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.

A065347 Positions of zeros in A065344, i.e., binomial(2n,n) mod ((n+1)*(n+2)) = 0.

Original entry on oeis.org

3, 5, 8, 9, 11, 12, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 29, 32, 33, 35, 36, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90
Offset: 1

Views

Author

Labos Elemer, Oct 30 2001

Keywords

Crossrefs

Programs

  • Mathematica
    ris = {}; Do[If[Mod[Binomial[2 n, n], (n + 1) (n + 2)] == 0, AppendTo[ris, n]], {n, 100}]; ris (* Bruno Berselli, Jan 06 2014 *)
  • PARI
    isok(k) = { binomial(2*k, k) % ((k + 1)*(k + 2)) == 0 } \\ Harry J. Smith, Oct 17 2009

A065349 Positions of zeros in A065346.

Original entry on oeis.org

19, 41, 42, 43, 49, 50, 53, 54, 55, 58, 67, 71, 74, 75, 95, 97, 98, 99, 102, 103, 123, 135, 138, 139, 145, 149, 153, 154, 155, 159, 163, 165, 166, 167, 168, 169, 170, 173, 174, 175, 178, 179, 180, 183, 184, 185, 191, 193, 194, 195, 197, 198, 199, 200, 201, 202
Offset: 1

Views

Author

Labos Elemer, Oct 30 2001

Keywords

Crossrefs

Programs

  • Mathematica
    ris = {}; Do[If[Mod[Binomial[2 n, n], (n + 1) (n + 2) (n + 3) (n + 4)] == 0, AppendTo[ris, n]], {n, 250}]; ris (* Bruno Berselli, Jan 06 2014 *)
  • PARI
    isok(k) = { binomial(2*k, k) % ((k + 1)*(k + 2)*(k + 3)*(k + 4)) == 0 } \\ Harry J. Smith, Oct 17 2009

A002503 Numbers k such that binomial(2*k,k) is divisible by (k+1)^2.

Original entry on oeis.org

5, 14, 27, 41, 44, 65, 76, 90, 109, 125, 139, 152, 155, 169, 186, 189, 203, 208, 209, 219, 227, 230, 237, 265, 275, 298, 307, 311, 314, 321, 324, 329, 344, 377, 413, 419, 428, 434, 439, 441, 449, 458, 459, 467, 475
Offset: 1

Views

Author

Keywords

Comments

From Amiram Eldar, Mar 28 2021: (Start)
Balakram (1929) proved that:
1) This sequence is infinite.
2) If m is an even perfect number (A000396) then m-1 is a term.
3) If m = p*q - 1, where p and q are primes, and (3/2)*p < q < 2*p, then m is a term.
4) m is a term if and only if Sum_{k>=1} floor(2*m/p^k) >= 2 * Sum_{k>=1} floor((m+1)/p^k), for all primes p. (End)

References

  • Hoon Balakram, On the values of n which make (2n)!/(n+1)!(n+1)! an integer, J. Indian Math. Soc., Vol. 18 (1929), pp. 97-100.
  • Thomas Koshy, Catalan numbers with applications, Oxford University Press, 2008, pp. 69-70.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Positions of zeros in A065350.
Equals A067348(n+2)/2 - 1.

Programs

  • Haskell
    import Data.List (elemIndices)
    a002503 n = a002503_list !! (n-1)
    a002503_list = map (+ 1) $ elemIndices 0 a065350_list
    -- Reinhard Zumkeller, Sep 16 2014
    
  • Mathematica
    Select[Range[500],Divisible[Binomial[2#,#],(#+1)^2]&] (* Harvey P. Dale, May 21 2012 *)
  • PARI
    isok(n) = binomial(2*n, n) % (n+1)^2 == 0; \\ Michel Marcus, Jan 11 2016

Formula

A065350(a(n)) = 0. - Reinhard Zumkeller, Sep 16 2014

Extensions

Balakram reference corrected by T. D. Noe, Jan 16 2007

A065345 a(n) = binomial(2*n,n) mod ((n+1)*(n+2)*(n+3)).

Original entry on oeis.org

2, 6, 20, 70, 252, 420, 552, 0, 1100, 1144, 0, 1456, 1400, 2040, 2448, 0, 3420, 0, 0, 0, 6072, 5520, 0, 5850, 13104, 0, 12992, 17980, 22320, 27280, 5984, 7854, 7140, 15540, 0, 36556, 13832, 0, 45920, 24682, 0, 0, 0, 0, 51888, 0, 23520, 0, 0, 0, 0, 94446, 0, 0, 0
Offset: 1

Views

Author

Labos Elemer, Oct 30 2001

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[2 n, n], (n + 1) (n + 2) (n + 3)], {n, 60}] (* Harvey P. Dale, Feb 21 2012 *)
  • PARI
    a(n) = { binomial(2*n, n) % ((n + 1)*(n + 2)*(n + 3)) } \\ Harry J. Smith, Oct 17 2009

A065346 a(n) = binomial(2*n, n) mod ((n+1)*(n+2)*(n+3)*(n+4)).

Original entry on oeis.org

2, 6, 20, 70, 252, 924, 3432, 990, 14300, 16588, 17472, 39676, 4760, 18360, 46512, 29070, 30780, 87780, 0, 191268, 273240, 322920, 140400, 58500, 190008, 350784, 402752, 611320, 81840, 649264, 41888, 164934, 264180, 295260, 1316016, 694564
Offset: 1

Views

Author

Labos Elemer, Oct 30 2001

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[2 n, n], (n + 1) (n + 2) ((n + 3) (n + 4))], {n, 100}] (* Bruno Berselli, Jan 06 2014 *)
  • PARI
    a(n) = { binomial(2*n, n) % ((n + 1)*(n + 2)*(n + 3)*(n + 4)) } \\ Harry J. Smith, Oct 17 2009

A065348 Positions of zeros in A065345.

Original entry on oeis.org

8, 11, 16, 18, 19, 20, 23, 26, 35, 38, 41, 42, 43, 44, 46, 48, 49, 50, 51, 53, 54, 55, 56, 58, 59, 60, 64, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 80, 83, 86, 89, 92, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 107, 110, 113, 116, 119, 123, 124, 128, 130, 131, 134
Offset: 1

Views

Author

Labos Elemer, Oct 30 2001

Keywords

Crossrefs

Programs

  • Mathematica
    ris = {}; Do[If[Mod[Binomial[2 n, n], (n + 1) (n + 2) (n + 3)] == 0,
    AppendTo[ris, n]], {n, 150}]; ris (* Bruno Berselli, Jan 06 2014 *)
  • PARI
    isok(k) = { binomial(2*k, k) % ((k + 1)*(k + 2)*(k + 3)) == 0 } \\ Harry J. Smith, Oct 17 2009

A065350 a(n) = binomial(2*n, n) mod (n+1)^2.

Original entry on oeis.org

2, 6, 4, 20, 0, 42, 40, 72, 20, 110, 120, 156, 56, 0, 208, 272, 108, 342, 200, 378, 176, 506, 432, 600, 260, 459, 0, 812, 840, 930, 928, 396, 476, 490, 360, 1332, 608, 1131, 1200, 1640, 0, 1806, 880, 0, 920, 2162, 864, 2352, 1100, 1224, 208, 2756, 1296, 2145
Offset: 1

Views

Author

Labos Elemer, Oct 30 2001

Keywords

Comments

a(A002503(n)) = 0. - Reinhard Zumkeller, Sep 16 2014

Crossrefs

Programs

  • Haskell
    a065350 n = a065350_list !! (n-1)
    a065350_list = zipWith mod (tail a000984_list) (drop 2 a000290_list)
    -- Reinhard Zumkeller, Sep 16 2014
  • Mathematica
    Table[Mod[Binomial[2 n, n], (n + 1)^2], {n, 100}] (* Bruno Berselli, Jan 06 2014 *)
  • PARI
    a(n) = { binomial(2*n, n) % (n + 1)^2 } \\ Harry J. Smith, Oct 17 2009
    

A065352 Smallest m such that C(2m,m) is divisible by (m+n)!/m!.

Original entry on oeis.org

1, 3, 8, 19, 42, 153, 216, 375, 950, 3565, 4068, 12273, 12274, 31729, 122352, 131023, 458222, 522221, 1046508, 3145451, 6291178, 12320745, 16769000, 56623079, 113246182, 267780069, 469745636, 671088611, 1879015394, 2146959329, 6442418144, 16642932703, 16911433694, 60129279965, 206091288540
Offset: 1

Views

Author

Labos Elemer, Oct 31 2001

Keywords

Comments

For n=1 see Catalan numbers A000108.
Heuristically one can observe that a(n) + n + 1 has a 'high' valuation of 2. For n = 17..25 we have 2^8|(a(n) + n + 1). - David A. Corneth, Mar 28 2021
Since (m+n)!/m! = C(m+n,m) * n!, Kummer's theorem implies that A000120(a(n)) >= A007814(n!) = A011371(n) = n - A000120(n), and a(n) >= 2^(n-1). - Max Alekseyev, Sep 24 2024

Examples

			n=4: a(4)=19 means that C(38,19)=35345263800 is divisible by (19+1)(19+2)(19+3)(19+4)=23!/19!=20*21*22*23=215520; the quotient is 166315. Smaller (<19) central binomial coefficients are not divisible by such a product of 4 successive terms; the corresponding quotients for n = 1, 2, 3, 4, 5,... are 1, 1, 13, 166315, 9120910752273999,...
		

Crossrefs

Programs

  • Mathematica
    Do[m = 1; While[Not[Divisible[Binomial[2*m,m],(m+n)!/m!]], m++]; Print[m], {n, 1, 16}] (* Vaclav Kotesovec, Sep 05 2019 *)
  • PARI
    \\ See Corneth link

Formula

C(2m, m)=A*((m+1)(m+2)...(m+n-1)(m+n)); a(n) is the smallest such m belonging to n: a(n)=Min(m; Mod(A000984(m), (m+n)!/m!)=0)

Extensions

More terms from Naohiro Nomoto, Apr 21 2002
a(16)-a(17) from Vaclav Kotesovec, Sep 06 2019
a(18)-a(25) from David A. Corneth, Mar 28 2021
a(26)-a(31) from David A. Corneth confirmed and terms a(32) onward added by Max Alekseyev, Sep 24 2024
Showing 1-8 of 8 results.