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 1-10 of 19 results. Next

A064033 Product of non-unitary divisors of binomial(n, floor(n/2)) or a(n) = 1 if all divisors are unitary. See A046098.

Original entry on oeis.org

1, 1, 1, 1, 1, 20, 1, 1, 15876, 1016255020032, 1, 728933458176, 8670998958336, 19247673071478783248355557376, 1714723915100625, 752711194884611945703392100000000, 1, 31226235883841773375939805209600000000, 1, 1357651828905889565182743230460164655087616
Offset: 1

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Author

Labos Elemer, Sep 13 2001

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_] := n^((DivisorSigma[0, n] - 2^PrimeNu[n]) / 2); Table[f[Binomial[n, Floor[n/2]]], {n, 1, 20}] (* Amiram Eldar, Jul 22 2024 *)
  • PARI
    a(n) = apply(x -> x^((numdiv(x) - 2^omega(x))/2), binomial(n, n\2)); \\ Amiram Eldar, Jul 22 2024

Formula

a(n) = A061538(A001405(n)).

Extensions

a(18)-a(20) from Amiram Eldar, Jul 22 2024

A048278 Positive numbers n such that the numbers binomial(n,k) are squarefree for all k = 0..n.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 23
Offset: 1

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Author

Keywords

Comments

It has been shown by Granville and Ramaré that the sequence is complete.
These are all the positive integers m that, when m is represented in binary, contain no composites represented in binary as substrings. - Leroy Quet, Oct 30 2008
This is a number-theoretic sequence, so it automatically assumes that n is positive. To quote Granville and Ramaré, "From Theorem 2 it is evident that there are only finitely many rows of Pascal's Triangle in which all of the entries are squarefree. In section 2 we show that this only occurs in rows 1, 2, 3, 5, 7, 11 and 23 (a result proved by Erdős long ago)." - N. J. A. Sloane, Mar 06 2014
See also comment in A249441. - Vladimir Shevelev, Oct 29 2014
This sequence is equivalent to: Positive integers n such that Fibonacci(n+1) divides n!. This comment depends on the finiteness of A019532. - Altug Alkan, Mar 31 2016

Examples

			n=11: C[11,k] = 1, 11, 55, 165, 330, 462, ... are all squarefree (or 1).
		

Crossrefs

Programs

  • Maple
    select(n -> andmap(t -> numtheory:-issqrfree(binomial(n,t)),[$1..floor(n/2)]),[$1..100]); # Robert Israel, Oct 29 2014
  • Mathematica
    Do[m = Prime[n]; k = 2; While[k < m/2 + .5 && Union[ Transpose[ FactorInteger[ Binomial[m, k]]] [[2]]] [[ -1]] < 2, k++ ]; If[k >= m/2 + .5, Print[ Prime[n]]], {n, 1, PrimePi[10^6]} ]
    Select[Range[10^3], Function[n, AllTrue[Binomial[n, Range@ n], SquareFreeQ]]] (* Michael De Vlieger, Apr 01 2016, Version 10 *)
  • PARI
    is(n)=for(k=0,n\2,if(!issquarefree(binomial(n,k)),return(0))); 1 \\ Charles R Greathouse IV, Mar 06 2014

Formula

Integers n>0 in set difference between union (A000225, A055010) and A249452. - Vladimir Shevelev, Oct 30 2014
a(n) = A018253(n+1) - 1. - Altug Alkan, Apr 26 2016

Extensions

Edited by Ralf Stephan, Aug 03 2004

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

A048277 Number of (not necessarily distinct) nonsquarefree numbers among C(n,k), k=0..n.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 1, 0, 6, 8, 5, 0, 9, 4, 3, 2, 15, 12, 17, 12, 13, 12, 11, 0, 21, 22, 19, 26, 25, 18, 25, 20, 31, 30, 27, 28, 35, 30, 25, 28, 37, 30, 29, 18, 29, 38, 27, 6, 47, 48, 49, 48, 47, 36, 51, 50, 55, 52, 49, 38, 53, 36, 23, 56, 63, 62, 61, 60, 61, 54, 59, 54, 71, 66, 57
Offset: 0

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Author

Keywords

Comments

Number of nonsquarefree numbers (A013929) on row n of Pascal's triangle (A007318). - Antti Karttunen, Nov 05 2014

Examples

			a(13) = 4 because C(13,5) = C(13,8) = 3^2*11*13 and C(13,6) = C(13,7) = 2^2*3*11*13.
If n=20, then C[ 20, k ] is divisible by a square for 13 values of k, i.e. for k = 1, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 19, so a[ 20 ] = 13.
		

Crossrefs

Programs

  • Maple
    seq(nops(remove(numtheory:-issqrfree,[seq(binomial(n,k),k=0..n)])),n=0..100); # Robert Israel, Nov 05 2014
  • Mathematica
    f[ n_ ] := (c = 0; k = 1; While[ k < n, If[ Union[ Transpose[ FactorInteger[ Binomial[ n, k ] ] ] [ [ 2 ] ] ] [ [ -1 ] ] > 1, c++ ]; k++ ]; c); Table[ f[ n ], {n, 0, 75} ]
    Table[(1 + n) - Length[Select[Binomial[n, Range[0, n]], SquareFreeQ[#] &]], {n, 0, 100}] (* Vincenzo Librandi, Nov 06 2014 *)
  • PARI
    a(n) = sum(k=0, n, !issquarefree(binomial(n, k))); \\ Michel Marcus, Mar 05 2014
    
  • PARI
    A048277(n) = sum(k=0,n\2,((0==moebius(binomial(n,k)))*(if(k<(n/2),2,1))));
    for(n=0, 8192, write("b048277.txt", n, " ", A048277(n))); \\ b-file was computed with this program. - Antti Karttunen, Nov 05 2014

Formula

From Antti Karttunen, Nov 05 2014: (Start)
a(n) = 1 + n - A048276(n).
Also, for all n >= 0:
a(n) >= A249732(n).
a(n) >= A249733(n).
(End)

Extensions

Definition corrected by Michel Marcus, Mar 05 2014

A056057 The largest square which divides n-th central binomial coefficient.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 9, 36, 1, 4, 4, 4, 9, 9, 1, 4, 1, 4, 4, 4, 1, 4, 100, 100, 900, 900, 36, 144, 9, 9, 9, 36, 25, 100, 100, 100, 9, 36, 4, 4, 4, 4, 900, 3600, 225, 900, 1764, 1764, 1764, 1764, 196, 784, 4, 4, 4, 16, 4, 16, 16, 16, 441, 441, 49, 196, 49, 196, 36, 36, 1, 4
Offset: 1

Views

Author

Labos Elemer, Jul 26 2000

Keywords

Crossrefs

Programs

  • Mathematica
    Table[First@ Select[Reverse@ Divisors@ Binomial[n, Floor[n/2]], IntegerQ@ Sqrt@ # &], {n, 72}] (* Michael De Vlieger, Feb 18 2017 *)
    a[n_] := Times @@ (First[#]^(2*Floor[Last[#]/2]) & /@ FactorInteger[Binomial[n, Floor[n/2]]]); Array[a, 100] (* Amiram Eldar, Sep 06 2020 *)

Formula

a(n) = A008833(A001405(n)).
a(A046098(n)) = 1.

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

A056058 Squarefree part of the n-th central binomial coefficient.

Original entry on oeis.org

1, 2, 3, 6, 10, 5, 35, 70, 14, 7, 462, 231, 429, 858, 715, 1430, 24310, 12155, 92378, 46189, 88179, 176358, 1352078, 676039, 52003, 104006, 22287, 44574, 2154410, 1077205, 33393355, 66786710, 129644790, 64822395, 181502706, 90751353
Offset: 1

Views

Author

Labos Elemer, Jul 26 2000

Keywords

Comments

a(3387) has 1001 decimal digits. - Michael De Vlieger, Jul 14 2017

Crossrefs

Cf. A048633 for another version.

Programs

Formula

a(n) = A007913(A001405(n)).
For squarefree central binomial coefficients (A046098), a(n)=A001405(n).

A048276 a(n) = number of squarefree numbers among C(n,k), k=0..n.

Original entry on oeis.org

1, 2, 3, 4, 3, 6, 6, 8, 3, 2, 6, 12, 4, 10, 12, 14, 2, 6, 2, 8, 8, 10, 12, 24, 4, 4, 8, 2, 4, 12, 6, 12, 2, 4, 8, 8, 2, 8, 14, 12, 4, 12, 14, 26, 16, 8, 20, 42, 2, 2, 2, 4, 6, 18, 4, 6, 2, 6, 10, 22, 8, 26, 40, 8, 2, 4, 6, 8, 8, 16, 12, 18, 2, 8, 18, 4, 6, 14, 18, 34, 2, 2, 4, 6, 4, 10, 12, 16, 4
Offset: 0

Views

Author

Keywords

Comments

The only odd numbers are at n = 0, 2, 4, and 8. So this sequence is essentially twice A238337. - T. D. Noe, Mar 07 2014

Examples

			If n=20, then C(20, k) is squarefree for k = 0,2,4,8,12,16,18,20, that is, for 8 cases of k, so a(20)=8.
		

Crossrefs

Programs

  • Maple
    A048276 := proc(n)
        local a,k ;
        a := 0 ;
        for k from 0 to n do
            if issqrfree(binomial(n,k)) then
                a := a+1 ;
            end if;
        end do:
        a ;
    end proc:
    seq(A048276(n),n=0..40) ; # R. J. Mathar, Jan 18 2018
  • Mathematica
    Table[Length[Select[Binomial[n, Range[0, n]], SquareFreeQ[#] &]], {n, 0, 100}]
  • PARI
    a(n) = sum(k=0, n, issquarefree(binomial(n, k))); \\ Michel Marcus, Dec 19 2013

Formula

a(n) = n+1-A048277(n). - R. J. Mathar, Jan 18 2018

A056651 Numbers k such that binomial(k,floor(k/2)) has no non-unitary square divisors: all of their square divisors are unitary ones.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 31, 32, 35, 36, 37, 39, 40, 41, 43, 47, 48, 49, 55, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 75, 79, 80, 95, 96, 97, 111, 129, 130, 131, 132, 133, 143, 144, 151, 161, 163, 167, 191, 192, 193
Offset: 1

Views

Author

Labos Elemer, Aug 09 2000

Keywords

Comments

This property is weaker than "squarefreedom", but shows how central binomial coefficients are "poor of squares".
Numbers k such that binomial(k,floor(k/2)) is cubefree (A004709). - Amiram Eldar, Jul 22 2024

Examples

			223 is a term because x = binomial(223,111) has 35 prime divisors. 33 arises at power 1. Only 2 and 13 has powers 2 > 1. So square divisors of x are {1, 4, 169, 676} ={s}. All of them are also unitary divisors since GCD(s,x/s) = 1 holds for them.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 11000], AllTrue[FactorInteger[Binomial[#, Floor[#/2]]][[;;, 2]], #1 <= 2 &] &] (* Amiram Eldar, Jul 22 2024 *)
  • PARI
    is(n) = if(n <= 1, 1, vecmax(factor(binomial(n, floor(n/2)))[, 2]) < 3); \\ Amiram Eldar, Jul 22 2024

A081391 Numbers k such that the central binomial coefficient C(2*k,k) has only one prime divisor whose exponent equals one.

Original entry on oeis.org

3, 6, 7, 8, 9, 10, 11, 12, 16, 21, 22, 28, 29, 30, 31, 36, 37, 54, 55, 57, 58, 110, 171, 784, 786
Offset: 1

Views

Author

Labos Elemer, Mar 27 2003

Keywords

Comments

Numbers k such that C(2*k,k) has one non-unitary prime divisor.
Numbers k for which A081387(k) = 1.
No more terms through 10^6; conjecture: no terms after 786. - Jon E. Schoenfield, Jul 29 2017

Examples

			For k = 786, C(1572,786) = 2*2*2*2*m, where m is a squarefree product of 169 primes.
		

Crossrefs

Programs

Showing 1-10 of 19 results. Next