A262944 -id:A262944 - cp's OEIS frontend

A270566 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.

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

Offset: 0

Zhi-Wei Sun, Mar 19 2016

nonn

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.

Conjecture verified for n up to 10^11. - Mauro Fiorentini, Jul 24 2023

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Cf. A001318, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559.

(* From Zhi-Wei Sun, Start *)
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}]
(* From Zhi-Wei Sun, End *)
A270566[n_] := Length@Solve[x >= 0 && n == x^4 + y*(3 y + 1)/2 + z*(7 z + 1)/2, {x, y, z}, Integers];
Array[A270566, 25, 0] (* JungHwan Min, Mar 19 2016 *)

A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 20, 22, 70, 87, 167, 252, 388, 562, 636, 658, 873, 2598, 14979, 18892, 20824.
(ii) Each n = 0,1,2,... can be written as x*P(x) + y^2 + z*(3z-1)/2 with x and y nonnegative integers, and z an integer, where P(x) is either of the polynomials x^3+2, x^3+3, x^3+2x+8, x^3+x^2+4x+2, x^3+x^2+7x+6.
(iii) Any nonnegative integer can be expressed as x*(x^3+3) + y*(5y+4) + z*(3z-1)/2, where x is an nonnegative integer, and y and z are integers.
See also A270516 for a similar conjecture.

			a(20) = 1 since 20 = 1^4 + 1^3 + 4^2 + (-1)*(3*(-1)-1)/2.
a(22) = 1 since 22 = 0^4 + 0^3 + 0^2 + 4*(3*4-1)/2.
a(873) = 1 since 873 = 5^4 + 5^3 + 11^2 + (-1)*(3*(-1)-1)/2.
a(2598) = 1 since 2598 =  4^4 + 4^3 + 4^2 + 39*(3*39-1)/2.
a(14979) = 1 since 14979 = 1^4 + 1^3 + 51^2 + 91*(3*91-1)/2.
a(18892) = 1 since 18892 = 3^4 + 3^3 + 137^2 + (-3)*(3*(-3)-1)/2.
a(20824) = 1 since 20824 = 1^4 + 1^3 + 115^2 + (-71)*(3*(-71)-1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516.

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

A262941 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is an even square or twice a square.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 3, 1, 3, 3, 6, 3, 4, 4, 4, 4, 3, 4, 2, 3, 3, 4, 3, 2, 5, 3, 4, 3, 6, 5, 6, 4, 2, 3, 2, 4, 4, 4, 5, 3, 3, 1, 3, 5, 6, 6, 4, 3, 3, 4, 1, 5, 4, 3, 4, 3, 4, 3, 4, 4, 5, 3, 5, 4, 5, 4, 4, 3, 2, 4, 6, 3, 4, 6, 4, 5, 2, 7, 7, 4, 3, 3, 5, 4, 5, 6, 6, 5, 2, 6, 4, 8

Offset: 1

Zhi-Wei Sun, Oct 04 2015

nonn

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as x^4 + 2^k*y^2 + z*(z+1)/2, where k is 1 or 2, and x,y,z are integers with z > 0.
This has been verified for n up to 2*10^6. We also guess that a(n) = 1 only for n = 1, 2, 13, 16, 50, 59, 239, 493, 1156, 1492, 1984, 3332.
See also A262944, A262945, A262954, A262955, A262956 for similar conjectures.

			a(1)    = 1 since    1 = 0^4 +    0^2 +  1*2/2  with  0 even.
a(2)    = 1 since    2 = 1^4 +    0^2 +  1*2/2  with  0 even.
a(13)   = 1 since   13 = 1^4 + 2* 1^2 +  4*5/2.
a(16)   = 1 since   16 = 1^4 +    0^2 +  5*6/2  with  0 even.
a(50)   = 1 since   50 = 1^4 +    2^2 +  9*10/2 with  2 even.
a(59)   = 1 since   59 = 0^4 +    2^2 + 10*11/2 with  2 even.
a(239)  = 1 since  239 = 0^4 + 2* 2^2 + 21*22/2 with  2 even.
a(493)  = 1 since  493 = 2^4 +   18^2 + 17*18/2 with 18 even.
a(1156) = 1 since 1156 = 1^4 + 2*24^2 +  2*3/2  with 24 even.
a(1492) = 1 since 1492 = 2^4 + 2* 7^2 + 52*53/2.
a(1984) = 1 since 1984 = 5^4 +   18^2 + 45*46/2 with 18 even.
a(3332) = 1 since 3332 = 5^4 +   52^2 +  2*3/2  with 52 even.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and aT_x+by^2+f(z), arXiv:1502.03056 [math.NT], 2015.

Cf. A000217, A000290, A000583, A254885, A262813, A262827, A262944, A262945, A262954, A262955, A262956.

SQ[n_]:=IntegerQ[Sqrt[n/2]]||IntegerQ[Sqrt[n/4]]
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A270559 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.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 18 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000578, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.

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}]

A270516 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.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

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.

			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).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Cf. A000217, A000578, A001082, A001318, A262813, A262815, A262816, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270533.

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}]

A262954 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y(y+1)/2 has the form z^2 or 8z^2.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 13, 89, 449, 524, 1049, 2164, 14969, 51334.
(ii) For any positive integer n, there are integers x and y > 0 such that n - x^4 - T(y) has the form T(z) or 4*T(z), where T(k) refers to the triangular number k*(k+1)/2.
(iii) For every n = 1,2,3,... there are integers x and y > 0 such that n - x^4 - T(y) has the form T(z) or 2*z^2.
(iv) For {c,d} = {1,2} and n > 0, there are integers x and y > 0 such that n - 2*x^4 - T(y) has the form c*T(z) or d*z^2.
(v) For each n = 1,2,3,... there are integers x and y > 0 such that n - 4*x^4 - T(y) has the form 2*T(z) or z^2.
See also A262941, A262944, A262945, A262954, A262955 and A262956 for similar conjectures.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + 0^2.
a(9) = 1 since 9 = 0^4 + 1*2/2 + 8*1^2.
a(13) = 1 since 13 = 1^4 + 2*3/2 + 3^2.
a(89) = 1 since 89 = 2^4 + 1*2/2 + 8*3^2.
a(449) = 1 since 449 = 0^4 + 22*23/2 + 14^2.
a(524) = 1 since 524 = 3^4 + 29*30/2 + 8*1^2.
a(1049) = 1 since 1049 = 5^4 + 16*17/2 + 8*6^2.
a(2164) = 1 since 2164 = 1^4 + 34*35/2 + 8*14^2.
a(14969) = 1 since 14969 = 8^4 + 145*146/2 + 8*6^2.
a(51334) = 1 since 51334 = 5^4 + 313*314/2 + 8*14^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000583, A262941, A262944, A262945, A262954, A262955, A262956.

SQ[n_]:=IntegerQ[Sqrt[n]]||IntegerQ[Sqrt[n/8]]
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A262945 Number of ordered pairs (x,y) with x >= 0 and y >= 0 such that n - x^4 - 2*y^2 is a triangular number or a pentagonal number.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: a(n) > 0 for every n = 0,1,2,..., and a(n) = 1 only for the following 55 values of n: 0, 26, 27, 32, 41, 42, 50, 65, 80, 97, 112, 122, 130, 160, 196, 227, 239, 272, 322, 371, 612, 647, 736, 967, 995, 1007, 1106, 1127, 1205, 1237, 1240, 1262, 1637, 1657, 1757, 2912, 2987, 3062, 3107, 3524, 3647, 3902, 5387, 5587, 5657, 6047, 6107, 11462, 13427, 14717, 15002, 17132, 20462, 30082, 35750.

See also A262941, A262944, A262954 and A262955 for similar conjectures.

			a(26) = 1 since 26 = 2^4 + 2*0^2 + 4*5/2.
a(32) = 1 since 32 = 0^4 + 2*4^2 + 0*1/2.
a(41) = 1 since 41 = 1^4 + 2*3^2 + p_5(4), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(196) = 1 since 196 = 1^4 + 2*5^2 + p_5(10).
a(3524) = 1 since 3524 = 0^4 + 2*22^2 + 71*72/2.
a(3647) = 1 since 3647 = 0^4 + 2*34^2 + p_5(30).
a(6047) = 1 since 6047 = 5^4 + 2*39^2 + p_5(40).
a(6107) = 1 since 6107 = 0^4 + 2*1^2 + 110*111/2.
a(11462) = 1 since 11462 = 9^4 + 2*5^2 + 98*99/2.
a(13427) = 1 since 13427 = 7^4 + 2*0^2 + 148*149/2.
a(14717) = 1 since 14717 = 8^4 + 2*72^2 + 22*23/2.
a(15002) = 1 since 15002 = 0^4 + 2*86^2 + 20*21/2.
a(17132) = 1 since 17132 = 3^4 + 2*30^2 + p_5(101).
a(20462) = 1 since 20462 = 0^4 + 2*26^2 + 195*196/2.
a(30082) = 1 since 30082 = 11^4 + 2*63^2 + 122*123/2.
a(35750) = 1 since 35750 = 0^4 + 2*44^2 + 252*253/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000217, A000290, A000326, A000583, A262941, A262944, A262954, A262955.

SQ[n_]:=IntegerQ[Sqrt[8n+1]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1, 6]==0)
Do[r=0;Do[If[SQ[n-x^4-2y^2],r=r+1],{x, 0, n^(1/4)},{y,0,Sqrt[(n-x^4)/2]}];Print[n, " ", r];Continue,{n,0,100}]

A262956 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a square minus 1.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: a(n) > 0 for all n > 0. In other words, for any positive integer n, either n or n + 1 can be written as the sum of a fourth power, a square and a positive triangular number.

We also guess that a(n) = 1 only for n = 1, 89, 244, 464, 5243, 14343.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + 0^2.
a(89) = 1 since 89 = 2^4 + 4*5/2 + 8^2 - 1.
a(244) = 1 since 244 = 2^4 + 2*3/2 + 15^2.
a(464) = 1 since 464 = 2^4 + 22*23/2 + 14^2 - 1.
a(5243) = 1 since 5243 = 0^4 + 50*51/2 + 63^2 - 1.
a(14343) = 1 since 14343 = 2^4 + 163*164/2 + 31^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000583, A262941, A262944, A262945, A262954, A262955, A262959.

SQ[n_]:=IntegerQ[Sqrt[n]]||IntegerQ[Sqrt[n+1]]
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A262955 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a pentagonal number (A000326) or twice a pentagonal number.

Original entry on oeis.org

1, 2, 3, 3, 2, 3, 3, 3, 2, 1, 4, 4, 3, 2, 3, 5, 4, 3, 3, 3, 4, 5, 5, 4, 3, 5, 6, 5, 5, 3, 6, 4, 4, 4, 1, 4, 5, 7, 6, 2, 6, 3, 3, 3, 5, 8, 5, 4, 3, 5, 4, 4, 4, 5, 5, 5, 7, 4, 3, 3, 7, 3, 3, 2, 2, 8, 5, 6, 2, 3, 5, 7, 6, 2, 1, 4, 4, 3, 6, 7, 6, 3, 5, 4, 3, 2, 6, 6, 6, 4, 6, 8

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 10, 35, 75, 134, 415, 515, 1465, 2365, 3515, 4140.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(10) = 1 since 10 = 0^4 + 4*5/2 + p_5(0).
a(35) = 1 since 35 = 1^4 + 4*5/2 + 2*p_5(3).
a(75) = 1 since 75 = 2^4 + 5*6/2 + 2*p_5(4).
a(134) = 1 since 134 = 2^4 + 1*2/2 + p_5(9).
a(415) = 1 since 415 = 0^4 + 21*22/2 + 2*p_5(8).
a(515) = 1 since 515 = 0^4 + 6*7/2 + 2*p_5(13).
a(1465) = 1 since 1465 = 5^4 + 35*36/2 + p_5(12).
a(2365) = 1 since 2365 = 5^4 + 8*9/2 + 2*p_5(24).
a(3515) = 1 since 3515 = 5^4 + 51*52/2 + 2*p_5(23).
a(4140) = 1 since 4140 = 1^4 + 90*91/2 + 2*p_5(4).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000217, A000326, A000583, A262941, A262944, A262945, A262954.

PenQ[n_]:=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0)
PQ[n_]:=PenQ[n]||PenQ[n/2]
Do[r=0;Do[If[PQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A270594 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).

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 19 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
B. K. Oh and Z.-W. Sun, Mixed sums of squares and triangular numbers (III), J. Number Theory 129(2009), 964-969.
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), no. 7, 1367-1396.
Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.

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

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}]

cp's OEIS Frontend

A270566 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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A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

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A262941 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is an even square or twice a square.

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A270559 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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A270516 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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A262954 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 has the form z^2 or 8*z^2.

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A262945 Number of ordered pairs (x,y) with x >= 0 and y >= 0 such that n - x^4 - 2*y^2 is a triangular number or a pentagonal number.

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A262956 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a square minus 1.

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A270566 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.

A270516 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.

A262954 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y(y+1)/2 has the form z^2 or 8z^2.