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.

A366630 a(n) = phi(6^n+1), where phi is Euler's totient function (A000010).

Original entry on oeis.org

1, 6, 36, 180, 1296, 6000, 41472, 230496, 1580800, 8359200, 58579200, 310968900, 2175102720, 10971642240, 76065091200, 351048600000, 2811459796992, 14508487949472, 88870766837760, 522016066337712, 3564233663616000, 17479898551382400, 128060205344805888
Offset: 0

Views

Author

Sean A. Irvine, Oct 14 2023

Keywords

Crossrefs

Programs

  • Mathematica
    EulerPhi[6^Range[0, 22] + 1] (* Paul F. Marrero Romero, Oct 17 2023 *)
  • PARI
    {a(n) = eulerphi(6^n+1)}

Formula

a(n) = A000010(A062394(n)). - Paul F. Marrero Romero, Oct 17 2023

A366606 Number of divisors of 4^n+1.

Original entry on oeis.org

2, 2, 2, 4, 2, 6, 4, 8, 2, 16, 4, 8, 8, 16, 4, 48, 4, 16, 16, 16, 4, 64, 8, 32, 8, 64, 8, 64, 8, 8, 16, 32, 4, 64, 12, 96, 32, 32, 16, 768, 8, 32, 32, 32, 16, 1536, 4, 16, 8, 64, 64, 512, 4, 16, 64, 96, 32, 256, 8, 128, 64, 64, 16, 1024, 4, 768, 128, 64, 16
Offset: 0

Views

Author

Sean A. Irvine, Oct 14 2023

Keywords

Examples

			a(3)=4 because 4^3+1 has divisors {1, 5, 13, 65}.
		

Crossrefs

Programs

  • Maple
    a:=n->numtheory[tau](4^n+1):
    seq(a(n), n=0..100);
  • Mathematica
    DivisorSigma[0,4^Range[0,100]+1] (* Paolo Xausa, Oct 14 2023 *)
  • PARI
    a(n) = numdiv(4^n+1);
    
  • Python
    from sympy import divisor_count
    def A366606(n): return divisor_count((1<<(n<<1))+1) # Chai Wah Wu, Oct 14 2023

Formula

a(n) = sigma0(4^n+1) = A000005(A052539(n)).
a(n) = A046798(2*n). - Max Alekseyev, Jan 08 2024

A366629 Sum of the divisors of 6^n+1.

Original entry on oeis.org

3, 8, 38, 256, 1298, 9792, 52136, 338580, 1778436, 11889152, 62367272, 414625216, 2178461956, 15224775552, 80673299432, 611106029568, 2830769440776, 19344856702976, 115255634181184, 696800841097536, 3748220725527432, 27388329197137920, 135183433256806480
Offset: 0

Views

Author

Sean A. Irvine, Oct 14 2023

Keywords

Examples

			a(3)=256 because 6^3+1 has divisors {1, 7, 31, 217}.
		

Crossrefs

Programs

  • Maple
    a:=n->numtheory[sigma](6^n+1):
    seq(a(n), n=0..100);
  • Mathematica
    DivisorSigma[1, 6^Range[0, 30] + 1] (* Paolo Xausa, Jul 03 2024 *)

Formula

a(n) = sigma(6^n+1) = A000203(A062394(n)).

A366616 Number of divisors of 5^n+1.

Original entry on oeis.org

2, 4, 4, 12, 4, 8, 8, 16, 8, 32, 16, 32, 8, 16, 8, 96, 8, 16, 32, 32, 16, 576, 16, 16, 16, 32, 24, 320, 8, 16, 128, 32, 16, 384, 64, 128, 64, 32, 16, 192, 32, 64, 64, 64, 8, 512, 8, 32, 32, 128, 128, 768, 32, 32, 64, 128, 128, 384, 8, 64, 64, 64, 16, 24576, 16
Offset: 0

Views

Author

Sean A. Irvine, Oct 14 2023

Keywords

Examples

			a(3)=12 because 5^3+1 has divisors {1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126}.
		

Crossrefs

Programs

  • Maple
    a:=n->numtheory[tau](5^n+1):
    seq(a(n), n=0..100);
  • Mathematica
    DivisorSigma[0, 5^Range[0, 70] + 1] (* Paolo Xausa, Apr 20 2025 *)
  • PARI
    a(n) = numdiv(5^n+1);

Formula

a(n) = sigma0(5^n+1) = A000005(A034474(n)).

A366627 Number of distinct prime divisors of 6^n + 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 3, 3, 2, 3, 3, 2, 2, 4, 3, 5, 3, 3, 6, 3, 3, 5, 4, 3, 4, 4, 4, 5, 5, 4, 9, 2, 3, 6, 3, 10, 5, 4, 3, 9, 5, 4, 7, 2, 3, 7, 5, 2, 7, 5, 6, 8, 4, 5, 10, 7, 6, 7, 3, 2, 6, 3, 2, 9, 3, 8, 11, 5, 5, 6, 7, 4, 5, 6, 4, 10, 5, 5, 10, 6, 6, 8, 4, 5, 8
Offset: 0

Views

Author

Sean A. Irvine, Oct 14 2023

Keywords

Crossrefs

Programs

  • Mathematica
    PrimeNu[6^Range[0,84] + 1] (* Paul F. Marrero Romero, Nov 11 2023 *)
  • PARI
    for(n = 0, 100, print1(omega(6^n + 1), ", "))

Formula

a(n) = omega(6^n+1) = A001221(A062394(n)).

A366637 Number of divisors of 7^n+1.

Original entry on oeis.org

2, 4, 6, 8, 4, 16, 24, 16, 8, 16, 32, 16, 32, 16, 12, 64, 8, 8, 48, 16, 16, 128, 48, 8, 16, 32, 24, 32, 64, 8, 512, 32, 16, 128, 48, 1024, 256, 16, 12, 256, 64, 64, 96, 512, 32, 2048, 96, 8, 64, 2048, 640, 128, 32, 64, 384, 3072, 256, 256, 96, 64, 512, 8, 48
Offset: 0

Views

Author

Sean A. Irvine, Oct 15 2023

Keywords

Examples

			a(4)=4 because 7^4+1 has divisors {1, 2, 1201, 2402}.
		

Crossrefs

Programs

  • Maple
    a:=n->numtheory[tau](7^n+1):
    seq(a(n), n=0..100);
  • Mathematica
    DivisorSigma[0, 7^Range[0, 62] + 1] (* Paul F. Marrero Romero, Oct 16 2023 *)
  • PARI
    a(n) = numdiv(7^n+1);

Formula

a(n) = sigma0(7^n+1) = A000005(A034491(n)).

A366656 Number of divisors of 8^n+1.

Original entry on oeis.org

2, 3, 4, 8, 4, 12, 16, 12, 8, 20, 48, 24, 16, 12, 64, 64, 8, 48, 64, 24, 16, 64, 64, 24, 32, 96, 768, 192, 32, 24, 1536, 24, 8, 256, 512, 1536, 64, 96, 256, 64, 64, 96, 1024, 48, 128, 1280, 256, 96, 128, 96, 8192, 1024, 32, 48, 1024, 2304, 256, 192, 256, 192
Offset: 0

Views

Author

Sean A. Irvine, Oct 15 2023

Keywords

Examples

			a(4)=4 because 8^4+1 has divisors {1, 17, 241, 4097}.
		

Crossrefs

Programs

  • Maple
    a:=n->numtheory[tau](8^n+1):
    seq(a(n), n=0..100);
  • Mathematica
    DivisorSigma[0, 8^Range[0,59] + 1] (* Paul F. Marrero Romero, Nov 12 2023 *)
  • PARI
    a(n) = numdiv(8^n+1);

Formula

a(n) = sigma0(8^n+1) = A000005(A062395(n)).
a(n) = A046798(3*n). - Max Alekseyev, Jan 09 2024

A366665 Number of divisors of 9^n+1.

Original entry on oeis.org

2, 4, 4, 8, 8, 12, 8, 16, 4, 16, 8, 16, 64, 16, 16, 48, 4, 16, 16, 16, 32, 128, 32, 16, 16, 128, 16, 32, 64, 16, 128, 32, 4, 64, 32, 384, 256, 32, 64, 128, 32, 32, 1024, 128, 64, 384, 16, 16, 64, 512, 64, 256, 128, 64, 512, 192, 512, 512, 32, 8, 2048, 64, 16
Offset: 0

Views

Author

Sean A. Irvine, Oct 15 2023

Keywords

Examples

			a(2)=4 because 9^2+1 has divisors {1, 2, 41, 82}.
		

Crossrefs

Programs

  • Maple
    a:=n->numtheory[tau](9^n+1):
    seq(a(n), n=0..100);
  • Mathematica
    DivisorSigma[0, 9^Range[0,62] + 1] (* Paul F. Marrero Romero, Nov 13 2023 *)
  • PARI
    a(n) = numdiv(9^n+1);

Formula

a(n) = sigma0(9^n+1) = A000005(A062396(n)).
a(n) = A366577(2*n). - Max Alekseyev, Jan 08 2024
Showing 1-8 of 8 results.