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 1-9 of 9 results.

A056056 Square root of largest square dividing n-th central binomial coefficient.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 2, 2, 2, 3, 3, 1, 2, 1, 2, 2, 2, 1, 2, 10, 10, 30, 30, 6, 12, 3, 3, 3, 6, 5, 10, 10, 10, 3, 6, 2, 2, 2, 2, 30, 60, 15, 30, 42, 42, 42, 42, 14, 28, 2, 2, 2, 4, 2, 4, 4, 4, 21, 21, 7, 14, 7, 14, 6, 6, 1, 2, 2, 2, 10, 10, 70, 140, 7, 14, 126, 126, 6, 6, 30, 60
Offset: 1

Views

Author

Labos Elemer, Jul 26 2000

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sqrt@ Max@ Select[Divisors@ Binomial[n, Floor[n/2]], IntegerQ@ Sqrt@ # &], {n, 0, 86}] (* Michael De Vlieger, Jul 04 2016 *)
    a[n_] := Times @@ (First[#]^Floor[Last[#]/2] & /@ FactorInteger[Binomial[n, Floor[n/2]]]); Array[a, 100] (* Amiram Eldar, Sep 06 2020 *)
  • PARI
    a(n) = b = binomial(n, n\2); sqrtint(b/core(b)); \\ Michel Marcus, Dec 10 2013

Formula

a(n) = A000188(A001405(n)).

A056059 GCD of largest square and squarefree part of central binomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 6, 2, 1, 1, 1, 3, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 6, 3, 1, 1, 1, 2, 3, 6, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 3, 6, 6, 3, 1, 2, 2, 1, 2, 1, 3, 6, 1, 1, 1, 2, 1, 2
Offset: 1

Views

Author

Labos Elemer, Jul 26 2000

Keywords

Examples

			n=14, binomial(14,7) = 3432 = 8*3*11*13. The largest square divisor is 4, squarefree part is 858. So a(14) = gcd(4,858) = 2.
		

Crossrefs

Programs

  • Mathematica
    Table[GCD[First@ Select[Reverse@ Divisors@ #, IntegerQ@ Sqrt@ # &], Times @@ Power @@@ Map[{#1, Mod[#2, 2]} & @@ # &, FactorInteger@ #]] &@ Binomial[n, Floor[n/2]], {n, 80}] (* Michael De Vlieger, Feb 18 2017, after Zak Seidov at A007913 *)
  • PARI
    A001405(n) = binomial(n, n\2);
    A055229(n) = { my(c=core(n)); gcd(c, n/c); } \\ Charles R Greathouse IV, Nov 20 2012
    A056059(n) = A055229(A001405(n)); \\ Antti Karttunen, Jul 20 2017
    
  • Python
    from sympy import binomial, gcd
    from sympy.ntheory.factor_ import core
    def a001405(n): return binomial(n, n//2)
    def a055229(n):
        c = core(n)
        return gcd(c, n//c)
    def a(n): return a055229(a001405(n))
    print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Jul 20 2017

Formula

a(n) = A055229(A001405(n)), where A055229(n) = gcd(A008833(n), A007913(n)).

Extensions

Formula clarified by Antti Karttunen, Jul 20 2017

A056061 Number of square divisors of central binomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 4, 4, 8, 8, 4, 6, 2, 2, 2, 4, 2, 4, 4, 4, 2, 4, 2, 2, 2, 2, 8, 12, 4, 8, 8, 8, 8, 8, 4, 6, 2, 2, 2, 3, 2, 3, 3, 3, 4, 4, 2, 4, 2, 4, 4, 4, 1, 2, 2, 2, 4, 4, 8, 12, 2, 4, 12, 12, 4, 4, 8, 12, 12, 12, 4, 6, 8, 12, 12, 12, 8, 16, 8, 8, 6
Offset: 1

Views

Author

Labos Elemer Jul 26 2000

Keywords

Examples

			n=27: binomial(27,13) = 20058300, its largest square-divisor is 900=30^2 so a(27) = tau(30) = 8.
		

Crossrefs

Programs

  • Mathematica
    Table[Count[Divisors@ Binomial[n, Floor[n/2]], d_ /; IntegerQ@ Sqrt@ d], {n, 0, 84}] (* Michael De Vlieger, Feb 18 2017 *)
  • PARI
    a(n) = sumdiv(binomial(n, n\2), d, issquare(d)); \\ Michel Marcus, Feb 19 2017

Formula

A056060 The powerfree part of the central binomial coefficients.

Original entry on oeis.org

1, 2, 3, 6, 10, 5, 35, 70, 14, 7, 462, 231, 429, 429, 715, 1430, 24310, 12155, 92378, 46189, 88179, 88179, 1352078, 676039, 52003, 52003, 7429, 7429, 1077205, 1077205, 33393355, 66786710, 43214930, 21607465, 181502706, 90751353, 176726319, 176726319, 7658140490
Offset: 1

Views

Author

Labos Elemer, Jul 26 2000

Keywords

Examples

			n=14, binomial(14,7) = 3432 = 8*3*11*13. The largest square divisor is 4, and the squarefree part is 858. So GCD(4,858) = 2 and a(14) = 858/2 = 429.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Denominator[(b = Binomial[n, Floor[n/2]])/(Times @@ First /@ FactorInteger[b])^2]; Array[a, 36] (* Amiram Eldar, Sep 05 2020 *)

Formula

a(n) = A055231(A001405(n)).

Extensions

New name and more terms from Amiram Eldar, Sep 05 2020

A056175 Number of nonunitary prime divisors of the central binomial coefficient C(n, floor(n/2)) (A001405).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 2, 3, 3, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 3, 3, 2, 3, 3, 3, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 0, 1, 1, 1, 2, 2, 3, 3, 1, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 3, 3, 3, 3, 4, 3, 3, 2, 2, 2, 2, 2, 2, 2
Offset: 1

Views

Author

Labos Elemer, Jul 27 2000

Keywords

Comments

Number of prime divisors of the largest square dividing A001405(n). (A prime divisor is nonunitary iff its exponent exceeds 1.)

Examples

			For n=10, binomial(10, 5) = 252 = 2*2*3*3*7 has 3 prime divisors of which only one, p=7, is unitary, while 2 and 3 are not. So a(10)=2.
For n=256, binomial(256, 128) also has only 2 prime divisors (3 and 13) whose exponents exceed 1 (4 and 2, respectively), thus a(256)=2.
		

Crossrefs

Programs

  • Mathematica
    Table[Count[FactorInteger[Binomial[n, Floor[n/2]]][[All, -1]], e_ /; e > 1], {n, 105}] (* Michael De Vlieger, Mar 05 2017 *)
  • PARI
    a(n)=omega(core(binomial(n, n\2), 1)[2]) \\ Charles R Greathouse IV, Mar 09 2017

Formula

a(n) = A001221(A000188(A001405(n))).
a(n) = A001221(A056057(n)).

Extensions

Edited by Jon E. Schoenfield, Mar 05 2017

A056647 a(n) = A056623(A001405(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 9, 36, 1, 4, 4, 1, 9, 9, 1, 4, 1, 4, 4, 1, 1, 4, 100, 25, 100, 25, 9, 144, 9, 9, 1, 4, 25, 100, 100, 25, 9, 36, 4, 1, 4, 1, 25, 400, 225, 900, 1764, 441, 196, 49, 49, 784, 4, 1, 1, 16, 1, 16, 16, 4, 441, 441, 49, 196, 49, 196, 36, 9, 1, 4, 4, 1, 100, 25, 1225
Offset: 1

Views

Author

Labos Elemer, Aug 09 2000

Keywords

Comments

Previous name "Largest unitary square divisor of central binomial coefficient" was incorrect. See A376553 for the correct sequence with this name. - Amiram Eldar, Sep 28 2024

Examples

			a(28) = A056623(binomial(28,14)) = A056623(40116600) = 25.
		

Crossrefs

Formula

a(n) = A008833(A001405(n))/A055229(A001405(n))^2 = A056057(n)/A056059(n)^2.

Extensions

Incorrect name replaced with a formula by Amiram Eldar, Sep 28 2024

A056670 Largest non-unitary prime factor of A001405(n) = binomial(n, floor(n/2)), or 1 if no such prime exists.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 2, 2, 2, 3, 3, 1, 2, 1, 2, 2, 2, 1, 2, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 3, 3, 2, 2, 2, 2, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 2, 2, 2, 2, 2, 2, 2, 2, 7, 7, 7, 7, 7, 7, 3, 3, 1, 2, 2, 2, 5, 5, 7, 7, 7, 7, 7, 7, 3, 3, 5, 5, 5, 5, 3, 3, 7, 7, 7, 7, 7, 7, 5, 5, 3, 3, 3, 3, 3, 3, 3
Offset: 1

Views

Author

Labos Elemer, Aug 10 2000

Keywords

Comments

The largest prime divisor of A056057(n), the largest square divisor of binomial(n, floor(n/2)), or 1 if no such prime exists.

Examples

			For n = 28: binomial(28,14) = 2*2*2*3*3*3*5*5*17*19*23, so a(28) = 5.
For n = 342: binomial(342,171) = 32*F, where F is squarefree, so a(341) = 2.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{f = Select[FactorInteger[Binomial[n, Floor[n/2]]], Last[#] > 1 &]}, If[f == {}, 1, f[[-1, 1]]]]; Array[a, 100] (* Amiram Eldar, Oct 05 2024 *)

Formula

a(n) = A006530(A056057(n)).

A056648 a(n) = A034444(A056647(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 4, 2, 4, 2, 2, 4, 2, 2, 1, 2, 2, 4, 4, 2, 2, 4, 2, 1, 2, 1, 2, 4, 4, 8, 8, 4, 4, 2, 2, 4, 2, 1, 1, 2, 1, 2, 2, 2, 4, 4, 2, 4, 2, 4, 4, 2, 1, 2, 2, 1, 4, 2, 4, 8, 2, 4, 8, 4, 2, 1, 2, 4, 8, 4, 2, 4, 4, 8, 4, 4, 8, 16, 8, 4, 4, 2, 1, 2, 2, 1
Offset: 1

Views

Author

Labos Elemer, Aug 09 2000

Keywords

Comments

Previous name, "Number of unitary square divisors of central binomial coefficient", was incorrect. See A376555 for the correct sequence with this name. - Amiram Eldar, Sep 28 2024

Examples

			a(28) = A034444(A056647(28)) = A034444(25) = 2.
		

Crossrefs

Programs

  • Mathematica
    A008833[n_] := First[Select[Reverse[Divisors[n]], IntegerQ[Sqrt[#]] &, 1]]; A055229[n_] := With[{sf = Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n])}, GCD[sf, n/sf]]; Table[2^(PrimeNu[ Sqrt[A008833[Binomial[n, Floor[n/2]]]]/A055229[Binomial[n, Floor[n/2]]]]), {n, 1, 25}] (* G. C. Greubel, May 20 2017 *)

Formula

a(n) = 2^r, where r = A001221(A000188(A001405(n))/A055229(A001405(n))).

Extensions

Incorrect name replaced with a formula by Amiram Eldar, Sep 28 2024

A056649 a(n) = A056061(n) - A034444(A056647(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 4, 6, 2, 2, 0, 0, 1, 2, 0, 0, 0, 2, 0, 0, 0, 1, 0, 1, 6, 8, 0, 0, 0, 4, 4, 6, 2, 2, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 4, 4, 0, 0, 4, 8, 2, 3, 6, 8, 4, 8, 2, 2, 4, 4, 8, 8, 0, 0, 0, 4, 2, 4, 3, 4, 2, 3, 4
Offset: 1

Views

Author

Labos Elemer, Aug 09 2000

Keywords

Comments

Previous name, "Number of non-unitary square divisors of central binomial coefficient", was incorrect. See A376556 for the correct sequence with this name. - Amiram Eldar, Sep 28 2024

Examples

			a(28) = A056061(28) - A034444(A056647(28)) = A056061(28) - A034444(25) = 8 - 2 = 6.
		

Crossrefs

Programs

  • Mathematica
    A056061[n_] := Count[Divisors@Binomial[n, Floor[n/2]], d_ /; IntegerQ@Sqrt@d]; A008833[n_] := First[Select[Reverse[Divisors[n]], IntegerQ[Sqrt[#]] &, 1]]; A055229[n_] := With[{sf = Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n])}, GCD[sf, n/sf]];
    Table[A056061[n] - 2^(PrimeNu[Sqrt[A008833[Binomial[n, Floor[n/2]]]]/ A055229[Binomial[n, Floor[n/2]]]]), {n, 1, 15}] (* G. C. Greubel, May 20 2017 *)

Formula

a(n) = A056061(n) - 2^r, where r = A001221(A000188(A001405(n))/A055229(A001405(n))).

Extensions

Incorrect name replaced with a formula by Amiram Eldar, Sep 28 2024
Showing 1-9 of 9 results.