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 11-18 of 18 results.

A270705 Number of ordered ways to write n as x^2*pen(x) + pen(y) + pen(z) with pen(x) = x*(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal").

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 21 2016

Keywords

Comments

Conjecture: (i) Any natural number can be written as a*x^2*pen(x) + b*pen(y) + c*pen(z) with x, y and z integers, provided that (a,b,c) is among the following ordered triples: (j,1,k) (j = 1,2; k = 1,2,3,4), (1,2,3), (3,1,4) and (4,1,3).
(ii) Every n = 0,1,2,... can be expressed as x^2*pen(x) + T(y) + T(z) with x, y and z integers, where T(m) denotes the triangular number m*(m+1)/2. Also, for each (a,b) = (1,2),(1,4),(2,2), any natural number can be written as a*x^2*T(x) + b*T(y) + T(z) with x, y and z integers.
(iii) Each natural number can be written as x^2*P(x) + pen(y) + pen(z) with x, y and z integers, where P(x) is either of the following polynomials: a*T(x) (a = 1,2,3,4,5), x*(5x+3)/2, x*(3x+1), x*(3x+2), x*(7x+1)/2, x*(4x+1), x*(4x+3), x*(9x+5)/2, x*(5x+3), x*(11x+9)/2, x*(13x+5)/2, x*(17x+9)/2, 3x*(3x+2), x*(11x+2).
See also A270594 and A270706 for other similar conjectures.

Examples

			a(88) = 1 since 88 = 1^2*pen(1) + pen(-5) + pen(-6).

Links

Crossrefs

Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566, A270594, A270616, A270706.

Programs

  • Mathematica
    pen[x_]:=pen[x]=x(3x+1)/2
pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[pQ[n-pen[y]-x^2*pen[x]],r=r+1],{y,-Floor[(Sqrt[12n+1]+1)/6],(Sqrt[12n+1]-1)/6},{x,-1-Floor[(2(n-pen[y])/3)^(1/4)],(2(n-pen[y])/3)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

A270706 Number of ordered ways to write n as x^2*T(x) + y^2 + T(z), where x, y and z are integers with x nonzero, y positive and z nonnegative, and T(m) denotes the triangular number m*(m+1)/2.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 21 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 90, 438, 480, 7108.
(ii) Let pen(x) = x*(3x+1)/2. Any natural number can be written as a*f(x)*g(x) + f(y) + g(z) with x, y and z integers, whenever (a,f(x),g(x)) is among the following ordered triples: (1,T(x),x^2), (1,T(x),pen(x)), (1,T(x),x*(5x+1)/2), (1,T(x),x*(5x+3)/2), (1,T(x),x*(3x+j)) (j = 1,2), (1,pen(x),3*T(x)), (1,pen(x),x*(7x+j)/2) (j = 1,3,5), (1,pen(x),x*(4x+1)), (2,T(x),x^2), (2,T(x),pen(x)), (2,T(x),x(5x+j)/2) (j = 1,3), (2,T(x),x*(3x+j)) (j = 1,2), (2,2*T(x),pen(x)), (2,pen(x),x(7x+j)/2) (j = 3,5), (k,x^2,pen(x)) (k = 1,2,3,4,5,8,11).
(iii) Each natural number can be written as P(x,y,z) with x, y and z integers, where P(x,y,z) is either of the following polynomials: T(x)*x(5x+1)/2+T(y)+2*T(z), a*T(x)*pen(x)+pen(y)+pen(z) (a = 1,2,3,4), T(x)*pen(x)+pen(y)+3*pen(z), T(x)*pen(x)+pen(y)+4*pen(z), 2*T(x)*pen(x)+pen(y)+3*pen(z), pen(x)*x(5x+j)/2+pen(y)+3*pen(z) (j = 1,3), x(3x+2)*pen(x)+pen(y)+4*pen(z), pen(x)*x(7x+1)/2+pen(y)+pen(z), pen(x)*x(9x+7)/2+pen(y)+pen(z).
See also A270594 and A270705 for some other similar conjectures.

Examples

			a(1) = 1 since 1 = (-1)^2*T(-1) + 1^2 + T(0).
a(3) = 1 since 3 = 1^2*T(1) + 1^2 + T(1).
a(90) = 1 since 90 = 3^2*T(3) + 6^2 + T(0).
a(438) = 1 since 438 = 4^2*T(4) + 5^2 + T(22).
a(480) = 1 since 480 = 1^2*T(1) + 17^2 + T(19).
a(7108) = 1 since 1^2*T(1) + 69^2 + T(68).

Links

Crossrefs

Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566, A270594, A270616, A270705.

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^3*(x+1)/2],r=r+1],{y,1,Sqrt[n]},{x,-1-Floor[(2(n-y^2))^(1/4)],(2(n-y^2))^(1/4)}];Print[n," ",r];Continue,{n,1,90}]

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

Links

Crossrefs

Cf. A000583, A001082, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270920.

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.

Links

Crossrefs

Cf. A000040, A000217, A000290, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270966.

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.

Links

Crossrefs

Cf. A000040, A000217, A000290, A001318, A160326, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270928.

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}]

A280153 Number of ways to write n as x^3 + 2*y^3 + z^2 + 4^k, where x is a positive integer and y,z,k are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 27 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 7, 10, 17, 24, 36, 38, 52, 62, 115, 136, 185, 990.
(ii) Each positive integer n can be written as x^2 + 2*y^2 + z^3 + 8^k with x,y,z,k nonnegative integers.
(iii) Let a,b,c be positive integers with b <= c. Then any positive integer can be written as a*x^3 + b*y^2 + c*z^2 + 4^k with x,y,z,k nonnegative integers, if and only if (a,b,c) is among the following triples: (1,1,1), (1,1,2), (1,1,3), (1,1,5), (1,1,6), (1,2,3), (1,2,5), (2,1,1), (2,1,2), (2,1,3), (2,1,6), (4,1,2), (5,1,2), (8,1,2), (9,1,2).
We have verified that a(n) > 0 for all n = 2..10^6, and that part (ii) of the conjecture holds for all n = 1..10^6.
For any positive integer n, it is easy to see that at least one of n-1, n-8, n-64 is not of the form 4^k*(8m+7) with k and m nonnegative integers, thus, by the Gauss-Legendre theorem on sums of three squares, n = x^2 + y^2 + z^2 + 8^k for some nonnegative integers x,y,z and k < 3.
See also A280356 for a similar conjecture involving powers of two.

Examples

			a(7) = 1 since 7 = 1^3 + 2*1^3 + 0^2 + 4^1.
a(10) = 1 since 10 = 2^3 + 2*0^3 + 1^2 + 4^0.
a(17) = 1 since 17 = 1^3 + 2*0^3 + 0^2 + 4^2.
a(24) = 1 since 24 = 2^3 + 2*0^3 + 0^2 + 4^2.
a(36) = 1 since 36 = 2^3 + 2*1^3 + 5^2 + 4^0.
a(38) = 1 since 38 = 1^3 + 2*0^3 + 6^2 + 4^0.
a(52) = 1 since 52 = 3^3 + 2*0^3 + 3^2 + 4^2.
a(62) = 1 since 62 = 2^3 + 2*1^3 + 6^2 + 4^2.
a(115) = 1 since 115 = 2^3 + 2*3^3 + 7^2 + 4^1.
a(136) = 1 since 136 = 2^3 + 2*0^3 + 8^2 + 4^3.
a(185) = 1 since 185 = 3^3 + 2*3^3 + 10^2 + 4^1.
a(990) = 1 since 990 = 7^3 + 2*3^3 + 23^2 + 4^3.

Links

Crossrefs

Cf. A000290, A000302, A000578, A262813, A262827, A262857, A270533, A270559, A280356.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-4^k-x^3-2y^3],r=r+1],{k,0,Log[4,n]},{x,1,(n-4^k)^(1/3)},{y,0,((n-4^k-x^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,1,80}]

A271365 Number of ordered ways to write n as u^2 + v^3 + x^4 + y^5 + z^6, where u is a positive integer, and v, x, y, z are nonnegative integers.

Original entry on oeis.org

1, 4, 6, 5, 5, 6, 4, 1, 2, 7, 9, 6, 4, 3, 1, 1, 6, 12, 10, 4, 3, 3, 1, 1, 6, 12, 11, 7, 6, 4, 2, 4, 9, 12, 8, 5, 10, 12, 6, 2, 5, 9, 8, 8, 10, 6, 2, 3, 8, 13, 10, 8, 11, 8, 2, 1, 6, 10, 8, 7, 6, 2, 2, 7, 15, 20, 14, 9, 13, 11
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 05 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 15, 16, 23, 24, 56. Moreover, the only positive integers not represented by u^2+v^3+x^4+y^5 (u > 0 and v,x,y >= 0) are 8, 15, 23, 55, 62, 71, 471, 478, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471.
Note that 1/2+1/3+1/4+1/5+1/6 = 29/20 < 3/2.

Examples

			a(1) = 1 since 1 = 1^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(8) = 1 since 8 = 2^2 + 1^3 + 1^4 + 1^5 + 1^6.
a(15) = 1 since 15 = 2^2 + 2^3 + 1^4 + 1^5 + 1^6.
a(16) = 1 since 16 = 4^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(23) = 1 since 23 = 2^2 + 1^3 + 2^4 + 1^5 + 1^6.
a(24) = 1 since 24 = 4^2 + 2^3 + 0^4 + 0^5 + 0^6.
a(56) = 1 since 56 = 4^2 + 2^3 + 0^4 + 2^5 + 0^6.

Links

Crossrefs

Cf. A000290, A000578, A000583, A000584, A001014, A262813, A262824, A262827, A262857, A266968, A270488, A270516, A270533, A270559, A270566, A270920, A271026, A271106, A271325.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x1^6-x2^5-x3^4-x4^3],r=r+1],{x1,0,n^(1/6)},{x2,0,(n-x1^6)^(1/5)},{x3,0,(n-x1^6-x2^5)^(1/4)},{x4,0,(n-x1^6-x2^5-x3^4)^(1/3)}];Print[n," ",r];Continue,{n,1,70}]

A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 06 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 23, and a(n) = 1 only for n = 0, 3, 4, 7, 11, 12, 15, 19, 23, 30, 78, 203, 219.
(ii) Any natural number can be written as the sum of two squares and three fourth powers.

Examples

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^3 + 1^3 with 0 even.
a(4) = 1 since 4 = 0^2 + 2^2 + 0^3 + 0^3 with 0 and 2 even.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^3 + 1^3 with 2 even.
a(11) = 1 since 11 = 0^2 + 3^2 + 1^3 + 1^3 with 0 even.
a(12) = 1 since 12 = 0^2 + 2^2 + 0^3 + 2^3 with 0 and 2 even.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^3 + 1^3 with 2 even.
a(19) = 1 since 19 = 1^2 + 4^2 + 1^3 + 1^3 with 4 even.
a(30) = 1 since 30 = 2^2 + 5^2 + 0^3 + 1^3 with 2 even.
a(78) = 1 since 78 = 2^2 + 3^2 + 1^3 + 4^3 with 2 even.
a(203) = 1 since 203 = 7^2 + 10^2 + 3^3 + 3^3 with 10 even.
a(219) = 1 since 219 = 8^2 + 8^2 + 3^3 + 4^3 with 8 even.

Links

Crossrefs

Cf. A000290, A000578, A000583, A262813, A262824, A262827, A262857, A262880, A270488, A270559, A270969, A271325.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^3-y^3-u^2]&&(Mod[u*Sqrt[n-x^3-y^3-u^2],2]==0),r=r+1],{x,0,(n/2)^(1/3)},{y,x,(n-x^3)^(1/3)},{u,0,((n-x^3-y^3)/2)^(1/2)}];Print[n," ",r];Continue,{n,0,80}]
Previous Showing 11-18 of 18 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.