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.

Previous Showing 21-24 of 24 results.

A270921 Number of ordered ways to write n as x*(3x+2) + y*(5y+1)/2 - z^4, where x and y are integers, and z is a nonnegative integer with z^4 <= n.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 25 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 0, 5, 6, 12, 13, 14, 112, 193, 194, 200, 242, 333, 345, 376, 492, 528, 550, 551, 613, 797, 1178, 1195, 1222, 1663, 3380, 3635, 6508, 8755, 9132, 12434, 20087.
Compare this conjecture with the conjecture in A270566.

Examples

			a(5) = 1 since 5 = 1*(3*1+2) + 0*(5*0+1)/2 - 0^4.
a(6) = 1 since 6 = 1*(3*1+2) + (-1)*(5*(-1)+1)/2 - 1^4.
a(13) = 1 since 13 = 1*(3*1+2) + (-2)*(5*(-2)+1)/2 - 1^4.
a(376) = 1 since 376 = 0*(3*0+2) + (-16)*(5*(-16)+1)/2 - 4^4.
a(9132) = 1 since 9132 = (-13)*(3*(-13)+2) + 59*(5*59+1)/2 - 3^4.
a(12434) = 1 since 12434 = (-21)*(3*(-21)+2) + 78*(5*78+1)/2 - 8^4.
a(20087) = 1 since 20087 = 19*(3*19+2) + 87*(5*87+1)/2 - 0^4.
5, 6, 12, 13, 14, 112, 193, 194, 200, 242, 333, 345, 376, 492, 528, 550, 551, 613, 797, 1178, 1195, 1222, 1663, 3380, 3635, 6508, 8755, 9132, 12434, 20087
		

Crossrefs

Programs

  • Mathematica
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[40n+1]]&&(Mod[Sqrt[40n+1],10]==1||Mod[Sqrt[40n+1],10]==9)
    Do[r=0;Do[If[pQ[n+x^4-y(3y+2)],r=r+1],{x,0,n^(1/4)},{y,-Floor[(Sqrt[3(n+x^4)+1]+1)/3],(Sqrt[3(n+x^4)+1]-1)/3}];Print[n," ",r];Continue,{n,0,80}]

A270928 Number of ways to write n = x*(x-1)/2 + y*(y-1)/2 + z*(z-1)/2, where 0 < x <= y <= z, and one of x, y, z is prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 26 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for n > 0 with the only exception n = 15^2 = 225. Also, a(n) = 1 only for n = 1, 2, 4, 5, 6, 8, 11, 14, 15, 18, 20, 29, 36, 50, 53, 60, 62, 96, 117, 119, 218, 411, 540, 645, 1125, 1590, 2346, 4068.
(ii) Any positive integer can be written as p*(p-1)/2 + x*(x-1)/2 + P(y) with p prime and x and y integers, where the polynomial P(y) is either of the following ones: y*(y-1), y*(3*y+1)/2, y*(5*y+j)/2 (j = 1,3), y*(3*y+j) (j = 1,2), y*(7*y+3)/2, y*(9*y+j)/2 (j = 1,5,7), y*(5*y+j) (j = 1,3), y*(11*y+9)/2, 2*y*(3*y+j) (j = 1,2), y*(7*y+3).
(iii) Any positive integer can be written as p*(p-1)/2 + P(x,y) with p prime and x and y integers, where the polynomial P(x,y) is either of the following ones: a*x*(x-1)/2+y*(3*y+1)/2 (a = 2,3,4), x*(x-1)+y*(5*y+3)/2, b*x^2+y*(3*y+1)/2 (b = 1,2,3), x^2+y*(5*y+j)/2 (j = 1,3), x^2+y*(3*y+1), x^2+y*(7*y+j)/2 (j = 1,3,5), x^2+y*(4*y+1).
(iv) Every positive integer can be written as p*(p-1)/2+x*(3*x+1)/2+y*(3*y+1)/2 with p prime, x an nonnegative integer and y an integer. Also, for each r = 1,3, any positive integer n can be written as p*(p-1)/2+x*(3*x-1)/2+y*(5*y+r)/2, where p is a prime, and x and y are integers with x nonnegative.
Note that Gauss proved a classical assertion of Fermat which states that any natural number is the sum of three triangular numbers.
See also A270966 for a similar conjecture involving (p-1)^2 with p prime.
The conjecture that a(n) > 0 except for n = 225 appeared as Conjecture 1.2(i) of the author's JNT paper in the links.

Examples

			a(1) = 1 since 1 = 1*(1-1)/2 + 1*(1-1)/2 + 2*(2-1)/2 with 2 prime.
a(4) = 1 since 4 = 1*(1-1)/2 + 2*(2-1)/2 + 3*(3-1)/2 with 2 and 3 prime.
a(29) = 1 since 29 = 1*(1-1)/2 + 2*(2-1)/2 + 8*(8-1)/2 with 2 prime.
a(50) = 1 since 50 = 2*(2-1)/2 + 7*(7-1)/2 + 8*(8-1)/2 with 2 and 7 prime.
a(119) = 1 since 119 = 8*(8-1)/2 + 9*(9-1)/2 + 11*(11-1)/2 with 11 prime.
a(411) = 1 since 411 = 16*(16-1)/2 + 16*(16-1)/2 + 19*(19-1)/2 with 19 prime.
a(1125) = 1 since 1125 = 3*(3-1)/2 + 34*(34-1)/2 + 34*(34-1)/2 with 3 prime.
a(1590) = 1 since 1590 = 7*(7-1)/2 + 37*(37-1)/2 + 43*(43-1)/2 with 7, 37 and 43 prime.
a(2346) = 1 since 2346 = 6*(6-1)/2 + 16*(16-1)/2 + 67*(67-1)/2 with 67 prime.
a(4068) = 1 since 4068 = 7*(7-1)/2 + 34*(34-1)/2 + 84*(84-1)/2 with 7 prime.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
    Do[r=0;Do[If[TQ[n-x(x-1)/2-y(y-1)/2]&&(PrimeQ[x]||PrimeQ[y]||PrimeQ[(Sqrt[8(n-x(x-1)/2-y(y-1)/2)+1]+1)/2]),r=r+1],{x,1,(Sqrt[8n/3+1]+1)/2},{y,x,(Sqrt[8(n-x(x-1)/2)/2+1]+1)/2}];Print[n," ",r];Continue,{n,1,70}]

A270966 Number of ways to write n as x^2 + y^2 + z*(3z+1)/2, where x, y and z are integers with 0 <= x <= y such that x or y has the form p-1 with p prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 27 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 49, 608.
(ii) Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Any positive integer can be written as (p-1)^2+P(x,y) with p prime and x and y integral, where the polynomial P(x,y) is either of the following ones: T(x)+2*pen(y), 2*T(x)+pen(y), T(x)+y*(5y+1)/2, T(x)+y*(9y+5)/2, pen(x)+y*(5y+j)/2 (j = 1,3), pen(x)+y*(7y+k)/2 (k = 3,5), pen(x)+y*(4y+j) (j = 1,3), pen(x)+y*(5y+r) (r = 1,2,3,4), pen(x)+2y*(3y+i) (i = 1,2), pen(x)+6*pen(y), x*(5x+1)/2+y*(3y+2), x*(5x+1)/2+y*(9y+7)/2, x*(5x+3)/2+y*(3y+i) (i = 1,2), x*(5x+3)/2+y*(9y+5)/2.
See also A270928 for a similar conjecture involving T(p-1) = p*(p-1)/2 with p prime.

Examples

			a(1) = 1 since 1 = 0^2 + (2-1)^2 + 0*(3*0+1)/2 with 2 prime.
a(12) = 2 since 12 = (2-1)^2 + 2^2 + 2*(2*3+1)/2 = (2-1)^2 + 3^2 + 1*(3*1+1)/2 with 2 prime.
a(49) = 1 since 49 = (2-1)^2 + 6^2 + (-3)*(3*(-3)+1)/2 with 2 prime.
a(608) = 1 since 608 = (7-1)^2 + 14^2 + (-16)*(3*(-16)+1)/2 with 7 prime.
		

Crossrefs

Programs

  • Mathematica
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
    Do[r=0;Do[If[(PrimeQ[x+1]||PrimeQ[y+1])&&pQ[n-x^2-y^2],r=r+1],{x,0,Sqrt[n/2]},{y,x,Sqrt[n-x^2]}];Print[n," ",r];Continue,{n,1,70}]

A306240 Number of ways to write n as x^9 + y^3 + z*(z+1) + w*(w+1), where x,y,z,w are nonnegative integers with x <= 2 and z <= w.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 31 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 11, 17, 18, 47, 108, 109, 234, 359. Also, any nonnegative integer can be written as x^6 + y^3 + z*(z+1) + w*(w+1), where x,y,z,w are nonnegative integers with x <= 2.
We have verified a(n) > 0 for all n = 0..2*10^7.

Examples

			a(11) = 1 with 11 = 1^9 + 2^3 + 0*1 + 1*2.
a(18) = 1 with 18 = 0^9 + 0^3 + 2*3 + 3*4.
a(109) = 1 with 109 = 1^9 + 4^3 + 1*2 + 6*7.
a(234) = 1 with 234 = 0^9 + 6^3 + 2*3 + 3*4.
a(359) = 1 with 359 = 1^9 + 2^3 + 10*11 + 15*16.
a(1978) = 3 with 1978 = 2^9 + 2^3 + 26*27 + 27*28 = 2^9 + 6^3 + 19*20 + 29*30 = 2^9 + 6^3 + 24*25 + 25*26.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[4n+1]];
    tab={};Do[r=0;Do[If[TQ[n-x^9-y^3-z(z+1)],r=r+1],{x,0,Min[2,n^(1/9)]},{y,0,(n-x^9)^(1/3)},{z,0,(Sqrt[2(n-x^9-y^3)+1]-1)/2}];tab=Append[tab,r],{n,0,100}];Print[tab]
Previous Showing 21-24 of 24 results.