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Search: a266968 -id:a266968
Displaying 1-4 of 4 results found. page 1
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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. +10
4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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.
LINKS
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 29 2016
STATUS
approved
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. +10
4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,6
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.
LINKS
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 30 2016
STATUS
approved
A271076 Number of ordered ways to write n as u^5 + v^4 + x^3 + 2*y^3 + 3*z^3, where u, v , x, y and z are nonnegative integers with v > 0. +10
2
1, 2, 2, 3, 3, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 4, 4, 4, 4, 3, 1, 1, 3, 4, 4, 5, 4, 2, 2, 2, 5, 4, 3, 5, 2, 1, 1, 2, 5, 4, 6, 5, 2, 3, 2, 4, 5, 4, 3, 3, 3, 2, 2, 4, 5, 4, 5, 5, 1, 2, 3, 3, 5, 2, 5, 5, 3, 3, 3, 3, 3, 4, 4, 1, 1, 2, 3, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 9, 10, 11, 13, 14, 15, 16, 23, 24, 38, 39, 61, 76, 77, 104, 118, 188, 214, 229.
(ii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^4+x^3+2*y^3+5*z^3, u^5+2*v^4+x^3+2*y^3+3*z^3, u^5+3*v^4+x^3+2*y^3+3*z^3, 2*u^5+v^4+x^3+y^3+4*z^3, 2*u^5+v^4+x^3+2*y^3+4*z^3, 3*u^5+v^4+x^3+2*y^3+4*z^3, 5*u^5+v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+3*z^3, u^4+2*v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+6*z^3, u^4+2*v^4+x^3+3*y^3+4*z^3, u^4+2*v^4+x^3+4*y^3+5*z^3, u^4+2*v^4+x^3+4*y^3+6*z^3, u^4+2*v^4+x^3+4*y^3+10*z^3, u^4+3*v^4+x^3+2*y^3+3*z^3, u^4+3*v^4+x^3+2*y^3+4*z^3, u^4+3*v^4+x^3+2*y^3+6*z^3, u^4+4*v^4+x^3+y^3+2*z^3, u^4+4*v^4+x^3+2*y^3+3*z^3, u^4+4*v^4+x^3+2*y^3+4*z^3, u^4+5*v^4+x^3+2*y^3+4*z^3, u^4+6*v^4+x^3+2*y^3+3*z^3, u^4+7*v^4+x^3+2*y^3+3*z^3, u^4+9*v^4+x^3+2*y^3+4*z^3, 2*u^4+4*v^4+x^3+2*y^3+3*z^3,2*u^4+6*v^4+x^3+2*y^3+4*z^3, 3*u^4+6*v^4+x^3+2*y^3+4*z^3.
(iii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^3+x^3+2*y^3+4*z^3, u^5+v^3+2*x^3+3*y^3+c*z^3 (c = 6,9), a*u^5+v^3+2*x^3+4*y^3+5*z^3 (a = 1,2,6), b*u^5+v^3+2*x^3+4*y^3+6*z^3 (b = 2,3), u^5+v^3+2*x^3+4*y^3+d*z^3 (d = 9,13).
The listed polynomials in part (ii), together with u^5+v^4+x^3+2*y^3+3*z^3, should essentially exhaust all those polynomials P(u,v,x,y,z) = s*u^k+t*v^j+a*x^3+b*y^3+c*z^3 with s,t,a,b,c positive integers and k >= j > 3, such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
There are also finitely many (but quite a lot) polynomials P(u,v,x,y,z) of the form m*u^4+a*v^3+b*x^3+c*y^3+d*z^3 with a,b,c,d and m positive integers such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A267826, A271099 and A271237 for related comments.
Conjectures (i), (ii) and (iii) verified for n up to 10^11 for all polynomials. - Mauro Fiorentini, Sep 20 2023
LINKS
EXAMPLE
a(16) = 1 since 16 = 0^5 + 2^4 +0^3 + 2*0^3 + 3*0^3.
a(104) = 1 since 104 = 0^5 + 2^4 + 4^3 + 2*0^3 + 3*2^3.
a(188) = 1 since 188 = 2^5 + 1^4 + 3^3 + 2*4^3 + 3*0^3.
a(229) = 1 since 229 = 1^5 + 3^4 + 4^3 + 2*1^3 + 3*3^3.
MATHEMATICA
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0; Do[If[CQ[n-u^5-v^4-3z^3-2y^3], r=r+1], {u, 0, (n-1)^(1/5)}, {v, 1, (n-u^5)^(1/4)}, {z, 0, ((n-u^5-v^4)/3)^(1/3)}, {y, 0, ((n-u^5-v^4-3z^3)/2)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 07 2016
STATUS
approved
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. +10
1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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.
LINKS
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 05 2016
STATUS
approved
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