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.

A069474 First differences of A069473.

Original entry on oeis.org

540, 2100, 5460, 11340, 20460, 33540, 51300, 74460, 103740, 139860, 183540, 235500, 296460, 367140, 448260, 540540, 644700, 761460, 891540, 1035660, 1194540, 1368900, 1559460, 1766940, 1992060, 2235540, 2498100, 2780460, 3083340
Offset: 0

Views

Author

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 26 2002

Keywords

Crossrefs

Equals 60 * A005898(n+1).

Programs

  • Mathematica
    Differences[Table[(n + 1)^6 - n^6, {n, 0, 30}], 2] (* Harvey P. Dale, Dec 27 2011 *)

Formula

a(n) = 120*n^3 + 540*n^2 + 900*n + 540.
G.f.: 60*(9 - x + 5*x^2 - x^3)/(1 - x)^4. [Bruno Berselli, Feb 25 2015]

Extensions

Offset changed from 1 to 0 and added a(0)=540 by Bruno Berselli, Feb 25 2015

A069476 First differences of A069475, successive differences of (n+1)^6-n^6.

Original entry on oeis.org

1800, 2520, 3240, 3960, 4680, 5400, 6120, 6840, 7560, 8280, 9000, 9720, 10440, 11160, 11880, 12600, 13320, 14040, 14760, 15480, 16200, 16920, 17640, 18360, 19080, 19800, 20520, 21240, 21960, 22680, 23400, 24120, 24840, 25560, 26280, 27000
Offset: 0

Views

Author

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 26 2002

Keywords

Crossrefs

Programs

  • Mathematica
    Table[720 n + 1800, {n, 0, 40}] (* Bruno Berselli, Feb 25 2015 *)

Formula

a(n) = 720*n + 1800 = 360*A020735(n+1).
G.f.: 360*(5 - 3*x)/(1 - x)^2. [Bruno Berselli, Feb 25 2015]

Extensions

Offset changed from 1 to 0 and added a(0)=1800 by Bruno Berselli, Feb 25 2015

A069475 First differences of A069474, successive differences of (n+1)^6-n^6.

Original entry on oeis.org

1560, 3360, 5880, 9120, 13080, 17760, 23160, 29280, 36120, 43680, 51960, 60960, 70680, 81120, 92280, 104160, 116760, 130080, 144120, 158880, 174360, 190560, 207480, 225120, 243480, 262560, 282360, 302880, 324120, 346080, 368760, 392160, 416280
Offset: 0

Views

Author

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 26 2002

Keywords

Crossrefs

Programs

Formula

a(n) = 360*n^2 + 1440*n + 1560 = 120*A056107(n+2).
G.f.: 120*(13 - 11*x + 4*x^2)/(1 - x)^3. - Bruno Berselli, Feb 25 2015

Extensions

Offset changed from 1 to 0 and added a(0)=1560 by Bruno Berselli, Feb 25 2015

A353890 a(n) is the period of the binary sequence {b(m)} defined by b(m) = 1 if (m+1)^n - m^n and (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise.

Original entry on oeis.org

1, 1, 5, 11, 91, 1247, 3485, 263017, 852841, 1241058127, 74966255, 243641132605417, 181556731572385303, 718802057694183783881, 6582662048285, 943422576750791493013356207217, 487331778345355477261, 607088607861933740557075591887834842297
Offset: 2

Views

Author

Samuel Harkness, May 09 2022

Keywords

Comments

For any n, consecutive n-th powers will never share a divisor > 1, so now consider the second differences. Specifically, each m > 0, define the binary sequence {b(m)} as follows: b(m) = 1 if the first difference (m+1)^n - m^n and the second difference (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise. I conjecture that {b(m)} is periodic with period a(n).
If m^n mod p == (m+1)^n mod p == (m+2)^n mod p, then p is in the prime factorization of a(n).
All primes p >= 5 belong to a prime factorization for a(n). p will always belong to the prime factorization of n=p-1 due to Fermat's Little Theorem.
I conjecture that the greatest prime factor for any prime n >= 5 is phi(2^n+1)/2 + 1 = Jacobsthal(n). n*A069925 + 1 = A001045(n).
I conjecture that all prime factors "f" are f=n*k+1, unless n is composite, in which case additionally all prime factors for any divisor of n will also be included in the prime factorization for a(n).

Examples

			For n=2 and n=3, the first and second differences are coprime for all m. Each of their sequences {b(m)} consist only of 1's, which can be described trivially as [1] with a period of 1, so a(2) = a(3) = 1.
For n > 3, the first and second differences are coprime for some m values, but not for all. Each repeating periodic sequence {b(m)} begins at m=1, and can be used to predict what b(m) will be at any higher m value for that power n.
n=4 has the 5-term repeating sequence, beginning at m=1:
  [0 0 1 1 1], so a(4) = 5.
The sequence is repeating, so for example, f(41)..f(45) is also [0 0 1 1 1].
n=5 has the 11-term repeating sequence
  [1 1 0 1 1 0 1 1 1 1 1]
so a(5) = 11.
n=6 has the 91-term repeating sequence
  [0 0 0 0 0 0 1 0 0 0 0 1 1
   1 0 0 0 0 0 1 1 0 0 0 0 1
   1 1 0 0 0 0 1 1 1 0 0 0 0
   1 1 1 0 0 0 0 1 1 1 0 0 0
   0 1 1 1 0 0 0 0 1 1 1 0 0
   0 0 1 1 0 0 0 0 0 1 1 1 0
   0 0 0 1 0 0 0 0 0 0 1 1 1]
so a(6) = 91.
The period for higher n values has yet to be found. If they exist, it seems they would be quite large given the large expansion from 5, 11, to 91.
Example: the 233rd term in the sequence of values for n=6 is calculated by using m=233 and n=6. Define the first difference for the 233rd term as 234^6 - 233^6 = 4164782373647. The second difference for the 233rd term is 235^6 - 2*234^6 + 233^6 = 89948228762. The terms 4164782373647 and 89948228762 share a common factor, so the 233rd term of the sequence for 6th powered terms is denoted 0 (not coprime). Because the 6th powered terms repeat their tendency of being coprime or not every 91 terms, we could instead look at 233 mod 91 = 51, and from the table for n=6 above, the 51st term is 0.
		

Crossrefs

Programs

  • MATLAB
    See Links section.

Extensions

a(7)-a(19) from Jon E. Schoenfield, May 10 2022
Showing 1-4 of 4 results.