A262813 -id:A262813 - cp's OEIS frontend

A262944 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 pentagonal number.

1, 2, 2, 2, 2, 3, 4, 3, 1, 3, 5, 3, 2, 2, 5, 5, 3, 3, 5, 5, 3, 6, 6, 3, 3, 8, 6, 5, 5, 3, 7, 5, 5, 3, 4, 4, 8, 9, 3, 5, 7, 6, 3, 5, 5, 7, 5, 3, 4, 5, 6, 6, 9, 4, 5, 7, 7, 5, 4, 4, 7, 6, 1, 5, 5, 7, 7, 7, 1, 6, 10, 8, 6, 3, 4, 3, 6, 4, 6, 9, 5, 7, 9, 3, 5, 8, 9, 8, 3, 3, 11, 10, 6, 6, 8, 12, 5, 6, 4, 7

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, 63, 69, 489, 714, 1089.
(ii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is twice a square or twice a pentagonal number.
(iii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - 2*x^4 - y*(y+1)/2 is a square or a pentagonal number.
See also A262941 and A262945 for similar conjectures.

			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(9) = 1 since 9 = 1^4 + 2*3/2 + p_5(2).
a(63) = 1 since 63 = 0^4 + 7*8/2 + p_5(5).
a(69) = 1 since 69 = 2^4 + 7*8/2 + 5^2.
a(489) = 1 since 489 = 3^4 + 12*13/2 + p_5(15).
a(714) = 1 since 714 = 4^4 + 18*19/2 + p_5(14).
a(1089) = 1 since 1089 = 4^4 + 38*39/2 + p_5(8).

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, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945.

SQ[n_]:=IntegerQ[Sqrt[n]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1,6]==0)
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}]

A262857 Number of ordered ways to write n as w^3 + 2x^3 + y^2 + 2z^2, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

1, 2, 3, 4, 4, 3, 3, 2, 3, 5, 5, 6, 6, 3, 4, 1, 4, 6, 7, 10, 7, 5, 4, 2, 5, 8, 8, 9, 9, 6, 6, 2, 6, 10, 8, 13, 9, 6, 7, 5, 5, 8, 6, 9, 10, 6, 9, 4, 5, 9, 6, 13, 10, 7, 11, 6, 8, 10, 8, 10, 12, 9, 9, 7, 8, 13, 10, 16, 12, 6, 12, 8, 10, 13, 12, 13, 12, 8, 11, 7, 10, 16, 15, 17, 16, 6, 11, 7, 12, 16, 11, 16, 9, 10, 5, 6, 10, 15, 17, 18, 16

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: We have {a*w^3+b*x^3+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...} if (a,b,c,d) is among the following 63 quadruples:
(1,1,1,2),(1,1,2,4),(1,2,1,1),(1,2,1,2),(1,2,1,3),(1,2,1,4),(1,2,1,6),(1,2,1,13),(1,2,2,3),(1,2,2,4),(1,2,2,5),(1,3,1,1),(1,3,1,2),(1,3,1,3),(1,3,1,5),(1,3,1,6),(1,3,2,3),(1,3,2,4),(1,3,2,5),(1,4,1,1),(1,4,1,2),(1,4,1,3),(1,4,2,2),(1,4,2,3),(1,4,2,5),(1,5,1,1),(1,5,1,2),(1,6,1,1),(1,6,1,3),(1,7,1,2),(1,8,1,2),(1,9,1,2),(1,9,2,4),(1,10,1,2),(1,11,1,2),(1,11,2,4),(1,12,1,2),(1,14,1,2),(1,15,1,2),(2,3,1,1),(2,3,1,2),(2,3,1,3),(2,3,1,4),(2,4,1,1),(2,4,1,2),(2,4,1,6),(2,4,1,8),(2,4,1,10),(2,5,1,3),(2,6,1,1),(2,7,1,3),(2,8,1,1),(2,8,1,4),(2,10,1,1),(2,13,1,1),(3,4,1,2),(3,5,1,2),(3,7,1,2),(3,9,1,2),(4,5,1,2),(4,6,1,2),(4,8,1,2),(4,11,1,2).
Conjecture verified up to 10^11 for all quadruples. - Mauro Fiorentini, Jul 18 2023

			a(7) = 2 since 7 = 1^3 + 2*0^3 + 2^2 + 2*1^2 = 1^3 + 2*1^3 + 2^2 + 2*0^2.
a(15) = 1 since 15 = 1^3 + 2*1^3 + 2^2 + 2*2^2.

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

Cf. A000290, A000578, A262813, A262815, A262816, A262824, A262827.

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

A262824 Number of ordered ways to write n as w^2 + x^3 + 2y^3 + 3z^3, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: (i) For any m = 3, 4, 5, 6 and n >= 0, there are nonnegative integers w, x, y, z such that n = w^2 + x^3 + 2*y^3 + m*z^3.
(ii) For P(w,x,y,z) = w^2 + x^3 + 2*y^3 + z^4, w^2 + x^3 + 2*y^3 + 3*z^4, w^2 + x^3 + 2*y^3 + 6*z^4, 2*w^2 + x^3 + 4*y^3 + z^4, we have {P(w,x,y,z): w,x,y,z = 0,1,2,...} ={0,1,2,...}.
Conjectures (i) and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 22 2023
See also A262827 and A262857 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 0^3 + 2*0^3 + 3*0^3.
a(8) = 2 since 8 = 2^2 + 1^3 + 2*0^3 + 3*1^3 = 0^2 + 2^3 + 2*0^3 + 3*0^3.
a(23) = 1 since 23 = 2^2 + 0^3 + 2*2^3 + 3*1^3.
a(37) = 1 since 37 = 6^2 + 1^3 + 2*0^3 + 3*0^3.
a(72) = 1 since 72 = 8^2 + 2^3 + 2*0^3 + 3*0^3.

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

Cf. A000290, A000578, A262813, A262815, A262816, A262827, A262857.

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

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.

Original entry on oeis.org

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}]

A306227 Number of ways to write n as w + x^4 + y(y+1)/2 + z(z+1)/2, where w is 0 or 1, and x, y, z are nonnegative integers with x >= w and y < z.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jan 30 2019

nonn

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer can be written as the sum of two fourth powers one of which is 0 or 1, and two distinct triangular numbers.
We have verified a(n) > 0 for all n = 1..10^6. The conjecture implies that the set S = {x^4 + y*(y+1)/2: x,y = 0,1,2,...} is an additive basis of order two (i.e., the sumset S + S coincides with {0,1,2,...}).
See also A306225 for a similar conjecture.

			a(1) = 1 with 1 = 0 + 0^4 + 0*1/2 + 1*2/2.
a(2) = 1 with 2 = 0 + 1^4 + 0*1/2 + 1*2/2.
a(14) = 1 with 14 = 0 + 1^4 + 2*3/2 + 4*5/2.
a(3774) = 1 with 3774 = 1 + 5^4 + 52*53/2 + 59*60/2.
a(7035) = 1 with 7035 = 0 + 3^4 + 48*49/2 + 107*108/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000583, A262813, A262941, A262954, A262956, A262959, A270566, A306225.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[TQ[n-x-y^4-z(z+1)/2],r=r+1],{x,0,Min[1,(n-1)/2]},{y,x,(n-1-x)^(1/4)},{z,0,(Sqrt[4(n-1-x-y^4)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]

A306460 Number of ways to write n as x(2x-1) + y(y+1)/2 + z(z+1)(z+2)/6, where x,y,z are nonnegative integers with x > 0.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 17 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as the sum of a positive hexagonal number, a triangular number and a tetrahedral number.
We have verified a(n) > 0 for all n = 1..10^7.
Conjecture 2: Let c be 1 or 3. Then each n = 0,1,... can be written as c*x(x+1) + y*(y+1)/2 + z*(z+1)*(z+2)/6 with x,y,z nonnegative integers.
Conjecture 3: Let t(x) = x*(x+1)*(x+2)/6. Then each n = 0,1,... can be written as 2*t(w) + t(x) + t(y) + t(z) with w,x,y,z nonnegative integers.
We have verified Conjecture 3 for all n = 0..2*10^5. Clearly, Conjecture 3 implies Pollock's conjecture which states that any natural number is the sum of five tetrahedral numbers.

			a(3) = 1 with 3 = 1*(2*1-1) + 1*2/2 + 1*2*3/6.
a(14) = 1 with 14 = 1*(2*1-1) + 2*3/2 + 3*4*5/6.
a(75) = 1 with 75 = 5*(2*5-1) + 4*5/2 + 4*5*6/2.
a(349) = 1 with 349 = 5*(2*5-1) + 24*25/2 + 2*3*4/6.
a(369) = 1 with 369 = 4*(2*4-1) + 10*11/2 + 11*12*13/6.
a(495) = 1 with 495 = 8*(2*8-1) + 20*21/2 + 9*10*11/6.
a(642) = 1 with 642 = 16*(2*16-1) + 16*17/2 + 3*4*5/6.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Tetrahedral Number

Cf. A000217, A000292, A000384, A000797, A104246, A262813, A306462, A306471.

f[n_]:=f[n]=n(n+1)(n+2)/6;
TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[f[z]>=n, Goto[aa]]; Do[If[TQ[n-f[z]-x(2x-1)],r=r+1],{x,1,(Sqrt[8(n-f[z])+1]+1)/4}];Label[aa],{z,0,n}];tab=Append[tab,r],{n,1,100}];Print[tab]

A306462 Number of ways to write n as C(2w,2) + C(x+2,3) + C(y+3,4) + C(z+4,5), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!), w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 3, 1, 1, 4, 7, 6, 2, 2, 6, 8, 5, 1, 2, 9, 11, 5, 1, 4, 9, 12, 7, 2, 4, 10, 12, 7, 4, 6, 10, 11, 6, 5, 5, 10, 15, 8, 4, 7, 11, 14, 9, 4, 5, 11, 14, 6, 6, 10, 15, 12, 5, 7, 8, 11, 14, 7, 5, 6, 11, 14, 12, 11, 6, 11, 15, 12, 7, 9, 18, 21, 12, 5, 5, 15, 19, 11, 3, 5

Offset: 1

Zhi-Wei Sun, Feb 17 2019

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 14, 19.
We have verified a(n) > 0 for all n = 1..5*10^6.
See also A306471 and A306477 for similar conjectures.

			a(1) = 1 with 1 = C(2,2) + C(2,3) + C(3,4) + C(4,5).
a(4) = 1 with 4 = C(2,2) + C(3,3) + C(4,4) + C(5,5).
a(5) = 1 with 5 = C(2,2) + C(4,3) + C(3,4) + C(4,5).
a(14) = 1 with 14 = C(4,2) + C(3,3) + C(4,4) + C(6,5).
a(19) = 1 with 19 = C(6,2) + C(4,3) + C(3,4) + C(4,5).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000292, A000332, A000384, A000389, A000797, A262813, A306460, A306471, A306477.

f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==3;
tab={};Do[r=0;Do[If[f[5,z]>=n,Goto[cc]];Do[If[f[4,y]>=n-f[5,z],Goto[bb]];Do[If[f[3,x]>=n-f[5,z]-f[4,y],Goto[aa]];If[HQ[n-f[5,z]-f[4,y]-f[3,x]],r=r+1],{x,0,n-1-f[5,z]-f[4,y]}];Label[aa],{y,0,n-1-f[5,z]}];Label[bb],{z,0,n-1}];Label[cc];tab=Append[tab,r],{n,1,80}];Print[tab]

A306471 Number of ways to write n as C(2w+1,2) + C(x+2,3) + C(y+3,4) + C(z+4,5) with w,x,y,z nonnegative integers, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 3, 3, 2, 4, 6, 5, 4, 4, 5, 7, 8, 6, 4, 5, 8, 8, 5, 4, 6, 7, 10, 10, 6, 6, 12, 13, 8, 7, 7, 6, 11, 9, 4, 3, 8, 16, 12, 8, 9, 9, 13, 14, 10, 7, 9, 18, 12, 6, 5, 4, 11, 10, 4, 2, 5, 19, 21, 11, 9, 13, 20, 16, 9, 6, 8, 17, 17, 4, 2, 9, 20, 17, 6, 9, 9, 15, 23, 14, 9, 15

Offset: 0

Zhi-Wei Sun, Feb 17 2019

nonn

Conjecture 1: a(n) > 1 for all n > 0.
We have verified a(n) > 0 for all n = 0..5*10^6.
Conjecture 2: For each r = 0, 1, any positive integer can be written as w^2 + C(x,3) + C(y,4) + C(z,5), where w,x,y,z are nonnegative integers with w - r even.
See also A306462 and A306477 for similar conjectures.

			a(0) = 1 with 0 = C(1,2) + C(2,3) + C(3,4) + C(4,5).
a(3) = 2 with 3 = C(3,2) + C(2,3) + C(3,4) + C(4,5) = C(1,2) + C(3,3) + C(4,4) + C(5,5).
a(54) = 2 with 54 = C(3,2) + C(7,3) + C(6,4) + C(5,5) = C(3,2) + C(5,3) + C(7,4) + C(6,5).
a(69) = 1 with 69 = C(3,2) + C(5,3) + C(7,4) + C(7,5) = C(3,2) + C(5,3) + C(3,4) + C(8,5).

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

Cf. A000217, A000290, A000292, A000332, A000389, A000797, A014105, A262813, A306460, A306462, A306477.

f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1;
tab={};Do[r=0;Do[If[f[5,z]>n,Goto[cc]];Do[If[f[4,y]>n-f[5,z],Goto[bb]];Do[If[f[3,x]>n-f[5,z]-f[4,y],Goto[aa]];If[HQ[n-f[5,z]-f[4,y]-f[3,x]],r=r+1],{x,0,n-f[5,z]-f[4,y]}];Label[aa],{y,0,n-f[5,z]}];Label[bb],{z,0,n}];Label[cc];tab=Append[tab,r],{n,0,80}];Print[tab]

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}]

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

A262944 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 pentagonal number.

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A262857 Number of ordered ways to write n as w^3 + 2*x^3 + y^2 + 2*z^2, where w, x, y and z are nonnegative integers.

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A262824 Number of ordered ways to write n as w^2 + x^3 + 2*y^3 + 3*z^3, where w, x, y and z are nonnegative integers.

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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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A306227 Number of ways to write n as w + x^4 + y*(y+1)/2 + z*(z+1)/2, where w is 0 or 1, and x, y, z are nonnegative integers with x >= w and y < z.

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A306460 Number of ways to write n as x*(2x-1) + y*(y+1)/2 + z*(z+1)*(z+2)/6, where x,y,z are nonnegative integers with x > 0.

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A306462 Number of ways to write n as C(2w,2) + C(x+2,3) + C(y+3,4) + C(z+4,5), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!), w is a positive integer and x,y,z are nonnegative integers.

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A306471 Number of ways to write n as C(2w+1,2) + C(x+2,3) + C(y+3,4) + C(z+4,5) with w,x,y,z nonnegative integers, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A262857 Number of ordered ways to write n as w^3 + 2x^3 + y^2 + 2z^2, where w, x, y and z are nonnegative integers.

A262824 Number of ordered ways to write n as w^2 + x^3 + 2y^3 + 3z^3, where w, x, y and z are nonnegative integers.

A306227 Number of ways to write n as w + x^4 + y(y+1)/2 + z(z+1)/2, where w is 0 or 1, and x, y, z are nonnegative integers with x >= w and y < z.

A306460 Number of ways to write n as x(2x-1) + y(y+1)/2 + z(z+1)(z+2)/6, where x,y,z are nonnegative integers with x > 0.