A262827 -id:A262827 - cp's OEIS frontend

A270920 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.

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

Offset: 1

Zhi-Wei Sun, Mar 25 2016

nonn

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).

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
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, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270921.

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}]

A266003 Least nonnegative integer y such that n = x^4 - y^3 + z^2 for some nonnegative integers x and z, or -1 if no such y exists.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 139, 19, 1, 0, 0, 9, 2, 7, 3, 1, 0, 0, 2, 1, 0, 4, 3, 3, 1, 0, 0, 7, 2, 2, 19, 1, 0, 2, 6, 1, 0, 0, 3, 11, 1, 0, 2, 429, 2, 5, 11, 179, 1, 0, 0, 1, 0, 3, 3, 3, 2, 2, 3, 15, 5, 6, 7, 1, 0, 0, 4, 6337, 8, 16, 3, 5, 2, 2, 2, 31, 6, 2, 11, 1, 0, 0, 0, 17, 1, 0, 11, 5, 18, 1, 0, 621, 2, 2, 3, 3, 1, 0, 2, 1, 0

Offset: 0

Zhi-Wei Sun, Dec 19 2015

nonn

Conjecture: Any integer m can be written as x^4 - y^3 + z^2, where x, y and z are nonnegative integers.
I have verified this for all integers m with |m| <= 10^5.
See also A266004 for a related sequence.

			a(6) = 139 since 6 = 36^4 - 139^3 + 1003^2.
a(67) = 6337 since 67 = 676^4 - 6337^3 + 213662^2.
a(176) = 13449 since 176 = 140^4 - 13449^3 + 1559555^2.
a(2667) = 661^4 - 15655^3 + 1909401^2.
a(11019) = 71383 since 11019 = 4325^4 - 71383^3 + 3719409^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Checking the conjecture for m = 0..10^5

Cf. A000290, A000578, A000583, A262827, A266004.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[y=0;Label[bb];Do[If[SQ[n+y^3-x^4],Goto[aa]],{x,0,(n+y^3)^(1/4)}];y=y+1;Goto[bb];Label[aa];Print[n," ",y];Continue,{n,0,100}]

A266968 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.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 28 2016

nonn

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.
Compare this conjecture with the conjectures in A262813, A262827, A270559 and A271026.

			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.

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), 1367-1396.

Cf. A000217, A000326, A000578, A000583, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.

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}]

A280356 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.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jan 01 2017

nonn

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

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000079, A000290, A000578, A000583, A262827, A262956, A270533, A270559.

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}]

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}]

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

Zhi-Wei Sun, Mar 29 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
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.

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

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 + 3y^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

Zhi-Wei Sun, Mar 30 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
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.

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

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}]

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 13 2016

nonn

Conjecture: For positive integers a,b,c,d, any natural number can be written as a*x^2 + b*y^2 + c*z^3 + d*w^4 with x,y,z,w nonnegative integers, if and only if (a,b,c,d) is among the following 49 quadruples: (1,2,1,1), (1,3,1,1), (1,6,1,1), (2,3,1,1), (2,4,1,1), (1,1,2,1), (1,4,2,1), (1,2,3,1), (1,2,4,1), (1,2,12,1), (1,1,1,2), (1,2,1,2), (1,3,1,2), (1,4,1,2), (1,5,1,2), (1,11,1,2), (1,12,1,2), (2,4,1,2), (3,5,1,2), (1,1,4,2), (1,1,1,3), (1,2,1,3), (1,3,1,3), (1,2,4,3), (1,2,1,4), (1,3,1,4), (2,3,1,4), (1,1,2,4), (1,2,2,4), (1,8,2,4), (1,2,3,4), (1,1,1,5), (1,2,1,5), (2,3,1,5), (2,4,1,5), (1,3,2,5), (1,1,1,6), (1,3,1,6), (1,1,2,6), (1,2,1,8), (1,2,4,8), (1,2,1,10), (1,1,2,10), (1,2,1,11), (2,4,1,11), (1,2,1,12), (1,1,2,13), (1,2,1,14),(1,2,1,15).

See also A262824, A262827, A262857 and A273917 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(1) = 1 since 1 = 1^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(2) = 1 since 2 = 0^2 + 2*1^2 + 3*0^3 + 4*0^4.
a(14) = 1 since 14 = 3^2 + 2*1^2 + 3*1^3 + 4*0^4.
a(17) = 1 since 17 = 3^2 + 2*2^2 + 3*0^3 + 4*0^4.
a(59) = 1 since 59 = 3^2 + 2*5^2 + 3*0^3 + 4*0^4.
a(63) = 1 since 63 = 3^2 + 2*5^2 + 3*0^2 + 4*1^4.
a(287) = 1 since 287 = 11^2 + 2*9^2 + 3*0^2 + 4*1^4.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000290, A000578, A000583, A262824, A262827, A262857, A270969, A273429, A273915, A273917.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4w^4-3z^3-2y^2],r=r+1],{w,0,(n/4)^(1/4)},{z,0,((n-4w^4)/3)^(1/3)},{y,0,((n-4w^4-3z^3)/2)^(1/2)}];Print[n," ",r];Continue,{n,0,100}]

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

Zhi-Wei Sun, Jun 04 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016-2017.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. (See Remark 1.1.)
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. (See Conjecture 3.2.)

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

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}]

A266215 Positive integers x such that x^3 - 1 = y^4 + z^2 for some positive integers y and z.

Original entry on oeis.org

3, 13, 27, 147, 203, 5507, 15661, 16957, 21531, 29931, 38051, 47171, 57147, 84027, 85547, 90891, 167051, 273651, 337501, 392881, 421715, 566691, 609971, 698113, 914701, 1229283, 1435213, 1564573, 1786587, 1987571, 2523387, 2579377, 2716443, 3760347, 3778273

Offset: 1

Zhi-Wei Sun, Dec 24 2015

nonn

The conjecture in A266212 implies that this sequence has infinitely many terms.

			a(1) = 3 since 3^3 - 1 = 1^4 + 5^2.
a(2) = 13 since 13^3 - 1 = 6^4 + 30^2.
a(6) = 5507 since 5507^3 - 1 = 29^4 + 408669^2.
a(16) = 90891 since 90891^3 - 1 = 949^4 + 27387137^2.
a(35) = 3778273 since 3778273^3 - 1 = 85386^4 + 883654380^2.

Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000290, A000578, A000583, A262827, A266003, A266004, A266152, A266153, A266212.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
n=0;Do[Do[If[SQ[x^3-1-y^4],n=n+1;Print[n," ",x];Goto[aa]],{y,1,(x^3-1)^(1/4)}];Label[aa];Continue,{x,1,10^5}]

a(17)-a(35) from Lars Blomberg, Dec 30 2015

cp's OEIS Frontend

A270920 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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A266003 Least nonnegative integer y such that n = x^4 - y^3 + z^2 for some nonnegative integers x and z, or -1 if no such y exists.

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A266968 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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A280356 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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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).

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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.

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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.

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A272979 Number of ways to write n as x^2 + 2*y^2 + 3*z^3 + 4*w^4 with x,y,z,w nonnegative integers.

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

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.