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.

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A271026 Number of ordered ways to write n as x^7 + y^4 + z^3 + w*(3w+1)/2, where x, y, z are nonnegative integers, and w is an integer.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 29 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 47, 61, 62, 112, 175, 448, 573, 714, 1073, 1175, 1839, 2167, 8043, 13844.
(ii) Any natural number can be written as 3*x^6 + y^4 + z^3 + w*(3w+1)/2, where x, y, z are nonnegative integers and w is an integer.
(iii) For every a = 3, 4, 5, 9, 12, any natural number can be written as a*x^5 + y^4 + z^3 + w*(3w+1)/2, where x, y, z are nonnegative integers and w is an integer. Also, any natural number can be written as x^5 + 2*y^4 + 2*z^3 + w*(3w+1)/2 (or 3*x^5 + 2*y^4 + z^3 + w*(3w+1)/2), where x, y, z are nonnegative integers and w is an integer.
We have verified that a(n) > 0 for n up to 2*10^6.
See also A266968 for a related conjecture.

Examples

			a(47) = 1 since 47 = 1^7 + 2^4 + 2^3 + (-4)*(3*(-4)+1)/2.
a(61) = 1 since 61 = 1^7 + 1^4 + 2^3 + (-6)*(3*(-6)+1)/2.
a(62) = 1 since 62 = 0^7 + 0^4 + 3^3 + (-5)*(3*(-5)+1)/2.
a(112) = 1 since 112 = 1^7 + 3^4 + 2^3 + (-4)*(3*(-4)+1)/2.
a(175) = 1 since 175 = 1^7 + 3^4 + 1^3 + (-8)*(3*(-8)+1)/2.
a(448) = 1 since 448 = 2^7 + 4^4 + 4^3 + 0*(3*0+1)/2.
a(573) = 1 since 573 = 1^7 + 4^4 + 6^3 + 8*(3*8+1)/2.
a(714) = 1 since 714 = 2^7 + 4^4 + 0^3 + (-15)*(3*(-15)+1)/2.
a(1073) = 1 since 1073 = 0^7 + 2^4 + 10^3 + 6*(3*6+1)/2.
a(1175) = 1 since 1175 = 0^7 + 5^4 + 5^3 + (-17)*(3*(-17)+1)/2.
a(1839) = 1 since 1839 = 1^7 + 4^4 + 5^3 + 31*(3*31+1)/2.
a(2167) = 1 since 2167 = 1^7 + 5^4 + 11^3 + (-12)*(3*(-12)+1)/2.
a(8043) = 1 since 8043 = 1^7 + 2^4 + 20^3 + 4*(3*4+1)/2.
a(13844) = 1 since 13844 = 3^7 + 2^4 + 21^3 + (-40)*(3*(-40)+1)/2.

Links

Crossrefs

Cf. A000326, A000578, A000583, A000584, A001015, A001318, A262813, A262815, A262816, A262827, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920.

Programs

  • Mathematica
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[pQ[n-x^7-y^4-z^3],r=r+1],{x,0,n^(1/7)},{y,0,(n-x^7)^(1/4)},{z,0,(n-x^7-y^4)^(1/3)}];Print[n," ",r];Continue,{n,0,80}]

A271106 Number of ordered ways to write n as x^6 + 3*y^3 + z^3 + w*(w+1)/2, where x and y are nonnegative integers, and z and w are positive integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 30 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 6, 9, 13, 16, 20, 21, 24, 25, 44, 50, 51, 65, 84, 189, 290, 484, 616, 664, 680, 917, 1501, 1639, 3013.
Based on our computation, we also formulate the following general conjecture.
General Conjecture: Let T(w) = w*(w+1)/2. We have {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...} for any of the following polynomials P(x,y,z,w): x^3+y^3+c*z^3+T(w) (c = 2,3,4,6), x^3+y^3+c*z^3+2*T(w) (c = 2,3), x^3+b*y^3+3z^3+3*T(w) (b = 1,2), x^3+2y^3+3z^3+w(5w-1)/2, x^3+2y^3+3z^3+w(5w-3)/2, x^3+2y^3+c*z^3+T(w) (c = 2,3,4,5,6,7,12,20,21,34,35,40), x^3+2y^3+c*z^3+2*T(w) (c = 3,4,5,6,11), x^3+2y^3+c*z^3+w^2 (c = 3,4,5,6), x^3+2y^3+4z^3+w(3w-1)/2, x^3+2y^3+4z^3+w(3w+1)/2, x^3+2y^3+4z^3+w(2w-1), x^3+2y^3+6z^3+w(3w-1)/2, x^3+3y^3+c*z^3+T(w) (c = 3,4,5,6,10,11,13,15,16,18,20), x^3+3y^3+c*z^3+2*T(w) (c = 5,6,11), x^3+4y^3+c*z^3+T(w) (c = 5,10,12,16), x^3+4y^3+5z^3+2*T(w), x^3+5y^3+10z^3+T(w), 2x^3+3y^3+c*z^3+T(w) (c = 4,6), 2x^3+4y^3+8z^3+T(w), x^4+y^3+3z^3+w(3w-1)/2, x^4+y^3+c*z^3+T(w) (c = 2,3,4,5,7,12,13), x^4+y^3+c*z^3+2*T(w) (c = 2,3,4,5), x^4+y^3+2z^3+w^2, x^4+y^3+4z^3+2w^2, x^4+2y^3+c*z^3+T(w) (c = 4,5,12), x^4+2y^3+3z^3+2*T(w), 2x^4+y^3+2z^3+w(3w-1)/2, 2x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,6,10,11), 2x^4+y^3+c*z^3+2*T(w) (c = 2,3,4), 2x^4+2y^3+c*z^3+T(w) (c = 3,5), 3x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,11), 3x^4+y^3+2z^3+2*T(w), 3x^4+y^3+2z^3+w^2, 3x^4+y^3+2z^3+w(3w-1)/2, 4x^4+y^3+c*z^3+T(w) (c = 2,3,4,6), 4x^4+y^3+2z^3+2*T(w), 5x^4+y^3+c*z^3+T(w) (c = 2,4), a*x^4+y^3+2z^3+T(w) (a = 6,20,28,40), 6x^4+y^3+2z^3+2*T(w), 6x^4+y^3+2z^3+w^2, a*x^4+y^3+3z^3+T(w) (a = 6,8,11), 8x^4+2y^3+4z^3+T(w), x^5+y^3+c*z^3+T(w) (c = 2,3,4), x^5+2y^3+c*z^3+T(w) (c = 3,6,8), 2x^5+y^3+4z^3+T(w), 3x^5+y^3+2z^3+T(w), 5x^5+y^3+c*z^3+T(w) (c = 2,4), x^6+y^3+3z^3+T(w), x^7+y^3+4z^3+T(w), x^4+2y^4+z^3+w^2, x^4+2y^4+2z^3+T(w), x^4+b*y^4+z^3+T(w) (b = 2,3,4), 2x^4+3y^4+z^3+T(w), a*x^5+y^4+z^3+T(w) (a = 1,2), x^5+2y^4+z^3+T(w).
The polynomials listed in the general conjecture should exhaust all those polynomials P(x,y,z,w) = a*x^i+b*y^j+c*z^k+w*(s*w+/-t)/2 with {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...}, where a,b,c,s > 0, 0 <= t <= s, s == t (mod 2), i >= j >= k >= 3, a <= b if i = j, and b <= c if j = k.

Examples

			a(9) = 1 since 9 = 0^6 + 3*0^6 + 2^3 + 1*2/2.
a(24) = 1 since 24 = 1^6 + 3*0^6 + 2^3 + 5*6/2.
a(1501) = 1 since 1501 = 2^6 + 3*5^3 + 3^3 + 45*46/2.
a(1639) = 1 since 1639 = 0^6 + 3*6^3 + 1^3 + 44*45/2.
a(3013) = 1 since 3013 = 3^6 + 3*3^3 + 13^3 + 3*4/2.

Links

Crossrefs

Cf. A000217, A000290, A000326, A000578, A000583, A000584, A001014, A005449, A262813, A262824, A262827, A262857, A262880, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^6-3*y^3-z^3],r=r+1],{x,0,n^(1/6)},{y,0,((n-x^6)/3)^(1/3)},{z,1,(n-x^6-3y^3)^(1/3)}];Print[n," ",r];Continue,{n,0,70}]

A271325 Number of ordered ways to write n as x^3 + y^2 + z*(3z+1), where x, y and z are integers with x positive and y nonnegative.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 04 2016

Keywords

Comments

We guess that a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 6, 7, 8, 13, 18, 20, 23, 25, 44, 49, 55, 59, 121, 238.
Based on our computation, we propose the following general conjecture (which extends the conjectures in A262813 and A270469).
Conjecture: Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Every natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers, where P(x,y,z) is any of the following cubic polynomials: x^3+T(y)+z^2, a*x^3+T(y)+pen(z) (a = 1,2,3,4), x^3+T(y)+z*(5z+1)/2, x^3+T(y)+z*(3z+r) (r = 1,2), x^3+T(y)+z*(7z+3)/2, x^3+T(y)+z*(9z+j)/2 (j = 5,7), x^3+T(y)+z*(5z+r) (r = 2,3), x^3+T(y)+2z*(3z+r) (r = 1,2), x^3+T(y)+z*(6z+5), x^3+T(y)+z*(13z+j)/2 (j = 3,7,9), x^3+T(y)+z*(7z+k) (k = 2,6), a*x^3+y^2+pen(z) (a = 1,2,3,4), x^3+y^2+z*(5z+3)/2, x^3+y^2+2*pen(z), x^3+2*T(y)+pen(z), x^3+2*T(y)+z(5z+j)/2 (j = 1,3), a*x^3+2*T(y)+z*(3z+2) (a = 1,2,3), x^3+2*T(y)+z*(7z+3)/2, x^3+4*T(y)+pen(z), x^3+2y^2+pen(z), x^3+pen(y)+c*pen(z) (c = 1,2,3,4), x^3+b*pen(y)+z*(5z+j)/2 (b = 1,2; j = 1,3), x^3+pen(y)+z*(7z+k)/2 (k = 1,3,5), x^3+pen(y)+z*(4z+j) (j = 1,3), x^3+pen(y)+z*(9z+5)/2, a*x^3+pen(y)+z*(9z+r)/2 (a = 1,2; r = 1,7), x^3+pen(y)+z*(5z+r) (r = 1,2,3,4), a*x^3+pen(y)+z*(11z+9)/2 (a = 1,2), x^3+pen(y)+2z*(3z+2),x^3+pen(y)+z*(13z+11)/2, x^3+pen(y)+z*(7z+k) (k = 4,5,6), x^3+pen(y)+3z*(5z+3)/2, x^3+pen(y)+z*(15z+11)/2, x^3+pen(y)+z*(8z+7), x^3+pen(y)+z*(11z+7), x^3+2*pen(y)+z*(7z+j)/2 (j = 1,5), x^3+2*pen(y)+3*pen(z), x^3+2*pen(y)+z*(4z+1), x^3+2*pen(y)+z*(7z+2), x^3+y*(5y+j)/2+z*(7z+k)/2 (j = 1,3; k = 3,5), x^3+y*(5y+3)/2+z*(9z+7)/2, x^3+y*(3y+2)+z*(4z+1), x^3+y*(3y+2)+z*(5z+1)/2, x^3+y*(7y+3)/2+z*(7z+5)/2, 2x^3+T(y)+z*(5z+3)/2, 2x^3+T(y)+z*(3z+r) (r = 1,2), 2x^3+T(y)+z*(5z+4), 2x^3+2*T(y)+z*(5z+3)/2, 2x^3+3*T(y)+pen(z), 2x^3+y^2+2*pen(z), 2x^3+pen(y)+pen(z), a*x^3+pen(y)+3*pen(z) (a = 2,3,4), a*x^3+pen(y)+z*(7z+5)/2 (a = 2,3,4), 2x^3+pen(y)+z*(5z+k) (k = 1,3), 2x^3+y*(5y+3)/2+z*(7z+5)/2, 2x^3+2*pen(y)+z*(3z+2), 2x^3+2*pen(y)+z*(7z+5)/2, 2x^3+y*(3y+2)+z*(4z+3), 3x^3+pen(y)+z*(7z+3)/2, 4x^3+y^2+z*(5z+1)/2, 4x^3+pen(y)+z*(4z+3).
The listed ternary polynomials in the conjecture should exhaust all those P(x,y,z) = a*x^3+y*(s*y+t)/2+z*(u*z+v)/2 with a,s,u > 0, 0 <= t <= s, 0 <= v <= u, s == t (mod 2), u == v (mod 2), and (s-2t)*(u-2v) nonzero, such that any natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers. Note that those numbers y*(2y+1) with y integral are just triangular numbers.
Conjecture verified for all polynomials up to 10^11. - Mauro Fiorentini, Aug 03 2023
See also A271106 for another general conjecture on universal sums.

Examples

			a(13) = 1 since 13 = 2^3 + 1^2 + 1*(3*1+1).
a(18) = 1 since 18 = 2^3 + 0^2 + (-2)*(3*(-2)+1).
a(20) = 1 since 20 = 1^3 + 3^2 + (-2)*(3*(-2)+1).
a(23) = 1 since 23 = 2^3 + 1^2 + 2*(3*2+1).
a(25) = 1 since 25 = 1^3 + 0^2 + (-3)*(3*(-3)+1).
a(44) = 1 since 44 = 2^3 + 6^2 + 0*(3*0+1).
a(49) = 1 since 49 = 1^3 + 2^2 + (-4)*(3*(-4)+1).
a(55) = 1 since 55 = 3^3+ 2^2 + (-3)*(3*(-3)+1).
a(59) = 1 since 59 = 2^3 + 7^2 + (-1)*(3*(-1)+1).
a(121) = 1 since 121 = 3^3 + 8^2 + 3*(3*3+1).
a(238) = 1 since 238 = 4^3 + 12^2 + 3*(3*3+1).

Links

Crossrefs

Cf. A000217, A000290, A000326, A000578, A001318, A160325, A160326, A262813, A262815, A262816, A270488, A270469, A270516, A270533, A270559, A270566, A271106.

Programs

  • Mathematica
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[12n+1]]
Do[r=0;Do[If[pQ[n-x^3-y^2],r=r+1],{x,1,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Label[aa];Continue,{n,1,70}]

A275169 Positive integers not in the form x^3 + 2*y^2 + z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

15, 21, 47, 53, 79, 85, 92, 111, 117, 120, 181, 183, 245, 309, 311, 335, 372, 373, 398, 405, 421, 437, 447, 501, 565, 573, 629, 636, 645, 655, 693, 757, 791, 807, 820, 821, 853, 869, 885, 888, 949, 967, 1013, 1045, 1077, 1141, 1205, 1223, 1269, 1271, 1303, 1461, 1555, 1591, 1613, 1653, 2087, 2101, 2255, 2421
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 18 2016

Keywords

Comments

Conjecture: The sequence has totally 174 terms as listed in the b-file the largest of which is 375565.
This implies the conjecture in A275150. We note that the sequence contains no term greater than 375565 and not exceeding 10^6.
See also A275168 for a similar conjecture.

Examples

			a(1) = 15 since 15 is the least positive integer not in the form x^3 + 2*y^2 + z^2 with x,y,z nonnegative integers.

Links

Crossrefs

Cf. A000290, A000578, A022551, A022552, A262813, A270488, A274274, A275083, A275150, A275168.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
n=0;Do[Do[If[SQ[m-x^3-2*y^2],Goto[aa]],{x,0,m^(1/3)},{y,0,Sqrt[(m-x^3)/2]}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,2421}]

A262880 Number of ordered ways to write n as w*(w+1)/2 + x^3 + y^3 + 2*z^3 with w > 0, 0 <= x <= y and z >= 0.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 04 2015

Keywords

Comments

Conjecture: (i) Any positive integer can be written as w*(w+1)/2 + x^3 + b*y^3 + c*z^3 with w > 0 and x,y,z >= 0, provided that (b,c) is among the following ordered pairs: (1,2),(1,3),(1,4),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(2,20),(2,21),(2,34),(3,3),(3,4),(3,5),(3,6),(4,10).
(ii) For (b,c) = (3,4),(3,6),(4,8), we have {w*(w+1)/2 + 2*x^3 + b*y^3 + c*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A262813, A262824 and A262857 for similar conjectures.

Examples

			a(2) = 1 since 2 = 1*2/2 + 0^3 + 1^3 + 2*0^3.
a(34) = 2 since 34 = 4*5/2 + 0^3 + 2^3 + 2*2^3 = 3*4/2 + 1^3 + 3^3 + 2*0^3.
a(41) = 1 since 41 = 3*4/2 + 2^3 + 3^3 + 2*0^3.
a(51) = 1 since 51 = 6*7/2 + 1^3 + 3^3 + 2*1^3.
a(104) = 1 since 104 = 5*6/2 + 2^3 + 3^3 + 2*3^3.

Links

Crossrefs

Cf. A000217, A000578, A262813, A262824, A262857.

Programs

  • Mathematica
    TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-y^3-2*z^3],r=r+1],{x,0,(n/2)^(1/3)},{y,x,(n-x^3)^(1/3)},{z,0,((n-x^3-y^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,1,100}]

A262887 Number of ordered ways to write n as x^3 + y^2 + pi(z^2) (x >= 0, y >= 0 and z > 0) with z-1 or z+1 prime, where pi(m) denotes the number of primes not exceeding m.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 04 2015

Keywords

Comments

Conjectures:
(i) a(n) > 0 for all n > 0. Also, each natural number can be written as x^2 + y^2 + pi(z^2) (0 <= x <= y and z > 0) with z-1 or z+1 prime.
(ii) Any integer n > 1 can be written as x^3 + y^2 + pi(z^2) with x >= 0, y >= 0 and z > 0 such that y or z is prime.
(iii) Any integer n > 1 can be written as x^3 + pi(y^2) + pi(z^2) (x >= 0, y > 0 and z > 0) with y or z prime. Also, each integer n > 1 can be written as x^2 + pi(p^2) + pi(q^2) (x >= 0 and p >= q > 0) with p prime.
Compare these conjectures with the conjectures in A262746.

Examples

			a(22) = 2 since 22 = 0^3 + 4^2 + pi(4^2) = 0^3 + 2^2 + pi(8^2) with 4+1 = 5 and 8-1 = 7 both prime.
a(24) = 1 since 24 = 2^3 + 4^2 + pi(1^2) with 1+1 = 2 prime.
a(40) = 2 since 40 = 0^3 + 6^2 + pi(3^2) = 3^3 + 3^2 + pi(3^2) with 3-1 = 2 prime.
a(57) = 1 since 57 = 2^3 + 7^2 + pi(1^2) with 1+1 = 2 prime.
a(73) = 1 since 73 = 4^3 + 3^2 + pi(1^2) with 1+1 = 2 prime.

Links

Crossrefs

Cf. A000040, A000290, A000578, A000720, A262311, A262746, A262785, A262813.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
f[n_]:=PrimePi[n^2]
Do[r=0;Do[If[f[k]>n,Goto[aa]];If[PrimeQ[k-1]==False&&PrimeQ[k+1]==False,Goto[bb]];Do[If[SQ[n-f[k]-x^3],r=r+1],{x,0,(n-f[k])^(1/3)}];Label[bb];Continue,{k,1,n}];Label[aa];Print[n," ",r];Continue, {n,1,100}]

A262982 Number of ordered ways to write n as x^4 + phi(y^2) + z*(z+1)/2 with x >= 0, y > 0 and z > 0, where phi(.) is Euler's totient function given by A000010.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 06 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 6, 20, 26, 40, 82, 89, 105, 305, 416, 470, 725, 6135, 25430, 90285.
Compare this with the conjectures in A262311, A262785 and A262813.

Examples

			a(2) = 1 since 1 = 0^4 + phi(1^2) + 1*2/2.
a(6) = 1 since 6 = 1^4 + phi(2^2) + 2*3/2.
a(20) = 1 since 20 = 2^4 + phi(1^2) + 2*3/2.
a(26) = 1 since 26 = 0^4 + phi(5^2) + 3*4/2.
a(40) = 1 since 40 = 0^4 + phi(6^2) + 7*8/2.
a(82) = 1 since 82 = 0^4 + phi(9^2) + 7*8/2.
a(89) = 1 since 89 = 3^4 + phi(2^2) + 3*4/2.
a(105) = 1 since 105 = 0^4 + phi(14^2) + 6*7/2.
a(305) = 1 since 305 = 4^4 + phi(12^2) + 1*2/2.
a(416) = 1 since 416 = 4^4 + phi(10^2) + 15*16/2.
a(470) = 1 since 470 = 2^4 + phi(12^2) + 28*29/2.
a(725) = 1 since 725 = 2^4 + phi(3^2) + 37*38/2.
a(6135) = 1 since 6135 = 6^4 + phi(81^2) + 30*31/2.
a(25430) = 1 since 25430 = 5^4 + phi(152^2) + 166*167/2.
a(90285) = 1 since 90285 = 16^4 + phi(212^2) + 73*74/2.

Links

Crossrefs

Cf. A000010, A000217, A000290, A000583, A002618, A262311, A262781, A262785, A262813, A262941, A262979.

Programs

  • Mathematica
    f[n_]:=EulerPhi[n^2]
TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[f[x]>n,Goto[aa]];Do[If[TQ[n-f[x]-y^4],r=r+1],{y,0,(n-f[x])^(1/4)}];Label[aa];Continue,{x,1,n}];Print[n," ",r];Continue,{n,1,100}]

A270616 Number of ordered ways to write n as the sum of a positive square, the square of a triangular number, and a generalized pentagonal number (A001318).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 20 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 29, 34, 5949, 10913.
See also A270566 and A270594 for more similar conjectures.
By the author's work in Sci. China Math. 58(2015), any natural number can be written as the sum of a triangular number, a square and a generalized pentagonal number.

Examples

			a(1) = 1 since 1 = 1^2 + (0*1/2)^2 + 0*(3*0+1)/2.
a(8) = 1 since 8 = 1^2 + (0*1/2)^2 + 2*(3*2+1)/2.
a(29) = 1 since 29 = 4^2 + (1*2/2)^2 + 3*(3*3-1)/2.
a(34) = 1 since 34 = 5^2 + (2*3/2)^2 + 0*(3*0+1)/2.
a(5949) = 1 since 5949 = 47^2 + (10*11/2)^2 + 22*(3*22-1)/2.
a(10913) = 1 since 10913 = 23^2 + (2*3/2)^2 +83*(3*83+1)/2.

Links

Crossrefs

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

Programs

  • Mathematica
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[pQ[n-x^2-(y(y+1)/2)^2],r=r+1],{x,1,Sqrt[n]},{y,0,(Sqrt[8*Sqrt[n-x^2]+1]-1)/2}];Print[n," ",r];Continue,{n,1,80}]

A273917 Number of ordered ways to write n as w^2 + 3*x^2 + y^4 + z^5, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 04 2016

Keywords

Comments

Conjectures:
(i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 7, 11, 12, 15, 19, 24, 27, 31, 34, 35, 43, 46, 47, 56, 70, 71, 72, 87, 88, 115, 136, 137, 147, 167, 168, 178, 207, 235, 236, 267, 286, 297, 423, 537, 747, 762, 1017.
(ii) Any positive integer n can be written as w^2 + x^4 + y^5 + pen(z), where w is a positive integer, x,y,z are nonnegative integers, and pen(z) denotes the pentagonal number z*(3*z-1)/2.
Conjectures a(n) > 0 and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 19 2023
See also A262813, A262857, A270566, A271106 and A271325 for some other conjectures on representations.

Examples

			a(1) = 1 since 1 = 1^2 + 3*0^2 + 0^4 + 0^5.
a(3) = 1 since 3 = 1^2 + 3*0^2 + 1^4 + 1^5.
a(7) = 1 since 7 = 2^2 + 3*1^2 + 0^4 + 0^5.
a(11) = 1 since 11 = 3^2 + 3*0^2 + 1^4 + 1^5.
a(12) = 1 since 12 = 3^2 + 3*1^2 + 0^4 + 0^5.
a(15) = 1 since 15 = 1^2 + 3*2^2 + 1^4 + 1^5.
a(19) = 1 since 19 = 4^2 + 3*1^2 + 0^4 + 0^5.
a(24) = 1 since 24 = 2^2 + 3*1^2 + 2^4 + 1^5.
a(27) = 1 since 27 = 5^2 + 3*0^2 + 1^4 + 1^5.
a(31) = 1 since 31 = 2^2 + 3*3^2 + 0^4 + 0^5.
a(34) = 1 since 34 = 1^2 + 3*0^2 + 1^4 + 2^5.
a(35) = 1 since 35 = 4^2 + 3*1^2 + 2^4 + 0^5.
a(43) = 1 since 43 = 4^2 + 3*3^2 + 0^4 + 0^5.
a(46) = 1 since 46 = 1^2 + 3*2^2 + 1^4 + 2^5.
a(47) = 1 since 47 = 2^2 + 3*3^2 + 2^4 + 0^5.
a(56) = 1 since 56 = 6^2 + 3*1^2 + 2^4 + 1^5.
a(70) = 1 since 70 = 5^2 + 3*2^2 + 1^4 + 2^5.
a(71) = 1 since 71 = 6^2 + 3*1^2 + 0^4 + 2^5.
a(72) = 1 since 72 = 6^2 + 3*1^2 + 1^4 + 2^5.
a(87) = 1 since 87 = 6^2 + 2*1^2 + 2^4 + 2^5.
a(88) = 1 since 88 = 2^2 + 3*1^2 + 3^4 + 0^5.
a(115) = 1 since 115 = 8^2 + 3*1^2 + 2^4 + 2^5.
a(136) = 1 since 136 = 10^2 + 3*1^2 + 1^4 + 2^5.
a(137) = 1 since 137 = 11^2 + 3*0^2 + 2^4 + 0^5.
a(147) = 1 since 147 = 12^2 + 3*1^2 + 0^4 + 0^5.
a(167) = 1 since 167 = 2^2 + 3*7^2 + 2^4 + 0^5.
a(168) = 1 since 168 = 2^2 + 3*7^2 + 2^4 + 1^5.
a(178) = 1 since 178 = 7^2 + 3*4^2 + 3^4 + 0^5.
a(207) = 1 since 207 = 10^2 + 3*5^2 + 0^4 + 2^5.
a(235) = 1 since 235 = 12^2 + 3*5^2 + 2^4 + 0^5.
a(236) = 1 since 236 = 12^2 + 3*5^2 + 2^4 + 1^5.
a(267) = 1 since 267 = 12^2 + 3*5^2 + 2^4 + 2^5.
a(286) = 1 since 286 = 4^2 + 3*3^2 + 0^4 + 3^5.
a(297) = 1 since 297 = 3^2 + 3*0^2 + 4^4 + 2^5.
a(423) = 1 since 423 = 11^2 + 3*10^2 + 1^4 + 1^5.
a(537) = 1 since 537 = 21^2 + 3*4^2 + 2^4 + 2^5.
a(747) = 1 since 747 = 11^2 + 3*0^2 + 5^4 + 1^5.
a(762) = 1 since 762 = 27^2 + 3*0^2 + 1^4 + 2^5.
a(1017) = 1 since 1017 = 27^2 + 3*0^2 + 4^4 + 2^5.

Links

Crossrefs

Cf. A000118, A000290, A000326, A000583, A000584, A262813, A262827, A262857, A270566, A270969, A271076, A271106, A273429, A273915.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-3*x^2-y^4-z^5],r=r+1],{x,0,Sqrt[(n-1)/3]},{y,0,(n-1-3x^2)^(1/4)},{z,0,(n-1-3x^2-y^4)^(1/5)}];Print[n," ",r];Continue,{n,1,80}]

A275083 Positive integers congruent to 0 or 1 modulo 4 that cannot be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

120, 312, 813, 2136, 2680, 3224, 4404, 5340, 6420, 10060, 11320, 11824, 14008, 15856, 26544, 28804, 34392, 47984
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 15 2016

Keywords

Comments

Conjecture: (i) The sequence has totally 18 terms as listed.
(ii) For each r = 2,3 there are infinitely many positive integers n == r (mod 4) not in the form x^3 + y^2 + z^2 with x,y,z nonnegative integers.
Our computation indicates that the sequence has no other terms below 10^6.
Let d be 2 or 6. Clearly, n-d is congruent to 0 or 1 modulo 4 if n is congruent to 2 or 3 modulo 4. So part (i) of the conjecture essentially implies that for each n = 0,1,2,... either n or n-d can be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

Examples

			a(1) = 120 since all those positive integers congruent to 0 or 1 modulo 4 and smaller than 120 can be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers but 120 (divisible by 4) cannot be written in this way.

Crossrefs

Cf. A000290, A000578, A022551, A022552, A262813, A262857, A272979, A274274.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
n=0;Do[If[Mod[m,4]>1,Goto[aa]];Do[If[SQ[m-x^3-y^2],Goto[aa]],{x,0,m^(1/3)},{y,0,Sqrt[(m-x^3)/2]}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,50000}]
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