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.

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A230140 Number of ways to write n = x + y + z with 0 < x <= y <= z such that 6*x-1, 6*y-1, 6*z-1 are among those primes p (terms of A230138) with p + 2 and 2*p - 5 also prime.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 2, 3, 1, 2, 2, 2, 3, 3, 4, 2, 3, 2, 3, 3, 3, 4, 2, 5, 2, 6, 3, 6, 5, 4, 5, 3, 5, 5, 8, 7, 6, 5, 6, 5, 5, 7, 6, 8, 4, 6, 5, 6, 7, 9, 8, 8, 5, 7, 6, 8, 10, 6, 10, 4, 8, 6, 6, 10, 6, 9, 5, 6, 5, 7, 7, 9, 6, 7, 8, 5, 10, 6, 9, 6, 6, 7, 4, 7, 7, 9, 6, 5, 5, 4, 6, 5, 6, 5, 5, 6, 4, 6, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 10 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 2, i.e., 6*n-3 with n > 2 can be expressed as a sum of three terms from A230138. Moreover, for any integer n > 12, there are three distinct positive integers x, y, z with x + y + z = n such that 6*x-1, 6*y-1, 6*z-1 are primes in A230138.
(ii) For each integer n > 12, there are three distinct positive integers x, y, z with x + y + z = n such that 6*x-1, 6*y-1, 6*z-1 are among those primes p with p + 2 and 2*p + 9 also prime.
Note that part (i) of this conjecture implies that there are infinitely many primes in A230138.
Indices k such that a(m)>a(k) for all m>k, are (2, 10, 26, 334, 439, 544, 551, 684, ...). The only sequence which has the first 5 terms within the 3 lines of data is A212067. (Certainly a coincidence.) - M. F. Hasler, Oct 10 2013

Examples

			a(10) = 1 since 10 = 2 + 3 + 5, and the three numbers 6*2-1=11, 6*3-1=17 and 6*5-1=29 are terms of A230138.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=PrimeQ[6n-1]&&PrimeQ[6n+1]&&PrimeQ[12n-7]
    a[n_]:=Sum[If[SQ[i]&&SQ[j]&&SQ[n-i-j],1,0],{i,1,n/3},{j,i,(n-i)/2}]
    Table[a[n],{n,1,100}]
  • PARI
    ip(x)=isprime(6*x-1) && isprime(6*x+1) && isprime(12*x-7); a(n)=sum(x=1,n\3,sum(y=x,ip(x)*(n-x)\2,ip(y) && ip(n-x-y))) \\ - M. F. Hasler, Oct 10 2013

A230141 Number of ways to write n = x + y + z with y <= z such that 6*x-1, 6*y-1, 6*z-1 are terms of A230138 and 6*(y+z)+1 is prime.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 4, 5, 3, 2, 3, 4, 4, 5, 6, 5, 3, 5, 4, 4, 2, 4, 6, 2, 3, 2, 6, 9, 8, 8, 5, 5, 4, 5, 10, 14, 10, 12, 6, 11, 7, 9, 13, 6, 11, 3, 9, 7, 8, 14, 6, 11, 4, 4, 8, 9, 15, 15, 7, 14, 3, 6, 13, 10, 19, 6, 6, 12, 5, 10, 8, 7, 16, 6, 10, 4, 7, 19, 11, 13, 3, 12, 5, 6, 13, 5, 12, 7, 8, 4, 5, 6, 10, 6, 4, 6, 4, 6, 7, 7
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 10 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2. Also, any integer n > 2 can be written as x + y + z (x, y, z > 0) such that 6*x-1, 6*y-1, 6*z-1 are terms of A230138 and 6*y*z-1 is prime.
This is a further refinement of the conjecture in A230140.
Note that if x + y + z = n then 6*n = (6*x-1) + (6*(y+z)+1). So a(n) > 0 implies Goldbach's conjecture for the even number 6*n.

Examples

			a(10) = 2 since 10 = 3 + 2 + 5 = 5 + 2 + 3, and 6*3-1 = 17, 6*2-1 = 11, 6*5-1 = 29 are terms of A230138, and 6*(2+5)+1 = 43 and 6*(2+3)+1 = 31 are also prime.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=PrimeQ[6n-1]&&PrimeQ[6n+1]&&PrimeQ[12n-7]
    a[n_]:=Sum[If[SQ[i]&&PrimeQ[6(n-i)+1]&&SQ[j]&&SQ[n-i-j],1,0],{i,1,n-2},{j,1,(n-i)/2}]
    Table[a[n],{n,1,100}]

A341408 Number of partitions of n into 3 nonprime parts.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 2, 3, 2, 4, 5, 5, 5, 7, 6, 9, 8, 11, 10, 13, 12, 16, 14, 19, 16, 22, 19, 26, 22, 29, 27, 33, 28, 39, 33, 42, 38, 47, 43, 53, 45, 58, 52, 63, 59, 70, 61, 77, 68, 83, 76, 91, 79, 98, 88, 105, 95, 115, 102, 121, 111, 130, 119, 141, 124, 148
Offset: 3

Views

Author

Ilya Gutkovskiy, Feb 12 2021

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n=0,
          `if`(t=0, 1, 0), `if`(i<1 or t<1, 0, b(n, i-1, t)+
          `if`(isprime(i), 0, b(n-i, min(n-i, i), t-1))))
        end:
    a:= n-> b(n$2, 3):
    seq(a(n), n=3..72);  # Alois P. Heinz, Feb 12 2021
  • Mathematica
    b[n_, i_, t_] := b[n, i, t] = If[n == 0,
         If[t == 0, 1, 0], If[i < 1 || t < 1, 0, b[n, i - 1, t] +
         If[PrimeQ[i], 0, b[n - i, Min[n - i, i], t - 1]]]];
    a[n_] := b[n, n, 3];
    a /@ Range[3, 72] (* Jean-François Alcover, Mar 28 2021, after Alois P. Heinz *)

A124868 Natural numbers that are not the sum of 3 distinct primes.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 17
Offset: 1

Views

Author

Alexander Adamchuk, Nov 11 2006

Keywords

Comments

(Conjectured) Every number n > 17 is the sum of 3 distinct primes. a(n) is the complement of A124867(n) = {10, 12, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, ...}, numbers that are the sum of 3 distinct primes.
A125688(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2006

Crossrefs

Cf. A124867 (numbers that are the sum of 3 distinct primes), A068307.

A218007 Number of partitions of n into at most three primes (including 1).

Original entry on oeis.org

1, 2, 3, 3, 4, 4, 5, 4, 5, 4, 5, 4, 6, 5, 7, 5, 7, 5, 8, 6, 9, 6, 9, 7, 10, 7, 10, 5, 10, 6, 12, 7, 13, 7, 12, 8, 14, 7, 14, 6, 15, 8, 17, 9, 17, 8, 18, 10, 19, 10, 19, 7, 20, 9, 21, 9, 20, 7, 21, 11, 25, 11, 24, 9, 26, 11, 27, 9, 24, 8, 28, 12, 30, 13, 29
Offset: 1

Views

Author

Frank M Jackson, Mar 26 2013

Keywords

Comments

The above sequence relies on the strong Goldbach's conjecture that any positive integer is the sum of at most three distinct terms from {1 union primes}.

Examples

			a(21)=9 as 21 = 1+1+19 = 2+19 = 1+3+17 = 2+2+17 = 1+7+13 = 3+5+13 = 3+7+11 = 5+5+11 = 7+7+7
		

Crossrefs

Programs

  • Mathematica
    primeQ[p0_] := If[p0==1, True, PrimeQ[p0]]; SetAttributes[primeQ, Listable]; goldbachcount[p1_] := (parts=IntegerPartitions[p1, 3]; count=0; n=1; While[n<=Length[parts], If[Intersection[Flatten[primeQ[parts[[n]]]]][[1]] == True, count++]; n++]; count); Table[goldbachcount[i], {i, 1, 100}]
    Table[Length[Select[#/.(1->2)&/@IntegerPartitions[n,3],AllTrue[#,PrimeQ]&]],{n,80}] (* Harvey P. Dale, Jan 11 2023 *)

A087916 Number of ordered ways to write 2n+1 as a sum of 3 odd primes.

Original entry on oeis.org

0, 0, 0, 0, 1, 3, 6, 7, 9, 12, 16, 18, 21, 27, 30, 30, 34, 36, 42, 46, 48, 48, 51, 63, 60, 64, 81, 75, 76, 87, 87, 90, 102, 105, 97, 117, 114, 105, 144, 129, 126, 159, 141, 145, 177, 162, 160, 195, 186, 153, 207, 201, 171, 237, 210, 187, 255, 234, 222, 279
Offset: 0

Views

Author

Ralf Stephan, Oct 18 2003

Keywords

Crossrefs

Cf. A007963 (unordered), A068307 (with 2).
Cf. A087917.

Programs

  • Mathematica
    nn = 100; lim = 2*nn + 17; ps = Prime[Range[2, nn + 1]]; t = Table[0, {lim}]; Do[s = i + j + k; If[s <= lim, t[[s]]++], {i, ps}, {j, ps}, {k, ps}]; Take[t, {1, lim, 2}] (* T. D. Noe, Apr 10 2014 *)
  • PARI
    for(n=0, 100, t=2*n+1; c=0; for(k=2, t, for(l=2, t, for(m=2, t, tt=prime(k)+prime(l)+prime(m); if(tt>2*n+1, break); if(tt==2*n+1, c=c+1)))); print1(c", "))

Extensions

Leading zeros added by T. D. Noe, Apr 10 2014

A230502 Number of ways to write n = (2-(n mod 2))*p + q + r with p <= q <= r such that p, q, r, p^2 - 2, q^2 - 2, r^2 - 2 are all prime.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 2, 1, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 2, 3, 3, 3, 2, 2, 2, 4, 3, 4, 2, 2, 3, 2, 2, 4, 2, 3, 4, 4, 3, 3, 3, 3, 4, 5, 4, 4, 3, 3, 5, 7, 5, 6, 5, 5, 5, 6, 3, 5, 5, 5, 5, 6, 4, 4, 4, 5, 6, 7, 5, 6, 4, 3, 5, 7, 5, 5, 7, 7, 6, 7, 4, 6, 6, 7, 7, 6, 4, 6, 4, 4, 8, 8, 6, 6, 7, 6, 6, 10
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 21 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 6.
This is stronger than Goldbach's weak conjecture which was finally proved by H. Helfgott in 2013. It also implies that there are infinitely many primes p with p^2 - 2 also prime.
Conjecture verified for n up to 10^9. - Mauro Fiorentini, Sep 22 2023

Examples

			a(10) = 1 since 10 = 2*2 + 3 + 3 with 2, 3, 2^2 - 2 = 2, 3^2 - 2 = 7 all prime.
a(19) = 2 since 19 = 3 + 3 + 13 = 5 + 7 + 7 with 3, 13, 5, 7, 3^2 - 2 = 7, 13^2 - 2 = 167, 5^2 - 2 = 23, 7^2 - 2 = 47 all prime.
		

Crossrefs

Programs

  • Mathematica
    pp[n_]:=PrimeQ[n^2-2]
    pq[n_]:=PrimeQ[n]&&pp[n]
    a[n_]:=Sum[If[pp[Prime[i]]&&pp[Prime[j]]&&pq[n-(2-Mod[n,2])Prime[i]-Prime[j]],1,0],{i,1,PrimePi[n/(4-Mod[n,2])]},{j,i,PrimePi[(n-(2-Mod[n,2])Prime[i])/2]}]
    Table[a[n],{n,1,100}]

A307727 Number of partitions of n into 3 prime powers (not including 1).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 3, 4, 5, 6, 6, 8, 7, 9, 9, 10, 10, 12, 11, 14, 13, 14, 13, 16, 13, 18, 15, 18, 16, 20, 18, 23, 20, 25, 23, 26, 22, 28, 23, 30, 23, 30, 23, 32, 26, 32, 27, 34, 28, 37, 28, 36, 29, 40, 31, 43, 28, 42, 32, 44, 32, 46, 32, 46, 35, 46, 35, 50, 34, 51, 37, 53, 36, 59, 36, 57, 41
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 24 2019

Keywords

Examples

			a(11) = 4 because we have [7, 2, 2], [5, 4, 2], [5, 3, 3] and [4, 4, 3].
		

Crossrefs

Programs

  • Maple
    f:= proc(n,k,pmax) option remember;
      local t,p,j;
      if n = 0 then return `if`(k=0, 1, 0) fi;
      if k = 0 then return 0 fi;
      if n > k*pmax then return 0 fi;
      t:= 0:
      for p in A246655 do
        if p > pmax then return t fi;
        t:= t + add(procname(n-j*p, k-j, min(p-1,n-j*p)),j=1..min(k,floor(n/p)))
      od;
      t
    end proc:
    seq(f(n,3,n),n=0..80) # Robert Israel, Apr 25 2019
  • Mathematica
    Array[Count[IntegerPartitions[#, {3}], _?(AllTrue[#, PrimePowerQ] &)] &, 81, 0]

Formula

a(n) = [x^n y^3] Product_{k>=1} 1/(1 - y*x^A246655(k)).
a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} [omega(i) * omega(j) * omega(n-i-j) == 1], where omega(n) is the number of distinct prime factors of n and [==] is the Iverson bracket. - Wesley Ivan Hurt, Apr 25 2019

A229969 Number of ways to write n = x + y + z with 0 < x <= y <= z such that all the six numbers 2*x-1, 2*y-1, 2*z-1, 2*x*y-1, 2*x*z-1, 2*y*z-1 are prime.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 4, 4, 3, 3, 3, 3, 2, 3, 3, 3, 3, 4, 2, 7, 4, 3, 5, 3, 2, 6, 3, 4, 3, 4, 5, 3, 4, 6, 6, 3, 5, 4, 5, 6, 9, 4, 8, 4, 7, 10, 2, 6, 12, 9, 1, 7, 7, 6, 12, 10, 3, 7, 8, 8, 9, 9, 5, 3, 7, 3, 7, 3, 9, 10, 8, 6, 11, 11, 13, 15, 6, 6, 10, 15, 11, 11, 13, 8, 12, 12, 7, 10, 8, 13, 12
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 04 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5. Moreover, any integer n > 6 can be written as x + y + z with x among 3, 4, 6, 10, 15 such that 2*y-1, 2*z-1, 2*x*y-1, 2*x*z-1, 2*y*z-1 are prime.
We have verified this conjecture for n up to 10^6. As (2*x-1)+(2*y-1)+(2*z-1) = 2*(x+y+z)-3, it implies Goldbach's weak conjecture which has been proved.
Zhi-Wei Sun also had some similar conjectures including the following (i)-(iii):
(i) Any integer n > 6 can be written as x + y + z (x, y, z > 0) with 2*x-1, 2*y-1, 2*z-1 and 2*x*y*z-1 all prime and x among 2, 3, 4. Also, each integer n > 2 can be written as x + y + z (x, y, z > 0) with 2*x+1, 2*y+1, 2*z+1 and 2*x*y*z+1 all prime and x among 1, 2, 3.
(ii) Each integer n > 4 can be written as x + y + z with x = 3 or 6 such that 2*y+1, 2*x*y*z-1 and 2*x*y*z+1 are prime.
(iii) Every integer n > 5 can be written as x + y + z (x, y, z > 0) with x*y-1, x*z-1, y*z-1 all prime and x among 2, 6, 10. Also, any integer n > 2 not equal to 16 can be written as x + y + z (x, y, z > 0) with x*y+1, x*z+1, y*z+1 all prime and x among 1, 2, 6.
See also A229974 for a similar conjecture involving three pairs of twin primes.

Examples

			a(10) = 2 since 10 = 2+2+6 = 3+3+4 with 2*2-1, 2*6-1, 2*2*2-1, 2*2*6 -1, 2*3-1, 2*4-1, 2*3*3-1, 2*3*4-1 all prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[2i-1]&&PrimeQ[2j-1]&&PrimeQ[2(n-i-j)-1]&&PrimeQ[2i*j-1]&&PrimeQ[2i(n-i-j)-1]&&PrimeQ[2j(n-i-j)-1],1,0],{i,1,n/3},{j,i,(n-i)/2}]
    Table[a[n],{n,1,100}]

A230224 Number of ways to write 2*n = p + q + r + s with p <= q <= r <= s such that p, q, r, s are primes in A230223.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 2, 4, 1, 3, 3, 3, 4, 4, 3, 5, 4, 5, 3, 6, 4, 6, 5, 5, 5, 7, 5, 9, 4, 6, 6, 8, 6, 9, 5, 7, 7, 10, 6, 8, 7, 8, 7, 9, 5, 10, 7, 11, 7, 7, 7, 11, 7, 10, 6, 10, 6, 13, 7, 9, 7, 11, 9, 11, 7, 9, 6, 14, 8, 12, 6, 13, 11, 12, 11, 13, 10, 16, 9, 14, 7, 14
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 12 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 17.

Examples

			a(21) = 1 since 2*21 = 7 + 7 + 11 + 17, and 7, 11, 17 are primes in A230223.
a(27) = 1 since 2*27 = 7 + 11 + 17 + 19, and 7, 11, 17, 19 are primes in A230223.
		

Crossrefs

Programs

  • Mathematica
    RQ[n_]:=n>5&&PrimeQ[3n-4]&&PrimeQ[3n-10]&&PrimeQ[3n-14]
    SQ[n_]:=PrimeQ[n]&&RQ[n]
    a[n_]:=Sum[If[RQ[Prime[i]]&&RQ[Prime[j]]&&RQ[Prime[k]]&&SQ[2n-Prime[i]-Prime[j]-Prime[k]],1,0],
    {i,1,PrimePi[n/2]},{j,i,PrimePi[(2n-Prime[i])/3]},{k,j,PrimePi[(2n-Prime[i]-Prime[j])/2]}]
    Table[a[n],{n,1,100}]
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