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 1-3 of 3 results.

A263992 Number of ordered ways to write n as x^2 + 2*y^2 + phi(z^2) (x >= 0, y >= 0 and z > 0) such that y or z has the form p-1 with p prime, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 31 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 6, 7, 22, 3447.
This is similar to the conjecture in A262311, and we have verified it for n up to 10^6.

Examples

			a(1) = 1 since 1 = 0^2 + 2*0^2 + phi(1^2) with 1 + 1 = 2 prime.
a(6) = 1 since 6 = 2^2 + 2*0^2 + phi(2^2) with 2 + 1 = 3 prime.
a(7) = 1 since 7 = 2^2 + 2*1^2 + phi(1^2) with 1 + 1 = 2 prime.
a(22) = 1 since 22 = 0^2 + 2*1^2 + phi(5^2) with 1 + 1 = 2 prime.
a(3447) = 1 since 3447 = 42^2 + 2*29^2 + phi(1^2) with 1 + 1 = 2 prime.

Links

Crossrefs

Cf. A000010, A000290, A006093, A262311, A263998.

Programs

  • Mathematica
    phi[n_]:=phi[n]=EulerPhi[n]
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[n-z*phi[z]<0,Goto[aa]];Do[If[SQ[n-z*phi[z]-2y^2]&&(PrimeQ[y+1]||PrimeQ[z+1]),r=r+1],{y,0,Sqrt[(n-z*phi[z])/2]}];Label[aa];Continue,{z,1,n}];Print[n," ",r];Continue,{n,1,100}]

A264010 Number of ways to write n as x^2 + y*(y+1) + z*(z+1)/2, where x, y and z are nonnegative integers such that y or y+1 is prime, and z or z+1 is prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 31 2015

Keywords

Comments

Conjectures: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 4, 5, 6, 10, 11, 15, 20, 29, 1125.
(ii) Any integer n > 2 can be written as x*(x+1) + y*(y+1)/2 + z*(z+1)/2, where x, y and z are nonnegative integers such that x or x+1 is prime, and y or y+1 is prime.
(iii) Any integer n > 7 can be written as x*(x+1) + y*(y+1)/2 + 3*z*(z+1)/2, where x, y and z are nonnegative integers such that y or y+1 is prime, and z or z+1 is prime.
It is known that any natural number can be written as x^2 + y*(y+1)+ z*(z+1)/2 (or x*(x+1) + y*(y+1)/2 + z*(z+1)/2, or x*(x+1) + y*(y+1)/2 + 3*z(z+1)/2) with x, y and z nonnegative integers.
See also A264025 for similar conjectures.

Examples

			a(5) = 1 since 5 = 0^2 + 1*2 + 2*3/2 with 2 prime.
a(6) = 1 since 6 = 1^2 + 1*2 + 2*3/2 with 2 prime.
a(10) = 1 since 10 = 1^2 + 2*3 + 2*3/2 with 2 prime.
a(11) = 1 since 11 = 2^2 + 2*3 + 1*2/2 with 2 prime.
a(15) = 1 since 15 = 0^2 + 3*4 + 2*3/2 with 3 prime.
a(20) = 1 since 20 = 2^2 + 2*3 + 4*5/2 with 2 and 5 both prime.
a(29) = 1 since 29 = 4^2 + 3*4 + 1*2/2 with 3 and 2 both prime.
a(1125) = 1 since 1125 = 33^2 + 5*6 + 3*4/2 with 5 and 3 both prime.

Links

Crossrefs

Cf. A000040, A000217, A000290, A262785, A263998, A264025.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[(PrimeQ[y]||PrimeQ[y+1])==False,Goto[aa]];Do[If[(PrimeQ[z]||PrimeQ[z+1])&&SQ[n-y(y+1)-z(z+1)/2],r=r+1],{z,1,(Sqrt[8(n-y(y+1))+1]-1)/2}];Label[aa];Continue,{y,1,(Sqrt[4n+1]-1)/2}];Print[n, " ", r];Continue, {n,1,100}]

A264025 Number of ways to write n as x^2 + y*(2*y+1) + z*(z+1)/2 where x, y and z are nonnegative integers with z or z+1 prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Nov 01 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 8, 9, 23, 30, 44, 48, 198, 219, 1344.
(ii) Any positive integer n not equal to 8 can be written as x*(2*x+1) + y*(y+1)/2 + z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime.
(iii) Any integer n > 1 can be written as x^2 + y*(y+1) + z*(z+1) (or 2*x^2 + y*(y+1)/2 + z*(z+1)), where x, y and z are nonnegative integers with z or z+1 prime.
(iv) Each integer n > 2 can be written as x^2 + y*(y+1)/2 + 3*z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime.
(v) Every n = 1,2,3,... can be written as 2*x^2 + y*(y+1)/2 + z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime. Also, any integer n > 4 can be written as 2*x^2 + y*(y+1) + z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime.
Note that the integers n*(2*n+1) = 2n*(2n+1)/2 (n = 0,1,2,...) are second hexagonal numbers.
See also A262785, A263998 and A264010 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0^2 + 0*(2*0+1) + 1*2/2 with 2 prime.
a(2) = 1 since 2 = 1^2 + 0*(2*0+1) + 1*2/2 with 2 prime.
a(3) = 1 since 3 = 0^2 + 0*(2*0+1) + 2*3/2 with 2 prime.
a(8) = 1 since 8 = 2^2 + 1*(2*1+1) + 1*2/2 with 2 prime.
a(9) = 1 since 9 = 0^2 + 1*(2*1+1) + 3*4/2 with 3 prime.
a(23) = 1 since 23 = 1^2 + 3*(2*3+1) + 1*2/2 with 2 prime.
a(30) = 1 since 30 = 3^2 + 0*(2*0+1) + 6*7/2 with 7 prime.
a(44) = 1 since 44 = 4^2 + 0*(2*0+1) + 7*8/2 with 7 prime.
a(48) = 1 since 48 = 3^2 + 4*(2*4+1) + 2*3/2 with 2 prime.
a(198) = 1 since 198 = 3^2 + 4*(2*4+1) + 17*18/2 with 17 prime.
a(219) = 1 since 219 = 6^2 + 7*(2*7+1) + 12*13/2 with 13 prime.
a(1344) = 1 since 1344 = 21^2 + 0*(2*0+1) + 42*43/2 with 43 prime.

Links

Crossrefs

Cf. A000040, A000217, A000290, A014105, A262785, A263998, A264010.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[(PrimeQ[z]||PrimeQ[z+1])==False,Goto[aa]];Do[If[SQ[n-z(z+1)/2-y(2y+1)],r=r+1],{y,0,(Sqrt[8(n-z(z+1)/2)+1]-1)/4}];Label[aa];Continue,{z,1,(Sqrt[8n+1]-1)/2}];Print[n, " ", r];Continue, {n,1,100}]
Showing 1-3 of 3 results.

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