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 1-4 of 4 results.

A094252 a(n) = partition(n) mod prime(n).

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 15, 3, 7, 13, 25, 3, 19, 6, 35, 19, 2, 19, 21, 59, 62, 54, 10, 62, 18, 12, 23, 80, 96, 67, 111, 96, 5, 78, 132, 8, 128, 98, 123, 143, 12, 141, 40, 98, 90, 88, 53, 93, 97, 187, 186, 47, 2, 117, 241, 34, 27, 51, 266, 108, 259, 115, 278, 30, 281, 227, 244, 141
Offset: 1

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Author

Gary W. Adamson, Apr 25 2004

Keywords

Comments

Indices n such that a(n)=0 (i.e., prime(n) divides partition(n)) are listed in A245662. - Max Alekseyev, Jul 27 2014

Examples

			a(10) = 13: partition(10) = 42, prime(10) = 29. 42 mod 29 = 13.

Links

Crossrefs

Cf. A000041, A093952, A245525.

Programs

  • Mathematica
    Table[ Mod[ PartitionsP[n], Prime[n]], {n, 70}] (* Robert G. Wilson v, Apr 28 2004 *)

Extensions

Edited by Robert G. Wilson v, Apr 28 2004

A245526 Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that L(2*n) == r (mod prime(n)), where L(k) denotes the Lucas number A000032(k).

Original entry on oeis.org

1, 1, -2, -2, 2, -3, -7, 3, 5, -11, -15, 8, -18, -14, 3, -12, 19, -18, 25, 14, 5, 21, 11, 7, -22, 3, 43, -40, -7, -53, 54, 23, 11, -12, -7, 41, 6, -13, -66, 71, -32, 18, 94, -20, -79, 7, -88, 12, 11, -73, 3, 29, -120, 50, 10, -60, -63, 34, 94, 47, -113, 131, -18, 128, 60, 57, 79, 22, -45, -68, 100, 100, 131, -171, 56, -166, 11, -153, -174, 10
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 25 2014

Keywords

Comments

Conjecture: a(n) is always nonzero, i.e., prime(n) never divides the Lucas number L(2*n).
We have verified this for all n = 1, ..., 2*10^6.
On Jul 26 2014, Bjorn Poonen (from MIT) found a counterexample with n = 14268177. - Zhi-Wei Sun, Jul 26 2014

Examples

			a(10) = -11 since L(2*10) = 15127 == -11 (mod prime(10)=29).

Links

Crossrefs

Cf. A000032, A000040, A245525.

Programs

  • Mathematica
    rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
a[n_]:=rMod[LucasL[2n],Prime[n]]
Table[a[n],{n,1,80}]

A245662 Positive integers n such that n-th prime A000040(n) divides n-th partition number A000041(n).

Original entry on oeis.org

1119414, 52734279
Offset: 1

Views

Author

Max Alekseyev, Jul 28 2014

Keywords

Comments

The indices of zeros in A094252 and A245525.
There are no other terms below 10^8.

Crossrefs

Cf. A000040, A000041, A094252, A245525

Programs

A245595 Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that S(n) == r (mod prime(n)), where S(n) is the large Schroeder number A006318(n).

Original entry on oeis.org

0, 0, 2, -1, -2, -1, 7, -5, -5, 11, 10, -11, 11, 12, 2, 17, -2, 19, -15, -26, 33, 17, -22, -11, 18, 8, 18, -27, 17, 51, -37, -34, 28, -4, 66, -37, -69, -58, 45, -81, -20, -86, -19, 17, -12, -30, 35, -32, 5, -11, -8, -45, 12, -111, -28, -71, 76, 59, 102, -25
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 27 2014

Keywords

Comments

Conjecture: (i) For any integer n > 2, the term a(n) is nonzero, i.e., prime(n) does not divide the large Schroeder number S(n).
(ii) For any integer n > 2, prime(n) does not divide the Bell number B(2*n) = A000110(2*n).
We have verified parts (i) and (ii) for n up to 440000 and 66000 respectively.
Conjecture (i) fails for the first time for n=20239789. In particular, a(20239789)=0. - Max Alekseyev, Oct 05 2015

Examples

			a(5) = -2 since S(5) = 394 == -2 (mod prime(5)=11).

Links

Crossrefs

Cf. A000040, A000110, A006318, A245525.

Programs

  • Mathematica
    rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
S[n_]:=Sum[Binomial[n+k,2k]*Binomial[2k,k]/(k+1),{k,0,n}]
a[n_]:=rMod[S[n],Prime[n]]
Table[a[n],{n,1,60}]
Showing 1-4 of 4 results.

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