, 12, 4, 8, 4, 10, 3, 10, 6, 8, 4, 16, 2, 14, 4, 7, 8, 12, 4, 12, 4, 10, 4, 20, 2, 16, 6, 8, 6, 12, 3, 18, 6, 12, 4, 16, 4, 20, 8, 10, 4, 16, 2, 16, 6, 8, 12, 16, 4, 16, 4, 16, 4, 30, 2, 15, 6, 8, 12, 16, 4, 24, 5, 12, 5, 16, 4, 16, 8, 10, 4, 30, 4
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 09 2007

Keywords

Comments

a(n) = #{d: d<=n and A005563(n+1) mod d = 0};
a(n)>1 for n>2, see A129297 for m such that a(m)=2: a(A129297(n)) = 2.
If a(6n) = 2 for n>=1, then 6n-1 and 6n+1 are twin primes see A129297. - Fred Daniel Kline, Jan 02 2014
For n>1, a(n) is the number of positive integers k such that k+n divides k*n+1. - Thomas Ordowski, Dec 01 2024

Examples

			a(100) = #{1,3,9,11,33,99} = 6.
		

Crossrefs

Programs

Formula

a(n) = A000005(n^2-1)/2 for n >= 2. - Robert Israel, Aug 03 2015

A129292 Number of divisors of n^4 - 1 that are not greater than n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 5, 3, 8, 3, 10, 4, 6, 4, 12, 4, 13, 4, 11, 6, 14, 3, 10, 6, 12, 6, 17, 3, 16, 7, 10, 9, 13, 4, 18, 7, 11, 4, 22, 3, 26, 8, 9, 7, 23, 5, 18, 7, 13, 6, 25, 4, 24, 8, 21, 6, 18, 3, 18, 10, 12, 14, 16, 4, 26, 8, 17, 7, 31, 5, 30, 6, 11, 13, 26, 7, 25, 6, 16, 10, 35, 4, 18, 11
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 09 2007

Keywords

Comments

a(n) = #{d: d<=n and A123865(n) mod d = 0};
a(n)>1 for n>2, see A129293 for m such that a(m)=2: a(A129293(n))=2.

Examples

			a(100) = #{1,3,9,11,33,73,99} = 7.
		

Crossrefs

Programs

A129295 Numbers m such that m^3 - 1 has no divisors d with 1 < d < m - 1.

Original entry on oeis.org

3, 4, 6, 8, 12, 14, 20, 24, 38, 54, 62, 80, 90, 110, 138, 150, 164, 168, 192, 194, 272, 278, 314, 332, 348, 398, 402, 434, 500, 572, 642, 644, 720, 728, 762, 798, 812, 860, 864, 878, 920, 992, 1020, 1022, 1070, 1092, 1098, 1118, 1130, 1182, 1202, 1230, 1260, 1308
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 09 2007

Keywords

Comments

Numbers m such that A129294(m) = #{1,m-1} = 2.
Essentially the same as A096175. Note that m^3 - 1 = (m - 1)*(m^2 + m + 1), so m - 1 must be prime. For m > 4, the smallest divisor > 1 of m^2 + m + 1 is no larger than sqrt(m^2 + m + 1) < m + 1 unless m^2 + m + 1 is also prime. Also note that gcd(m, m^2 + m 1 ) = gcd(m - 1, m^2 + m + 1) = 1, so m^2 + m + 1 must also be prime, making m^3 - 1 a semiprime. - Jianing Song, Aug 01 2018

Examples

			{1,11,157,1727} is the set of divisors of 12^3 - 1, therefore 12 is a term, since A129294(12) = #{1,11} = 2.
		

Crossrefs

Formula

a(n) = A096175(n-2) for n > 2. - Jianing Song, Aug 01 2018
Showing 1-4 of 4 results.