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 1-9 of 9 results.

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.

Links

Crossrefs

Cf. A000010, A000040, A000041, A049575, A232504, A233307, A233346, A233918, A234200, A234246, A234309, A234337, A234344, A234347, A234359, A234360, A234451, A234475

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}]

A234309 a(n) = |{2 < k <= n/2: 2^{phi(k)} + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 23 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For any integer n > 1, 2^k +2^{phi(n-k)} - 1 is prime for some 0 < k < n, and 2^{sigma(j)} + 2^{phi(n-j)} - 1 is prime for some 0 < j < n, where sigma(j) is the sum of all positive divisors of j.
As phi(k) is even for any k > 2, part (i) of the conjecture implies that there are infinitely many primes of the form 4^a + 4^b - 1 with a and b positive integers (cf. A234310). Note that any Mersenne prime greater than 3 has the form 2^{2a+1} - 1 = 4^a + 4^a - 1.

Examples

			a(6) = 1 since 2^{phi(3)} + 2^{phi(3)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(7) = 1 since 2^{phi(3)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(8) = 2 since 2^{phi(3)} + 2^{phi(5)} - 1 = 2^2 + 2^4 - 1 = 19 and 2^{phi(4)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 are both prime.

Links

Crossrefs

Cf. A000010, A000040, A000079, A000668, A233307, A233390, A233542, A233544, A233547, A233566, A233867, A233918, A234200, A234246, A234308, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361

Programs

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

A236998 a(n) = |{0 < k < n/2: phi(k)*phi(n-k) is a square}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 8.
(ii) If n > 20, then phi(k)*phi(n-k) + 1 is a square for some 0 < k < n/2.
(iii) If n > 1 is not among 4, 7, 60, 199, 267, then k*phi(n-k) is a square for some 0 < k < n.
We have verified part (i) of the conjecture for n up to 2*10^6.

Examples

			a(17) = 1 since phi(5)*phi(12) = 4*4 = 4^2.
a(24) = 1 since phi(4)*phi(20) = 2*8 = 4^2.
a(56) = 1 since phi(8)*phi(48) = 4*16 = 8^2.

Links

Crossrefs

Cf. A000010, A000290, A234246, A236567, A237016.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
p[n_,k_]:=SQ[EulerPhi[k]*EulerPhi[n-k]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,100}]

A237016 a(n) = |{0 < k < n: phi(k)*sigma(n-k) is a square}|, where phi(.) is Euler's totient function and sigma(j) is the sum of all positive divisors of j.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 1, 7, 17.
(ii) If n > 5, then phi(k)*sigma(n-k) + 1 is a square for some 0 < k < n.
(iii) If n > 309, then there is a positive integer k < n/2 such that sigma(k)*sigma(n-k) is a square.
See also A236998 for a similar conjecture.

Examples

			a(9) = 1 since phi(8)*sigma(1) = 4*1 = 2^2.
a(16) = 1 since phi(6)*sigma(10) = 2*18 = 6^2.
a(31) = 1 since phi(24)*sigma(7) = 8*8 = 8^2.
a(65) = 1 since phi(19)*sigma(46) = 18*72 = 36^2.

Links

Crossrefs

Cf. A000010, A000203, A000290, A234246, A236567, A236998.

Programs

  • Mathematica
    sigma[n_]:=DivisorSigma[1,n]
SQ[n_]:=IntegerQ[Sqrt[n]]
p[n_,k_]:=SQ[EulerPhi[k]*sigma[n-k]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A234808 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*n - p are both prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 2, 0, 3, 1, 2, 5, 2, 1, 5, 1, 2, 7, 2, 1, 4, 1, 2, 1, 4, 1, 4, 2, 4, 11, 4, 2, 3, 1, 5, 2, 3, 2, 6, 1, 5, 15, 4, 2, 9, 1, 6, 2, 5, 4, 6, 4, 4, 3, 8, 3, 6, 4, 7, 21, 2, 4, 7, 1, 7, 4, 6, 4, 6, 4, 8, 22, 7, 3, 13, 1, 10, 5, 3, 5, 7, 4, 9, 5, 10, 5, 8, 7, 7, 6, 8, 5, 6, 3, 8, 6, 7, 4, 8, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 30 2013

Keywords

Comments

Conjecture: a(n) > 0 except for n = 1, 8.
Clearly, this implies Goldbach's conjecture.

Examples

			a(3) = 1 since 2 + phi(1) = 3 and 2*3 - 3 = 3 are both prime.
a(20) = 1 since 11 + phi(9) = 17 and 2*20 - 17 = 23 are both prime.
a(22) = 1 since 1 + phi(21) = 13 and 2*22 - 13 = 31 are both prime.
a(24) = 1 since 9 + phi(15) = 17 and 2*24 - 17 = 31 are both prime.
a(76) = 1 since 67 + phi(9) = 73 and 2*76 - 73 = 79 are both prime.

Links

Crossrefs

Cf. A000010, A000040, A002372, A002375, A233547, A233918, A234200, A234246, A234360, A234470, A234475, A234514, A234567, A234615, A234694

Programs

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

A236567 a(n) = |{0 < k < n - 2: k + phi(n-k)/2 is a square}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For n > 31, there is a positive integer k < n - 2 with phi(k) + phi(n-k)/2 a square. If n > 70 is not equal to 150, then phi(k) + phi(n-k) is a square for some 0 < k < n.
(iii) If n > 5, then phi(k) + phi(n-k)/2 is a triangular number for some 0 < k < n - 2. For each n = 20, 21, ... there is a positive integer k < n with (phi(k) + phi(n-k))/2 a triangular number.

Examples

			a(8) = 1 since 1 + phi(7)/2 = 1 + 3 = 2^2.
a(11) = 1 since 8 + phi(3)/2 = 8 + 1 = 3^2.
a(78) = 1 since 40 + phi(38)/2 = 40 + 9 = 7^2.

Links

Crossrefs

Cf. A000010, A000290, A234246.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
p[n_,k_]:=SQ[k+EulerPhi[n-k]/2]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-3}]
Table[a[n],{n,1,100}]

A237523 a(n) = |{0 < k < n/2: phi(k*(n-k)) + 1 is a square}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 08 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 7, and a(n) = 1 only for n = 8, 9, 13, 15, 20, 132.
(ii) If n > 5 is not among 10, 15, 20, 60, 105, then phi(k*(n-k)) is a square for some 0 < k < n/2.
See also A237524 for a similar conjecture involving higher powers.

Examples

			a(8) = 1 since phi(3*5) + 1 = 8 + 1 = 3^2.
a(9) = 1 since phi(4*5) + 1 = 8 + 1 = 3^2.
a(13) = 1 since phi(3*10) + 1 = 8 + 1 = 3^2.
a(15) = 1 since phi(7*8) + 1 = 24 + 1 = 5^2.
a(20) = 1 since phi(6*14) + 1 = 24 + 1 = 5^2.
a(132) = 1 since phi(46*(132-46)) + 1 = 1848 + 1 = 43^2.

Links

Crossrefs

Cf. A000010, A000290, A234246, A236998, A237497, A237524.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
s[n_]:=SQ[EulerPhi[n]+1]
a[n_]:=Sum[If[s[k(n-k)],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,80}]

A234308 a(n) = |{0 < k <= n/2: phi(k^2)*phi(n-k) - 1 is a Sophie Germain prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 22 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 4.
(ii) If n > 3, then phi(k^2)*phi(n-k) - 1 and phi(k^2)*phi(n-k) + 1 are both prime for some 0 < k < n, and also phi(j)^2*phi(n-j) - 1 and phi(j)^2*phi(n-j) + 1 are both prime for some 0 < j < n.
(iii) If n > 9 is not equal to 14, then |phi(k) - phi(n-k)|/2 is prime for some 0 < k < n, and also |phi(j) - phi(n-j)| - 1 and |phi(j) - phi(n-j)| + 1 are both prime for some 0 < j < n.
(iv) If n > 5, then sigma(k)*phi(n-k) + 1 is a square for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.
Note that part (i) of the conjecture implies that there are infinitely many Sophie Germain primes. We have verified part (i) for n up to 3*10^6.

Examples

			a(5) = 1 since phi(2^2)*phi(3) - 1 = 3 is a Sophie Germain prime.
a(10) = 1 since phi(1^2)*phi(9) - 1 = 5 is a Sophie Germain prime.
a(12) = 1 since phi(6^2)*phi(6) - 1 = 23 is a Sophie Germain prime.
a(30) = 1 since phi(2^2)*phi(28) - 1 = 23 is a Sophie Germain prime.
a(60) = 1 since phi(4^2)*phi(56) - 1 = 191 is a Sophie Germain prime.
a(75) = 1 since phi(14^2)*phi(61) - 1 = 5039 is a Sophie Germain prime.
a(95) = 1 since phi(30^2)*phi(65) - 1 = 11519 is a Sophie Germain prime.
a(106) = 1 since phi(22^2)*phi(84) - 1 = 5279 is a Sophie Germain prime.
a(110) = 1 since phi(9^2)*phi(101) - 1 = 5399 is a Sophie Germain prime.
a(156) = 1 since phi(27^2)*phi(129) - 1 = 40823 is a Sophie Germain prime.

Links

Crossrefs

Cf. A000010, A000040, A014574, A005384, A233542, A233547, A233566, A233867, A233918, A234200, A234246

Programs

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

A236548 Number of ways to write n = k^2 + m with k > 0 and m > 0 such that phi(k^2) + prime(m) is prime, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 28 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 100.
(ii) For any integer n > 4, there is a positive integer k < n with phi(k)^2 + prime(n-k)^2 prime.
Note that phi(k^2) (k = 1, 2, 3, ...) are pairwise distinct and this can be easily proved by induction.

Examples

			a(16) = 1 since 16 = 3^2 + 7 with phi(3^2) + prime(7) = 6 + 17 = 23 prime.
a(611) = 1 since 611 = 22^2 + 127 with phi(22^2) + prime(127) = 220 + 709 = 929 prime.

Links

Crossrefs

Cf. A000010, A000040, A233867, A233918, A234246, A236541.

Programs

  • Mathematica
    p[n_,k_]:=PrimeQ[EulerPhi[k^2]+Prime[n-k^2]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,Sqrt[n-1]}]
Table[a[n],{n,1,100}]
Showing 1-9 of 9 results.

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