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 26 results. Next

A046063 Numbers k such that the k-th partition number A000041(k) is prime.

Original entry on oeis.org

2, 3, 4, 5, 6, 13, 36, 77, 132, 157, 168, 186, 188, 212, 216, 302, 366, 417, 440, 491, 498, 525, 546, 658, 735, 753, 825, 841, 863, 1085, 1086, 1296, 1477, 1578, 1586, 1621, 1793, 2051, 2136, 2493, 2502, 2508, 2568, 2633, 2727, 2732, 2871, 2912, 3027, 3098, 3168, 3342, 3542, 3641, 4118
Offset: 1

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Author

Keywords

Comments

The corresponding primes are given in A049575. - Joerg Arndt, May 09 2013

Crossrefs

Programs

  • Mathematica
    Select[ Range@3341, PrimeQ@ PartitionsP@# &] (* Robert G. Wilson v *)
  • PARI
    for(n=0,10^5,my(p=numbpart(n));if(isprime(p),print1(n,", "))); \\ Joerg Arndt, May 09 2013
    
  • Python
    from sympy import isprime, npartitions
    print([n for n in range(1, 5001) if isprime(npartitions(n))]) # Indranil Ghosh, Apr 10 2017

Extensions

b-file extended by Max Alekseyev, Jul 07 2009, Jun 14 2011, Jan 08 2012, May 19 2014

A234470 Number of ways to write n = k + m with k > 0 and m > 2 such that p(k + phi(m)/2) is prime, where p(.) is the partition function (A000041) and phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 4, 5, 5, 4, 4, 4, 2, 2, 3, 5, 4, 2, 4, 2, 3, 2, 3, 2, 3, 1, 0, 3, 1, 1, 2, 1, 2, 0, 1, 2, 1, 1, 4, 2, 1, 4, 2, 1, 2, 3, 3, 3, 1, 0, 4, 2, 4, 1, 1, 2, 2, 3, 2, 2, 0, 2, 2, 1, 2, 2, 1, 1, 2, 2, 4, 2, 1, 0, 1, 3, 1, 0, 2, 4, 3, 1, 6, 2, 2, 1, 2, 4, 3, 1, 2, 6, 2, 3, 2, 2, 2, 2, 3, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 26 2013

Keywords

Comments

Conjecture: a(n) > 0 if n > 3 is not among 27, 34, 50, 61, 74, 78, 115, 120, 123, 127.
This implies that there are infinitely many primes in the range of the partition function p(n).

Examples

			a(26) = 1 since 26 = 2 + 24 with p(2 + phi(24)/2) = p(6) = 11 prime.
a(54) = 1 since 54 = 27 + 27 with p(27 + phi(27)/2) = p(36) = 17977 prime.
a(73) = 1 since 73 = 1 + 72 with p(1 + phi(72)/2) = p(36) = 17977 prime.
a(110) = 1 since 110 = 65 + 45 with p(65 + phi(45)/2) = p(77) = 10619863 prime.
a(150) = 1 since 150 = 123 + 27 with p(123 + phi(27)/2) = p(132) = 6620830889 prime.
a(170) = 1 since 170 = 167 + 3 with p(167 + phi(3)/2) = p(168) = 228204732751 prime.
		

Crossrefs

Programs

  • Mathematica
    f[n_,k_]:=PartitionsP[k+EulerPhi[n-k]/2]
    a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-3}]
    Table[a[n],{n,1,100}]

A234569 Primes p with P(p-1) also prime, where P(.) is the partition function (A000041).

Original entry on oeis.org

3, 5, 7, 37, 367, 499, 547, 659, 1087, 1297, 1579, 2137, 2503, 3169, 3343, 4457, 4663, 5003, 7459, 9293, 16249, 23203, 34667, 39971, 41381, 56383, 61751, 62987, 72661, 77213, 79697, 98893, 101771, 127081, 136193, 188843, 193811, 259627, 267187, 282913, 315467, 320563, 345923, 354833, 459029, 482837, 496477, 548039, 641419, 647189
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 28 2013

Keywords

Comments

By the conjecture in A234567, this sequence should have infinitely many terms. It seems that a(n+1) < a(n) + a(n-1) for all n > 5.
The b-file lists all terms not exceeding the 500000th prime 7368787. Note that P(a(113)-1) is a prime having 2999 decimal digits.
See also A234572 for primes of the form P(p-1) with p prime.

Examples

			a(1) = 3 since P(2-1) = 1 is not prime, but P(3-1) = 2 is prime.
a(2) = 5 since P(5-1) = 5 is prime.
a(3) = 7 since P(7-1) = 11 is prime.
		

Crossrefs

Programs

  • Mathematica
    n=0;Do[If[PrimeQ[PartitionsP[Prime[k]-1]],n=n+1;Print[n," ",Prime[k]]],{k,1,10^6}]

A236103 Number of distinct partition numbers dividing n.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 1, 3, 4, 2, 1, 3, 1, 3, 3, 4, 1, 3, 2, 2, 2, 3, 1, 6, 1, 2, 3, 2, 3, 3, 1, 2, 2, 3, 1, 5, 1, 4, 4, 2, 1, 3, 2, 3, 2, 2, 1, 3, 3, 4, 2, 2, 1, 6, 1, 2, 3, 2, 2, 5, 1, 2, 2, 4, 1, 3, 1, 2, 4, 2, 4, 3, 1, 3, 2, 2, 1, 5, 2, 2, 2, 4, 1, 6
Offset: 1

Views

Author

Omar E. Pol, Jan 21 2014

Keywords

Examples

			For n = 20 the divisors of 20 are 1, 2, 4, 5, 10, 20 and three of them are also partition numbers: 1, 2, 5, so a(20) = 3.
For n = 42 the divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42 and five of them are also partition numbers: 1, 2, 3, 7, 42, so a(42) = 5.
		

Crossrefs

Programs

  • Mathematica
    p = {1}; Table[If[n >= Last@p, AppendTo[p, PartitionsP[1 + Length@p]]]; Length@Select[p, Mod[n, #] == 0 &], {n, 90}] (* Giovanni Resta, Jan 22 2014 *)

Formula

From Amiram Eldar, Jan 01 2024: (Start)
a(n) = Sum_{d|n} A167392(d).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A078506 = 2.510597... . (End)

A236108 Nonprimes whose proper divisors are partition numbers.

Original entry on oeis.org

4, 6, 9, 10, 14, 15, 21, 22, 25, 33, 35, 49, 55, 77, 121, 202, 303, 505, 707, 1111, 10201, 35954, 53931, 89885, 125839, 197747, 1815677, 21239726, 31859589, 53099315, 74339041, 116818493, 323172529, 1072606163, 13241661778, 19862492667, 33104154445, 46345816223, 72829139779
Offset: 1

Views

Author

Omar E. Pol, Jan 22 2014

Keywords

Comments

Known terms are squares of A049575 or products of 2 distinct terms of A049575. - Michel Marcus, Jan 25 2023
This conjecture holds for terms <= 10^16. - David A. Corneth, Jan 25 2023

Examples

			10 is in the sequence because 10 is a nonprime number and the proper divisors of 10 are 1, 2, 5, which are also partition numbers.
		

Crossrefs

Programs

  • Maple
    isA000041 := proc(n)
        local k,P;
        for k from 1 do
            P := combinat[numbpart](k) ;
            if P > n then
                return false;
            elif P = n then
                return true ;
            end if;
        end do:
    end proc:
    isA236108 := proc(n)
        local pdvs,d ;
        if n =1 or isprime(n) then
            return false;
        end if;
        pdvs := numtheory[divisors](n) minus {n} ;
        for d in pdvs do
            if not isA000041(d) then
                return false;
            end if;
        end do:
        return true;
    end proc:
    for n from 1 to 300000 do
        if isA236108(n) then
            printf("%d,",n) ;
        end if;
    end do: # R. J. Mathar, Jan 29 2014
  • Mathematica
    partitionNumbers = Table[PartitionsP[n], {n, 1, 1000}];
    Select[Range[2, 10000],
     If[! PrimeQ[#],
    ContainsOnly[Divisors[#][[2 ;; -2]], partitionNumbers]] &] (* Julien Kluge, Dec 03 2016 *)

Extensions

a(17)-a(26) from R. J. Mathar, Jan 29 2014
a(27)-a(32) from Jon E. Schoenfield, Feb 05 2014
a(33)-a(34) from Michel Marcus, Jan 24 2023
More terms from David A. Corneth, Jan 25 2023

A234572 Primes of the form P(p-1), where p is a prime and P(.) is the partition function (A000041).

Original entry on oeis.org

2, 5, 11, 17977, 790738119649411319, 2058791472042884901563, 27833079238879849385687, 8121368081058512888507057, 675004412390512738195023734124239, 1398703012615213588677365804960180341, 16193798232344933888778097136641377589301, 204931453786129197483756438132982529754356479553, 3019564607799532159016586951616642980389816614848623, 22757918197082858017617136646280039394687006502870793231847, 1078734573992480956821414895441907729656949308800686938161281
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 28 2013

Keywords

Comments

Though the primes in this sequence are very rare, by the conjecture in A234567 there should be infinitely many such primes.
See A234569 for a list of known primes p with P(p-1) also prime.

Examples

			a(1) = 2 since 2 = P(3-1) with 2 and 3 both prime.
a(2) = 5 since 5 = P(5-1) with 5 prime.
a(3) = 11 since 11 = P(7-1) with 7 and 11 both prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_]:= A234569(n)
    Table[PartitionsP[p[n]-1],{n,1,15}]

Formula

a(n) = A000041(A234569(n)-1).

A236102 Numbers whose divisors are partition numbers.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 15, 22, 77, 101, 17977, 10619863, 6620830889, 80630964769, 228204732751, 1171432692373, 1398341745571, 10963707205259, 15285151248481, 10657331232548839, 790738119649411319, 18987964267331664557, 74878248419470886233, 1394313503224447816939
Offset: 1

Views

Author

Omar E. Pol, Jan 21 2014

Keywords

Comments

By definition all terms are partition numbers.
All members of A049575 are in this sequence.
Conjecture: the only composite numbers in this sequence are 15, 22, and 77. - Jon E. Schoenfield, Feb 05 2014

Examples

			15 is in the sequence because the divisors of 15 are 1, 3, 5, 15, which are also partition numbers.
		

Crossrefs

Extensions

More terms from Jon E. Schoenfield, Feb 05 2014

A121062 Partition numbers mod 4.

Original entry on oeis.org

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

Views

Author

Jonathan Vos Post, Aug 10 2006

Keywords

Comments

P_n==0: 11, 15, 21, 26, 30, 55, 58, 59, 62, 66, 70, 74, 75, 78, 80, 84, 94, 96, 98, 100, ... A237278.
P_n==1: 0, 1, 4, 12, 13, 17, 18, 29, 32, 36, 37, 39, 43, 48, 49, 52, 61, 67, 69, 71, 73, ... A237279.
P_n==2: 2, 8, 9, 10, 19, 22, 25, 27, 28, 31, 34, 40, 42, 45, 46, 47, 50, 57, 64, 65, 79, ... A237280.
P_n==3: 3, 5, 6, 7, 14, 16, 20, 23, 24, 33, 35, 38, 41, 44, 51, 53, 54, 56, 60, 63, 68, ... A237281.

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
         (b(n, i-1)+b(n-i, min(n-i, i))) mod 4))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..104);  # Alois P. Heinz, Dec 20 2024
  • Mathematica
    f[n_] := Mod[PartitionsP@n, 4]; Table[f@n, {n, 0, 104}] (* Robert G. Wilson v *)
  • PARI
    a(n) = numbpart(n) % 4; \\ Michel Marcus, Jun 29 2016

Formula

a(n) = A000041(n) mod 4 = A010873(A000041(n)).
a(n) = A000025(n) mod 4. - John M. Campbell, Jun 29 2016

Extensions

More terms from Robert G. Wilson v, Aug 17 2006

A050811 Partition numbers rounded to nearest integer given by the Hardy-Ramanujan approximate formula.

Original entry on oeis.org

2, 3, 4, 6, 9, 13, 18, 26, 35, 48, 65, 87, 115, 152, 199, 258, 333, 427, 545, 692, 875, 1102, 1381, 1725, 2145, 2659, 3285, 4046, 4967, 6080, 7423, 9037, 10974, 13293, 16065, 19370, 23304, 27977, 33519, 40080, 47833, 56981, 67757, 80431, 95316
Offset: 1

Views

Author

Patrick De Geest, Oct 15 1999

Keywords

Comments

The mounting error seems to be approximately A035949(n-3), n >= 4. - Alonso del Arte, Jul 28 2011
This conjecture is false, for correct approximation see the formula below. - Vaclav Kotesovec, Apr 03 2017

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 95.

Crossrefs

Programs

  • Maple
    A050811:=n->round(exp(Pi*sqrt(2*n/3))/(4*n*sqrt(3))): seq(A050811(n), n=1..70); # Wesley Ivan Hurt, Sep 11 2015
  • Mathematica
    f[n_] := Round[ E^(Sqrt[2n/3] Pi)/(4Sqrt[3] n)]; Array[f, 45] (* Alonso del Arte, May 21 2011, corrected by Robert G. Wilson v, Sep 11 2015 *)
  • PARI
    a(n)=round(exp(Pi*sqrt(2*n/3))/(4*n*sqrt(3))) \\ Charles R Greathouse IV, May 01 2012
  • UBASIC
    input N:print round(#e^(pi(1)*sqrt(2*N/3))/(4*N*sqrt(3)))
    

Formula

a(n) = round(exp(Pi*sqrt(2*n/3))/(4*n*sqrt(3))). - Alonso del Arte, May 21 2011
a(n) - A000041(n) ~ (1/Pi + Pi/72) * exp(sqrt(2*n/3)*Pi) / (4*sqrt(2)*n^(3/2)) * (1 - (9 + Pi^2/48)*Pi/((72 + Pi^2)*sqrt(6*n))). - Vaclav Kotesovec, Apr 03 2017

Extensions

a(1) = 1 replaced by 2, a(2) = 2 replaced by 3. - Alonso del Arte, D. S. McNeil, Aug 07 2011

A051143 Numbers k such that the k-th prime is a partition number.

Original entry on oeis.org

1, 2, 3, 4, 5, 26, 2061, 702993, 307058572, 3350187739, 9088200428, 43794115173, 51932790219, 378210209388, 521301342188, 297064987225918, 19677201507658441, 437852535314831447, 1673669998972800207, 29252504332047744188, 42842701894337201916
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    PrimePi@Select[PartitionsP@Range@301, PrimeQ@# &] (* Robert G. Wilson v, Nov 14 2005 *)

Formula

a(n) = A000720(A049575(n)).

Extensions

a(16)-a(21) using Kim Walisch's primecount, from Amiram Eldar, Jul 26 2019
Showing 1-10 of 26 results. Next