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.

Previous Showing 11-16 of 16 results.

A309255 a(n) = n + 1 - Sum_{k=0..n} (Stirling1(n,k) mod 2).

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 3, 4, 7, 8, 7, 8, 9, 10, 7, 8, 15, 16, 15, 16, 17, 18, 15, 16, 21, 22, 19, 20, 21, 22, 15, 16, 31, 32, 31, 32, 33, 34, 31, 32, 37, 38, 35, 36, 37, 38, 31, 32, 45, 46, 43, 44, 45, 46, 39, 40, 49, 50, 43, 44, 45, 46, 31, 32, 63, 64, 63, 64, 65, 66, 63, 64, 69, 70, 67, 68, 69
Offset: 0

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Author

Ilya Gutkovskiy, Jul 19 2019

Keywords

Comments

Number of even entries in n-th row of triangle of Stirling numbers of the first kind (A048994).

Crossrefs

Programs

  • Mathematica
    Table[n + 1 - Sum[Mod[StirlingS1[n, k], 2], {k, 0, n}], {n, 0, 76}]
    nmax = 76; CoefficientList[Series[1/(1 - x)^2 - (1 + x) Product[(1 + 2 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]
  • PARI
    a(n) = n + 1 - sum(k=0, n, stirling(n, k, 1) % 2); \\ Michel Marcus, Jul 19 2019
    
  • PARI
    a(n) = n + 1 - 2^hammingweight(n\2); \\ Amiram Eldar, Jul 25 2023

Formula

G.f.: 1/(1 - x)^2 - (1 + x) * Product_{k>=0} (1 + 2*x^(2^(k+1))).
a(n) = n + 1 - 2^A000120(floor(n/2)).

A190906 a(n) = gcd(n! / floor(n/2)!^2, 3^n).

Original entry on oeis.org

1, 1, 1, 3, 3, 3, 1, 1, 1, 9, 9, 9, 3, 3, 3, 9, 9, 9, 1, 1, 1, 3, 3, 3, 1, 1, 1, 27, 27, 27, 9, 9, 9, 27, 27, 27, 3, 3, 3, 9, 9, 9, 3, 3, 3, 27, 27, 27, 9, 9, 9, 27, 27, 27, 1, 1, 1, 3, 3, 3, 1, 1, 1, 9, 9, 9, 3, 3, 3, 9, 9, 9, 1, 1, 1, 3, 3, 3, 1, 1, 1
Offset: 0

Views

Author

Peter Luschny, Jun 30 2011

Keywords

Crossrefs

Cf. A060632.

Programs

Formula

a(n) = gcd(A056040(n), 3^n).
a(n) <= n. - Charles R Greathouse IV, Jun 30 2011
From Johannes W. Meijer, Jun 30 2011: (Start)
a(3*n) = a(3*n+1) = a(3*n+2) = A010684(n)*a(n) for n > 1 with a(0) = a(1) = a(2) = 1.
a(9*n+3) = a(9*n+4) = a(9*n+5) = 3*a(n).
a(9*n) = a(9*n+1) = a(9*n+2) = a(9*n+6) = a(9*n+7) = a(9*n+8) = A010690(n)*a(n). (End)

A060643 Number of conjugacy classes in the symmetric group S_n that have even number of elements.

Original entry on oeis.org

0, 0, 1, 3, 5, 7, 11, 20, 28, 38, 52, 73, 97, 127, 168, 229, 295, 381, 486, 623, 788, 994, 1247, 1571, 1954, 2428, 3002, 3710, 4557, 5588, 6826, 8347, 10141, 12306, 14879, 17973, 21633, 26007, 31177, 37334, 44579, 53166, 63253, 75167, 89126, 105542, 124738
Offset: 1

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Author

Avi Peretz (njk(AT)netvision.net.il), Apr 17 2001

Keywords

Comments

The total number of conjugacy classes of S_n is the partition function p(n) (sequence A000041) and the number of conjugacy classes that have odd number of elements is given in A060632 so a(n) = A000041(n) - A060632(n) for n >= 1.

Crossrefs

Formula

a(n) = A000041(n) - A060632(n).

Extensions

More terms from Sean A. Irvine, Dec 05 2022

A160745 First differences of A160744.

Original entry on oeis.org

3, 3, 6, 6, 6, 6, 12, 12, 6, 6, 12, 12, 12, 12, 24, 24, 6, 6, 12, 12, 12, 12, 24, 24, 12, 12, 24, 24, 24, 24, 48, 48, 6, 6, 12, 12, 12, 12, 24, 24, 12, 12, 24, 24, 24, 24, 48, 48, 12, 12, 24, 24, 24, 24, 48, 48, 24, 24, 48, 48, 48, 48, 96, 96, 6, 6, 12, 12, 12, 12, 24, 24, 12, 12, 24
Offset: 1

Views

Author

Omar E. Pol, May 25 2009

Keywords

Crossrefs

Formula

a(n)=3*A060632(n). [From R. J. Mathar, Feb 03 2010]

Extensions

Terms beyond a(7) from R. J. Mathar, Feb 03 2010

A309014 a(n) = Sum_{k=0..n} (-1)^(n-k) * (Stirling2(n,k) mod 2).

Original entry on oeis.org

1, 1, 0, 1, 1, 2, 1, 1, 2, 3, 1, 2, 3, 3, 2, 1, 3, 4, 1, 3, 4, 5, 3, 2, 5, 5, 2, 3, 5, 4, 3, 1, 4, 5, 1, 4, 5, 7, 4, 3, 7, 8, 3, 5, 8, 7, 5, 2, 7, 7, 2, 5, 7, 8, 5, 3, 8, 7, 3, 4, 7, 5, 4, 1, 5, 6, 1, 5, 6, 9, 5, 4, 9, 11, 4, 7, 11, 10, 7, 3, 10, 11, 3, 8, 11, 13, 8, 5, 13, 12, 5
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 06 2019

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n - k) Mod[StirlingS2[n, k], 2], {k, 0, n}], {n, 0, 90}]
    nmax = 90; CoefficientList[Series[1 + x (1 + x^3) Product[(1 + x^(2^k) + x^(2^(k + 1))), {k, 1, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k) * (stirling(n,k,2) % 2)); \\ Michel Marcus, Jul 06 2019

Formula

G.f.: 1 + x * (1 + x^3) * Product_{k>=1} (1 + x^(2^k) + x^(2^(k+1))).
a(0) = 1; a(2*k+1) = A002487(k+1); a(2*k+2) = A002487(k).

A335110 a(n) = Sum_{k=0..n} (Stirling1(n,k) mod 2) * k.

Original entry on oeis.org

0, 1, 3, 5, 6, 8, 18, 22, 12, 14, 30, 34, 36, 40, 84, 92, 24, 26, 54, 58, 60, 64, 132, 140, 72, 76, 156, 164, 168, 176, 360, 376, 48, 50, 102, 106, 108, 112, 228, 236, 120, 124, 252, 260, 264, 272, 552, 568, 144, 148, 300, 308, 312, 320, 648, 664, 336, 344, 696, 712, 720
Offset: 0

Views

Author

Ilya Gutkovskiy, May 23 2020

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Mod[StirlingS1[n, k], 2] k, {k, 0, n}], {n, 0, 60}]
    Table[Floor[(3 n + 1)/2] 2^(DigitCount[Floor[n/2], 2, 1] - 1), {n, 0, 60}]
  • PARI
    a(n) = sum(k=0, n, (stirling(n,k,1) % 2) * k); \\ Michel Marcus, May 23 2020

Formula

G.f.: x * g(x) + (3/4) * x * (1 + x) * g'(x), where g(x) = Product_{k>=0} (1 + 2 * x^(2^(k + 1))).
a(n) = floor((3*n + 1)/2) * 2^(A000120(floor(n/2)) - 1).
Previous Showing 11-16 of 16 results.