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 10 results.

A233346 Primes of the form p(k)^2 + q(m)^2 with k > 0 and m > 0, where p(.) is the partition function (A000041), and q(.) is the strict partition function (A000009).

Original entry on oeis.org

2, 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 101, 109, 113, 137, 149, 157, 193, 229, 241, 349, 373, 509, 709, 733, 1033, 1049, 1213, 1249, 1453, 1493, 1669, 1789, 2141, 2237, 2341, 2917, 3037, 3137, 3361, 4217, 5801, 5897, 6029, 6073, 8821, 10301, 10937, 11057, 18229, 18289, 19249, 20173, 20341, 20389, 21017, 24001, 30977, 36913, 42793
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 07 2013

Keywords

Comments

Conjecture: The sequence contains infinitely many terms.
This follows from part (i) of the conjecture in A233307. Similarly, the conjecture in A232504 implies that there are infinitely many primes of the form p(k) + q(m) with k and m positive integers.

Examples

			a(1) = 2 since p(1)^2 + q(1)^2 = 1^2 + 1^2 = 2.
a(2) = 5 since p(1)^2 + q(3)^2 = 1^2 + 2^2 = 5.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    n=0
    Do[If[Mod[Prime[m]+1,4]>0,Do[If[PartitionsP[j]>=Sqrt[Prime[m]],Goto[aa],
    If[SQ[Prime[m]-PartitionsP[j]^2]==False,Goto[bb],Do[If[PartitionsQ[k]^2==Prime[m]-PartitionsP[j]^2,
    n=n+1;Print[n," ",Prime[m]];Goto[aa]];If[PartitionsQ[k]^2>Prime[m]-PartitionsP[j]^2,Goto[bb]];Continue,{k,1,2*Sqrt[Prime[m]]}]]];
    Label[bb];Continue,{j,1,Sqrt[Prime[m]]}]];
    Label[aa];Continue,{m,1,4475}]

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}]

A233307 a(n) = |{0 < k < n: p(k)^2 + q(n-k)^2 is prime}|, where p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 07 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Also, for any integer n > 4, p(k)*q(n-k) - 1 is prime for some 0 < k < n/2.
(ii) If n > 9, then prime(k)*p(n-k) + 1 is prime for some 0 < k < n. If n > 2, then prime(k)*q(n-k) - 1 is prime for some 0 < k < n, and also prime(k)*q(n-k) + 1 is prime for some 0 < k < n.
(iii) If n > 11, then prime(k) + p(n-k) is prime for some 0 < k < n. If n > 4, then prime(k) + q(n-k) is prime for some 0 < k < n, and also prime(k)^2 + q(n-k)^2 is prime for some 0 < k < n.

Examples

			a(5) = 1 since 5 = 1 + 4 with p(1)^2 + q(4)^2 = 1^2 + 2^2 = 5 prime.
a(6) = 1 since 6 = 3 + 3 with p(3)^2 + q(3)^2 = 3^2 + 2^2 = 13 prime.
a(19) = 1 since 19 = 3 + 16 with p(3)^2 + q(16)^2 = 3^2 + 32^2 = 1033 prime.
a(28) = 1 since 28 = 3 + 25 with p(3)^2 + q(25)^2 = 3^2 + 142^2 = 20173 prime.
a(29) = 1 since 29 = 6 + 23 with p(6)^2 + q(23)^2 = 11^2 + 104^2 = 10937 prime.
a(38) = 1 since 38 = 1 + 37 with p(1)^2 + q(37)^2 = 1^2 + 760^2 = 577601 prime.
a(75) = 1 since 75 = 13 + 62 with p(13)^2 + q(62)^2 = 101^2 + 13394^2 = 179409437 prime.
a(160) = 1 since 160 = 48 + 112 with p(48)^2 + q(112)^2 = 147273^2 + 1177438^2 = 1408049580373 prime.
a(210) = 1 since 210 = 71 + 139 with p(71)^2 + q(139)^2 = 4697205^2 + 8953856^2 = 102235272080761 prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[PartitionsP[k]^2+PartitionsQ[n-k]^2],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]

A234475 Number of ways to write n = k + m with 2 < k <= m such that q(phi(k)*phi(m)/4) + 1 is prime, where phi(.) is Euler's totient function and q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 26 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
This implies that there are infinitely many primes p with p - 1 a term of A000009.

Examples

			a(6) = 1 since 6 = 3 + 3 with q(phi(3)*phi(3)/4) + 1 = q(1) + 1 = 2 prime.
a(76) = 1 since 76 = 18 + 58 with q(phi(18)*phi(58)/4) + 1 = q(42) + 1 = 1427 prime.
a(197) = 1 since 197 = 4 + 193 with q(phi(4)*phi(193)/4) + 1 = q(96) + 1 = 317789.
a(356) = 1 since 356 = 88 + 268 with q(phi(88)*phi(268)/4) + 1 = q(1320) + 1 = 35940172290335689735986241 prime.
		

Crossrefs

Programs

  • Mathematica
    f[n_,k_]:=PartitionsQ[EulerPhi[k]*EulerPhi[n-k]/4]+1
    a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,3,n/2}]
    Table[a[n],{n,1,100}]

A233393 Primes of the form 2^k - 1 + q(m) with k > 0 and m > 0, where q(.) is the strict partition function (A000009).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 37, 41, 43, 47, 53, 61, 67, 71, 73, 79, 83, 101, 107, 109, 127, 131, 137, 139, 149, 157, 167, 173, 181, 191, 193, 199, 223, 229, 257, 263, 269, 271, 277, 293, 311, 331, 347, 349, 359, 383, 397, 421, 449, 463, 467, 479, 521, 523, 557, 587
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2013

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
This follows from the conjecture in A233390.

Examples

			a(1) = 2 since 2^1 - 1 + q(1) = 1 + 1 = 2.
a(2) = 3 since 2^1 - 1 + q(3) = 1 + 2 = 3.
a(3) = 5 since 2^2 - 1 + q(3) = 3 + 2 = 5.
		

Crossrefs

Programs

  • Mathematica
    Pow[n_]:=Pow[n]=Mod[n,2]==0&&2^(IntegerExponent[n,2])==n
    n=0
    Do[Do[If[Pow[Prime[m]-PartitionsQ[k]+1],
    n=n+1;Print[n," ",Prime[m]];Goto[aa]];If[PartitionsQ[k]>=Prime[m],Goto[aa]];Continue,{k,1,2*Prime[m]}];
    Label[aa];Continue,{m,1,110}]

A238509 a(n) = |{0 < k < n: p(n) + p(k) - 1 is prime}|, where p(.) is the partition function (A000041).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 27 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Also, if n > 2 is different from 8 and 25, then p(n) + p(k) + 1 is prime for some 0 < k < n.
(ii) If n > 7, then n + p(k) is prime for some 0 < k < n.
(iii) If n > 1, then p(k) + q(n) is prime for some 0 < k < n, where q(.) is the strict partition function given by A000009. If n > 2, then p(k) + q(n) - 1 is prime for some 0 < k < n. If n > 1 is not equal to 8, then p(k) + q(n) + 1 is prime for some 0 < k < n.

Examples

			a(11) = 1 since p(11) + p(10) - 1 = 56 + 42 - 1 = 97 is prime.
a(247) = 1 since p(247) + p(228) - 1 = 182973889854026 + 40718063627362 - 1 = 223691953481387 is prime.
		

Crossrefs

Programs

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

A233417 a(n) = |{0 < k <= n/2: q(k)*q(n-k) + 1 is prime}|, where q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 09 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Similarly, for any integer n > 5, there is a positive integer k < n with q(k)*q(n-k) - 1 prime.
(ii) Let n > 1 be an integer. Then p(k) + q(n-k)^2 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n is not equal to 8, then k^3 + q(n-k)^2 is prime for some 0 < k < n.

Examples

			a(14) = 1 since q(1)*q(13) + 1 = 1*18 + 1 = 19 is prime.
a(17) = 1 since q(4)*q(13) + 1 = 2*18 + 1 = 37 is prime.
a(27) = 1 since q(13)*q(14) + 1 = 18*22 + 1 = 397 is prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[PartitionsQ[k]*PartitionsQ[n-k]+1],1,0],{k,1,n/2}]
    Table[a[n],{n,1,100}]

A233360 Primes of the form L(k) + q(m) with k > 0 and m > 0, where L(k) is the k-th Lucas number (A000204), and q(.) is the strict partition function (A000009).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 101, 103, 107, 127, 131, 149, 151, 193, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 307, 337, 347, 349, 379, 397, 401, 419, 421, 449, 463, 487, 523, 541, 571, 643, 647, 661
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2013

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
This follows from the conjecture in A233359.

Examples

			a(1) = 2 since L(1) + q(1) = 1 + 1 = 2.
a(2) = 3 since L(1) + q(3) = 1 + 2 = 3.
a(3) = 5 since L(2) + q(3) = 3 + 2 = 5.
		

Crossrefs

Programs

  • Mathematica
    n=0
    Do[Do[If[LucasL[j]>=Prime[m],Goto[aa],
    Do[If[PartitionsQ[k]==Prime[m]-LucasL[j],
    n=n+1;Print[n," ",Prime[m]];Goto[aa]];If[PartitionsQ[k]>Prime[m]-LucasL[j],Goto[bb]];Continue,{k,1,2*(Prime[m]-LucasL[j])}]];
    Label[bb];Continue,{j,1,2*Log[2,Prime[m]]}];
    Label[aa];Continue,{m,1,125}]

A236419 a(n) = |{0 < k < n: r = phi(k) + phi(n-k)/6 + 1 and p(r) + q(r) are both prime}|, where phi(.) is Euler's totient function, p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 25 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 127.
We have verified this for n up to 30000.
The conjecture implies that there are infinitely many primes r with p(r) + q(r) also prime.

Examples

			a(15) = 1 since phi(1) + phi(14)/6 + 1 = 3 with p(3) + q(3) = 3 + 2 = 5 prime.
a(54) = 1 since phi(41) + phi(13)/6 + 1 = 43 with p(43) + q(43) = 63261 + 1610 = 64871 prime.
		

Crossrefs

Programs

  • Mathematica
    pq[n_]:=PrimeQ[n]&&PrimeQ[PartitionsP[n]+PartitionsQ[n]]
    f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/6+1
    a[n_]:=Sum[If[pq[f[n,k]],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]

A236442 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that A000009(k) + A047967(m) is prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 26 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 2.
(ii) If n > 2 is neither 18 nor 30, then n can be written as k + m with k > 0 and m > 0 such that A000009(k)^2 + A047967(m)^2 is prime.
(iii) Any integer n > 4 can be written as k + m with k > 0 and m > 0 such that A000009(k)*A047967(m) - 1 (or A000009(k)*A047967(m) + 1) is prime.

Examples

			a(15) = 1 since 15 = 13 + 2 with A000009(13) + A047967(13) = 18 + 1 = 19 prime.
a(18) = 1 since 18 = 3 + 15 with A000009(3) + A047967(15) = 2 + 149 = 151 prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_,k_]:=PrimeQ[PartitionsQ[k]+(PartitionsP[n-k]-PartitionsQ[n-k])]
    a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]
Showing 1-10 of 10 results.