Search: a270533 -id:a270533
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A270969
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Number of ways to write n as w^4 + x^2 + y^2 + z^2, where w, x, y and z are nonnegative integers with x <= y <= z.
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+10
30
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1, 2, 2, 2, 2, 2, 2, 1, 1, 3, 3, 2, 2, 2, 2, 1, 2, 4, 5, 4, 3, 3, 3, 1, 2, 5, 5, 5, 3, 3, 4, 1, 2, 5, 6, 4, 4, 4, 4, 2, 2, 6, 6, 4, 2, 5, 4, 1, 2, 5, 7, 6, 5, 4, 7, 3, 2, 6, 4, 4, 3, 4, 5, 2, 2, 6, 9, 6, 4, 6, 6, 1, 3, 6, 6, 7, 3, 5, 5, 1, 2
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OFFSET
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0,2
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COMMENTS
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Theorem: a(n) > 0 for all n = 0,1,2,.... In other words, any nonnegative integer can be written as the sum of a fourth power and three squares.
This is stronger than Lagrange's four-square theorem, and it can be proved by induction on n. It is easy to check that a(n) > 0 for all n = 0..16. Now let n be an integer greater than 16, and assume that a(m) > 0 for all m = 0..n-1. If 16|n, then n/16 can be written as w^4+x^2+y^2+z^2 with w,x,y,z integers, and hence n = (2w)^4+(4x)^2+(4y)^2+(4z)^2. If n == 8 (mod 16), then n is not of the form 4^k*(8q+7) and hence n = 0^4+x^2+y^2+z^2 for some integers x,y,z. If n == 4 (mod 8), then n-1^4 can be written as the sum of three squares. If n == 2 (mod 4), then n-0^4 is a sum of three squares. If n == 7 (mod 8), then n-1^4 can be written as the sum of three squares. If n is odd but not congruent to 7 modulo 8, then n-0^4 can be expressed as the sum of three squares.
We have a(n) = 1 if n has the form 16^k*q with k a nonnegative integer and q among 7, 8, 15, 23, 31, 47, 71, 79. In fact, if n = 16*m with m > 0, and 16*m = w^4+x^2+y^2+z^2 with w,x,y,z integers, then w,x,y,z are all even and hence m = (w/2)^4+(x/2)^2+(y/2)^2+(z/2)^2. Therefore a(16*m) = a(m) for all m > 0. It is easy to check that a(q) = 1 for every q = 7, 8, 15, 23, 31, 47, 71, 79.
For (a,b,c) = (1,1,2),(1,1,3),(1,1,4),(1,1,6),(1,2,2),(1,2,3),(1,2,4),(1,2,5), we are also able to show that any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.
Conjecture: For each triple (a,b,c) = (1,2,11),(1,2,12),(1,2,13),(2,3,5), any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.
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LINKS
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EXAMPLE
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a(7) = 1 since 7 = 1^4 + 1^2 + 1^2 + 2^2.
a(8) = 1 since 8 = 0^4 + 0^2 + 2^2 + 2^2.
a(15) = 1 since 15 = 1^4 + 1^2 + 2^2 + 3^2.
a(23) = 1 since 23 = 1^4 + 2^2 + 3^2 + 3^2.
a(31) = 1 since 31 = 1^4 + 1^2 + 2^2 + 5^2.
a(47) = 1 since 47 = 1^4 + 1^2 + 3^2 + 6^2.
a(71) = 1 since 71 = 1^4 + 3^2 + 5^2 + 6^2.
a(79) = 1 since 79 = 1^4 + 2^2 + 5^2 + 7^2.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-w^4-x^2-y^2], r=r+1], {w, 0, n^(1/4)}, {x, 0, Sqrt[(n-w^4)/3]}, {y, x, Sqrt[(n-w^4-x^2)/2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A270566
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Number of ordered ways to write n as x^4 + y*(3y+1)/2 + z*(7z+1)/2, where x, y and z are integers with x nonnegative.
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+10
23
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1, 2, 2, 2, 3, 5, 4, 2, 2, 2, 2, 2, 2, 2, 2, 5, 7, 4, 4, 4, 5, 5, 3, 3, 1, 3, 5, 4, 3, 3, 5, 8, 4, 3, 4, 6, 6, 2, 6, 4, 4, 5, 4, 3, 3, 4, 5, 1, 3, 3, 2, 6, 2, 4, 5, 8, 8, 4, 3, 5, 6, 6, 2, 1, 4, 3, 5, 3, 2, 3, 7, 8, 3, 5, 5, 4, 3, 4, 1, 1, 4
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OFFSET
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0,2
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COMMENTS
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Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 24, 47, 63, 78, 79, 143, 153, 325, 494, 949, 1079, 3328, 4335, 5609, 7949, 7967, 8888, 9665.
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LINKS
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EXAMPLE
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a(24) = 1 since 24 = 2^4 + (-2)*(3*(-2)+1)/2 + (-1)*(7*(-1)+1)/2.
a(78) = 1 since 78 = 1^4 + 7*(3*7+1)/2 + 0*(7*0+1)/2.
a(143) = 1 since 143 = 1^4 + 6*(3*6+1)/2 + (-5)*(7*(-5)+1)/2.
a(494) = 1 since 494 = 4^4 + (-7)*(3*(-7)+1)/2 + (-7)*(7*(-7)+1)/2.
a(949) = 1 since 949 = 4^4 + 0*(3*0+1)/2 + 14*(7*14+1)/2.
a(1079) = 1 since 1079 = 0^4 + 25*(3*25+1)/2 + 6*(7*6+1)/2.
a(3328) = 1 since 3328 = 0^4 + 38*(3*38+1)/2 + 18*(7*18+1)/2.
a(4335) = 1 since 4335 = 2^4 + 49*(3*49+1)/2 + 14*(7*14+1)/2.
a(5609) = 1 since 5609 = 0^4 + (-61)*(3*(-61)+1)/2 + 4*(7*4+1)/2.
a(7949) = 1 since 7949 = 3^4 + 43*(3*43+1)/2 + 38*(7*38+1)/2.
a(7967) = 1 since 7967 = 7^4 + (-61)*(3*(-61)+1)/2 + 2*(7*2+1)/2.
a(8888) = 1 since 8888 = 0^4 + (-77)*(3*(-77)+1)/2 + 3*(7*3+1)/2.
a(9665) = 1 since 9665 = 3^4 + 73*(3*73+1)/2 + 21*(7*21+1)/2.
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MATHEMATICA
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pQ[n_] := pQ[n] = IntegerQ[Sqrt[24 n + 1]];
Do[r = 0; Do[If[pQ[n - x^4 - y (7 y + 1)/2], r = r + 1], {x, 0, n^(1/4)}, {y, -Floor[(Sqrt[56 (n - x^4) + 1] + 1)/14], (Sqrt[56 (n - x^4) + 1] - 1)/14}]; Print[n, " ", r]; Continue, {n, 0, 80}]
A270566[n_] := Length@Solve[x >= 0 && n == x^4 + y*(3 y + 1)/2 + z*(7 z + 1)/2, {x, y, z}, Integers];
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CROSSREFS
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Cf. A001318, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A270559
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Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.
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+10
19
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1, 1, 2, 2, 2, 2, 3, 1, 3, 4, 2, 5, 2, 3, 4, 2, 3, 4, 5, 1, 4, 3, 3, 4, 3, 4, 5, 5, 3, 6, 5, 3, 3, 6, 2, 4, 6, 3, 9, 4, 2, 3, 4, 3, 7, 6, 3, 6, 2, 4, 2, 6, 5, 7, 6, 4, 5, 3, 6, 4, 11, 1, 5, 9, 3, 6, 5, 3, 8, 8
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OFFSET
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1,3
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.
(ii) a(n) = 1 only for n = 1, 2, 8, 20, 62, 97, 296, 1493, 4283, 4346, 5433.
In contrast, the author conjectured in A262813 that any positive integer can be expressed as the sum of a nonnegative cube, a square and a positive triangular number.
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LINKS
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EXAMPLE
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a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.
a(2) = 1 since 2 = (-1)^4 + (-1)^3 + 1^2 + 1*2/2.
a(8) = 1 since 8 = 1^4 + 1^3 + 0^2 + 3*4/2.
a(20) = 1 since 20 = (-2)^4 + (-2)^3 + 3^2 + 2*3/2.
a(62) = 1 since 62 = (-2)^4 + (-2)^3 + 3^2 + 9*10/2.
a(97) = 1 since 97 = 1^4 + 1^3 + 2^2 + 13*14/2.
a(296) = 1 since 296 = (-4)^4 + (-4)^3 + 7^2 + 10*11/2.
a(1493) = 1 since 1493 = (-2)^4 + (-2)^3 + 0^2 + 54*55/2.
a(4283) = 1 since 4283 = (-6)^4 + (-6)^3 + 50^2 + 37*38/2.
a(4346) = 1 since 4346 = (-3)^4 + (-3)^3 + 49^2 + 61*62/2.
a(5433) = 1 since 5433 = (-8)^4 + (-8)^3 + 14^2 + 57*58/2.
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MATHEMATICA
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TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[x!=0&&TQ[n-y^2-x^4-x^3], r=r+1], {y, 0, Sqrt[n]}, {x, -1-Floor[(n-y^2)^(1/4)], (n-y^2)^(1/4)}]; Print[n, " ", r]; Continue, {n, 1, 10000}]
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CROSSREFS
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Cf. A000217, A000290, A000578, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A270516
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Number of ordered ways to write n = x^3*(x+1) + y*(y+1)/2 + z*(3z+2), where x and y are nonnegative integers, and z is an integer.
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+10
18
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1, 2, 2, 3, 2, 2, 3, 2, 4, 2, 3, 4, 1, 3, 1, 2, 3, 3, 3, 2, 2, 3, 4, 3, 5, 3, 4, 2, 4, 4, 3, 5, 2, 5, 2, 5, 5, 2, 5, 5, 3, 4, 3, 5, 4, 5, 7, 2, 4, 1, 5, 2, 4, 3, 2, 5, 3, 6, 3, 3, 5, 6, 2, 5, 2, 4, 5, 4, 8, 3, 4, 5, 1, 5, 3, 1, 4, 3, 5, 4, 5
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OFFSET
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0,2
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and the only values of n > 1428 with a(n) = 1 are 2205, 2259, 3556, 4107, 4337, 5387, 9331, 16561, 22237, 27569, 63947, 78610.
(ii) Any natural number can be written as x*(x^3+2) + y*(y+1)/2 + z*(3z+1), where x and y are nonnegative integers, and z is an integer.
(iii) Every n = 0,1,2,... can be written as x*(x^3+x^2+6) + y*(y+1)/2 + z*(3z+2) (or x*(x^3+x^2+4x+1) + y*(y+1)/2 + z*(3z+1)), where x and y are nonnegative integers, and z is an integer.
See also A270533 for a similar conjecture.
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LINKS
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EXAMPLE
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a(72) = 1 since 72 = 2^3*3 + 5*6/2 + 3*(3*3+2).
a(75) = 1 since 75 = 0^3*1 + 4*5/2 + (-5)*(3*(-5)+2).
a(5387) = 1 since 5387 = 7^3*8 + 2*3/2 + (-30)*(3*(-30)+2).
a(9331) = 1 since 9331 = 8^3*9 + 2*3/2 + (-40)*(3*(-40)+2).
a(16561) = 1 since 16561 = 1^3*2 + 101*102/2 + (-62)*(3*(-62)+2).
a(22237) = 1 since 22237 = 6^3*7 + 104*105/2 + 71*(3*71+2).
a(27569) = 1 since 27569 = 2^3*3 + 49*50/2 + (-94)*(3*(-94)+2).
a(63947) = 1 since 63947 = 0^3*1 + 173*174/2 + (-128)*(3*(-128)+2).
a(78610) = 1 since 78610 = 16^3*17 + 52*53/2 + 50*(3*50+2).
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MATHEMATICA
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OQ[x_]:=OQ[x]=IntegerQ[Sqrt[3x+1]]
Do[r=0; Do[If[OQ[n-y(y+1)/2-x^3*(x+1)], r=r+1], {y, 0, (Sqrt[8n+1]-1)/2}, {x, 0, (n-y(y+1)/2)^(1/4)}]; Print[n, " ", r]; Continue, {n, 0, 80}]
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CROSSREFS
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Cf. A000217, A000578, A001082, A001318, A262813, A262815, A262816, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270533.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A270920
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Number of ordered ways to write n as the sum of a positive triangular number, a positive square, and a fifth power whose absolute value does not exceed n.
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+10
8
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1, 2, 2, 3, 3, 3, 4, 2, 2, 5, 5, 3, 2, 3, 4, 4, 3, 4, 6, 3, 2, 4, 3, 3, 5, 5, 3, 3, 4, 5, 6, 7, 2, 2, 4, 6, 9, 9, 7, 6, 3, 5, 4, 4, 7, 8, 6, 3, 5, 7, 8, 7, 7, 6, 6, 5, 4, 5, 7, 7, 5, 5, 6, 9, 5, 3, 5, 4, 9, 11, 10, 6, 2, 6, 4, 3, 6, 7, 5, 5
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OFFSET
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1,2
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 112, 770, 801, 1593, 1826, 2320, 2334, 2849, 7561.
(ii) Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Any natural number n can be written as P(x,y) + z^5, where x, y and z are integers with |z^5| <= n, and the polynomial P(x,y) is either of the following ones: T(x)+2*T(y), T(x)+2*pen(y), x^2+pen(y), x^2+y(5y+1)/2, 2*T(x)+pen(y), pen(x)+pen(y), pen(x)+y(3y+j) (j = 1,2), pen(x)+6*T(y), pen(x)+y(7y+j)/2 (j = 1,3,5), pen(x)+y(4y+j) (j = 1,3), pen(x)+y(5y+j) (j = 1,2,3,4), pen(x)+y(13y+7)/2, x(5x+i)/2+y(3y+j) (i = 1,3; j = 1,2), x(5x+j)/2+y(7y+5)/2 (j = 1,3).
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LINKS
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EXAMPLE
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a(1) = 1 since 1 = 1*2/2 + 1^2 + (-1)^5 with |(-1)^5| <= 1.
a(112) = 1 since 112 = 10*11/2 + 5^2 + 2^5.
a(770) = 1 since 770 = 28*29/2 + 11^2 + 3^5.
a(801) = 1 since 801 = 45*46/2 + 3^2 + (-3)^5 with |(-3)^5| < 801.
a(1593) = 1 since 1593 = 49*50/2 + 20^2 + (-2)^5 with |(-2)^5| < 1593.
a(1826) = 1 since 1826 = 55*56/2 + 23^2 + (-3)^5 with |(-3)^5| < 1826.
a(2320) = 1 since 2320 = 5*6/2 + 48^2 + 1^5.
a(2334) = 1 since 2334 = 11*12/2 + 45^2 + 3^5.
a(2849) = 1 since 2849 = 70*71/2 + 11^2 + 3^5.
a(7561) = 1 since 7561 = 97*98/2 + 53^2 + (-1)^5 with |(-1)^5| < 7561.
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MATHEMATICA
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TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-(-1)^k*x^5-y^2], r=r+1], {k, 0, 1}, {x, 0, n^(1/5)}, {y, 1, Sqrt[n-(-1)^k*x^5]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
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CROSSREFS
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Cf. A000217, A000290, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270921.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A266968
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Number of ordered ways to write n as x^5+y^4+z^3+w*(w+1)/2, where x, y, z and w are nonnegative integers with z > 0 and w > 0.
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+10
5
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0, 0, 1, 2, 2, 2, 1, 1, 2, 2, 2, 3, 4, 2, 1, 2, 2, 2, 3, 3, 2, 1, 1, 4, 4, 2, 1, 2, 3, 4, 7, 5, 2, 2, 4, 3, 2, 5, 6, 5, 2, 1, 2, 4, 5, 5, 6, 4, 3, 4, 4, 1, 2, 4, 5, 5, 4, 4, 2, 3, 2, 4, 5, 4, 6, 5, 4, 3, 5, 6, 5, 4, 4, 3, 4, 5, 4, 3, 2, 5, 7
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OFFSET
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0,4
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 6, 7, 14, 21, 22, 26, 41, 51, 184, 189, 206, 225, 229, 526, 708.
(ii) Any natural number can be written as 2*x^5 + y^4 + z^3 + w*(w+1)/2 with x,y,z,w nonnegative integers. Also, each natural number can be written as x^5 + 2*y^4 + z^3 + w*(w+1)/2 with x,y,z,w nonnegative integers.
(iii) For each d = 1,2, every natural number can be written as x^5 + y^4 + z^3 + w*(3w+1)/d with x,y,z nonnegative integers and w an integer.
(iv) Any natural number can be written as x^4 + y^4 + z^3 + w*(3w+1)/2 with x,y,z nonnegative integers and w an integer.
Also, for each P(w) = w(3w+1)/2, w(7w+3)/2, we can write any natural number as x^4 + y^3 + z^3 + P(w) with x,y,z nonnegative integers and w an integer.
(v) Any natural number can be written as the sum of a nonnegative cube and three pentagonal numbers. Also, every n = 0,1,2,... can be expressed as the sum of two nonnegative cubes and two pentagonal numbers.
We have verified that a(n) > 1 for all n = 2..3*10^6.
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LINKS
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EXAMPLE
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a(2) = 1 since 2 = 0^5 + 0^4 + 1^3 + 1*2/2.
a(6) = 1 since 6 = 1^5 + 1^4 + 1^3 + 2*3/2.
a(7) = 1 since 7 = 0^5 + 0^4 + 1^3 + 3*4/2.
a(14) = 1 since 14 = 0^5 + 0^4 + 2^3 + 3*4/2.
a(21) = 1 since 21 = 1^5 + 2^4 + 1^3 + 2*3/2.
a(22) = 1 since 22 = 0^5 + 0^4 + 1^3 + 6*7/2.
a(26) = 1 since 26 = 1^5 + 2^4 + 2^3 + 1*2/2.
a(41) = 1 since 41 = 2^5 + 0^4 + 2^3 + 1*2/2.
a(51) = 1 since 51 = 2^5 + 1^4 + 2^3 + 4*5/2.
a(184) = 1 since 184 = 0^5 + 0^4 + 4^3 + 15*16/2.
a(189) = 1 since 189 = 1^5 + 2^4 + 1^3 + 18*19/2.
a(206) = 1 since 206 = 2^5 + 3^4 + 3^3 + 11*12/2.
a(225) = 1 since 225 = 0^5 + 3^4 + 2^3 + 16*17/2.
a(229) = 1 since 229 = 1^5 + 3^4 + 3^3 + 15*16/2.
a(526) = 1 since 526 = 3^5 + 1^4 + 6^3 + 11*12/2.
a(708) = 1 since 708 = 1^5 + 5^4 + 3^3 + 10*11/2.
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MATHEMATICA
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TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-x^5-y^4-z^3], r=r+1], {x, 0, n^(1/5)}, {y, 0, (n-x^5)^(1/4)}, {z, 1, (n-x^5-y^4)^(1/3)}]; Print[n, " ", r]; Continue, {n, 0, 80}]
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CROSSREFS
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Cf. A000217, A000326, A000578, A000583, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A280356
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Number of ways to write n as x^4 + y^3 + z^2 + 2^k, where x,y,z are nonnegative integers and k is a positive integer.
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+10
5
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0, 1, 3, 4, 4, 4, 3, 3, 5, 5, 4, 5, 6, 5, 2, 3, 7, 8, 7, 7, 8, 5, 1, 4, 9, 8, 5, 7, 8, 6, 3, 8, 14, 11, 7, 8, 7, 4, 4, 8, 13, 9, 4, 8, 8, 5, 4, 8, 11, 5, 5, 8, 8, 6, 4, 6, 9, 6, 6, 10, 6, 2, 3, 4, 10, 10, 9, 13, 12, 7, 2, 7, 11, 9, 7, 9, 6, 2, 3, 7
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OFFSET
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1,3
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 23, 1135, 6415, 6471.
(ii) If P(x,y) is one of the polynomials 3*x^4 + y^3 and x^6 + 3*y^2, then any positive integer n can be written as P(x,y) + z^2 + 2^k with x,y,z and k nonnegative integers.
We have verified that a(n) > 0 for all n = 2..2*10^7, and that part (ii) of the conjecture holds for all n = 1..10^7.
We also find finitely many polynomials of the form a*x^m + b*y^2 (including x^4 + y^2 and 10*x^5 + y^2) with a and b positive integers and m <= 5, for which it seems that any positive integer can be written as P(x,y) + z^2 + 2^k with x,y,z,k nonnegative integers.
See also A280153 for a similar conjecture involving powers of 4 or 8.
Qing-Hu Hou at Tianjin Univ. has verified that a(n) > 0 for all n = 2..10^9. In 2017, the author announced to offer US $234 as the prize for the first correct solution to his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Dec 30 2017
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LINKS
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EXAMPLE
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a(2) = 1 since 2 = 0^4 + 0^3 + 0^2 + 2^1.
a(23) = 1 since 23 = 2^4 + 1^3 + 2^2 + 2^1.
a(1135) = 1 since 1135 = 0^4 + 7^3 + 28^2 + 2^3.
a(6415) = 1 since 6415 = 1^4 + 13^3 + 11^2 + 2^12.
a(6471) = 1 since 6471 = 1^4 + 13^3 + 57^2 + 2^10.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
In[2]:= Do[r=0; Do[If[SQ[n-2^k-x^4-y^3], r=r+1], {k, 1, Log[2, n]}, {x, 0, (n-2^k)^(1/4)}, {y, 0, (n-2^k-x^4)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 80}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A270594
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Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).
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+10
4
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1, 2, 1, 2, 4, 2, 2, 4, 2, 3, 5, 2, 2, 3, 3, 4, 3, 2, 4, 5, 1, 2, 5, 1, 3, 7, 3, 2, 6, 5, 3, 6, 2, 2, 5, 4, 6, 4, 3, 5, 8, 2, 2, 6, 2, 5, 5, 1, 4, 9, 5, 3, 8, 5, 4, 8, 4, 3, 5, 5, 5, 6, 3, 6, 11, 2, 3, 9, 2, 5, 12, 2, 2, 9, 6, 3, 4, 4, 5, 6, 6, 6, 5, 5, 6, 11, 2, 4, 8, 1
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OFFSET
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1,2
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 21, 24, 48, 90, 138, 213, 283, 462, 468, 567, 573, 1998, 2068, 2488, 2687, 5208, 5547, 5638, 6093, 6492, 6548, 6717, 7538, 7731, 8522, 14763, 16222, 17143, 24958, 26148.
(ii) Let T(x) = x(x+1)/2, pen(x) = x(3x+1)/2 and hep(x) = x(5x+3)/2. Then any 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)^2+T(y)+z(5z+1)/2, T(x)^2+T(y)+z(3z+j) (j = 1,2), T(x)^2+y^2+pen(z), T(x)^2+pen(y)+hep(z), T(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), T(x)^2+pen(y)+z(4z+j) (j = 1,3), T(x)^2+pen(y)+z(5z+j) (j = 1,3,4), T(x)^2+pen(y)+z(11z+7)/2, T(x)^2+y(5y+1)/2+z(3z+2), T(x)^2+hep(y)+z(3z+2), pen(x)^2+T(y)+pen(z), pen(x)^2+T(y)+2*pen(z), pen(x)^2+T(y)+z(9z+7)/2, pen(x)^2+y^2+pen(z), pen(x)^2+2*T(y)+pen(z), pen(x)^2+pen(y)+3*T(z), pen(x)^2+pen(y)+2z^2, pen(x)^2+pen(y)+2*pen(z), pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+pen(y)+z(4z+3), pen(x)^2+pen(y)+z(9z+1)/2, pen(x)^2+pen(y)+3*pen(z), pen(x)^2+pen(y)+z(5z+j) (j = 1,2,3,4), pen(x)^2+pen(y)+z(11z+j)/2 (j = 7,9), pen(x)^2+pen(y)+z(7z+1), pen(x)^2+pen(y)+3*hep(z), pen(x)^2+y(5y+j)/2+z(3z+k) (j = 1,3; k = 1,2), pen(x)^2+hep(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+hep(y)+z(9z+5)/2, pen(y)^2+2pen(y)+z(3z+2), pen(x)^2+2*pen(y)+3*pen(z), (x(5x+1)/2)^2+2*T(y)+pen(z), (x(5x+1)/2)^2+pen(y)+z(7z+3)/2, (x(5x+1)/2)^2+pen(y)+z(4z+1), (x(5x+1)/2)^2+hep(y)+2*pen(z), hep(x)^2+T(y)+2*pen(z), hep(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), hep(x)^2+pen(y)+z(4z+1), hep(x)^2+pen(y)+z(5z+4), 4*pen(x)^2+T(y)+hep(z), 4*pen(x)^2+T(y)+2*pen(z), 4*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), (x(3x+2))^2+y^2+pen(z), (x(3x+2))^2+pen(y)+z(7z+j)/2 (j = 3,5), 2*T(x)^2+T(y)+z(3z+j) (j = 1,2), 2*T(x)^2+y^2+pen(z), 2*T(x)^2+2*T(y)+pen(z), 2*T(x)^2+pen(y)+z(7z+j)/2 (j = 1,5), 2*T(x)^2+pen(y)+z(5z+1), 2*pen(y)^2+T(y)+z(3z+2), 2*pen(x)^2+y^2+pen(z), 2*pen(x)^2+pen(y)+z(7z+3)/2, 2*pen(x)^2+pen(y)+z(4z+j) (j = 1,3), 2*pen(x)^2+pen(y)+z(5z+4), 2*pen(x)^2+pen(y)+z(7z+1), 2*pen(x)^2+hep(y)+2*pen(z), 2*hep(x)^2+pen(y)+z(7z+5)/2, 3*pen(x)^2+T(y)+z(3z+2), 3*pen(x)^2+y^2+pen(z), 3*pen(x)^2+2*T(y)+pen(z), 3*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), 3*pen(x)^2+pen(y)+z(4z+1), 6*pen(x)^2+pen(y)+z(7z+3)/2.
See also A270566 for a similar conjecture involving four powers.
It is known that any positive integer can be written as the sum of a triangular number, a square and an odd square.
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LINKS
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EXAMPLE
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a(21) = 1 since 21 = 1*2/2 + 4^2 + (1*(3*1+1)/2)^2.
a(24) = 1 since 24 = 5*6/2 + 3^2 + (0*(3*0-1)/2)^2.
a(468) = 1 since 468 = 0*1/2 + 18^2 + (3*(3*3-1)/2)^2.
a(7538) = 1 since 7538 = 64*65/2 + 47^2 + (6*(3*6+1)/2)^2.
a(7731) = 1 since 7731 = 82*83/2 + 62^2 + (4*(3*4-1)/2)^2.
a(8522) = 1 since 8522 = 127*128/2 + 13^2 + (3*(3*3+1)/2)^2.
a(14763) = 1 since 14763 = 164*165/2 + 33^2 + (3*(3*3-1)/2)^2.
a(16222) = 1 since 16222 = 168*169/2 + 45^2 + (1*(3*1-1)/2)^2.
a(17143) = 1 since 17143 = 182*183/2 + 21^2 + (2*(3*2+1)/2)^2.
a(24958) = 1 since 24958 = 216*217/2 + 39^2 + (1*(3*1-1)/2)^2.
a(26148) = 1 since 26148 = 10*11/2 + 142^2 + (7*(3*7+1)/2)^2.
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MATHEMATICA
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pQ[n_]:=pQ[n]=IntegerQ[n]&&IntegerQ[Sqrt[24n+1]]
Do[r=0; Do[If[pQ[Sqrt[n-x^2-y(y+1)/2]], r=r+1], {x, 1, Sqrt[n]}, {y, 0, (Sqrt[8(n-x^2)+1]-1)/2}]; Print[n, " ", r]; Continue, {n, 1, 90}]
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CROSSREFS
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Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A271026
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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.
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+10
4
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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Cf. A000326, A000578, A000583, A000584, A001015, A001318, A262813, A262815, A262816, A262827, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A271106
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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.
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+10
4
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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
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OFFSET
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0,6
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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Cf. A000217, A000290, A000326, A000578, A000583, A000584, A001014, A005449, A262813, A262824, A262827, A262857, A262880, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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