A160325 -id:A160325 - cp's OEIS frontend

A254677 Least positive integer m with A160325(m) = n.

2, 1, 5, 19, 15, 22, 37, 92, 71, 156, 136, 222, 206, 211, 257, 292, 506, 402, 577, 521, 632, 789, 682, 796, 742, 1006, 1046, 1192, 1346, 1482, 1312, 1507, 2021, 1522, 2172, 1977, 1962, 2007, 2161, 2479, 2502, 3047, 2761, 2326, 3097, 2876, 3316, 3216, 3421, 3386, 3902, 3652, 4406, 4356, 4587, 4492, 4342, 4917, 4811, 5472

Offset: 1

Zhi-Wei Sun, Feb 05 2015

nonn

Conjecture: a(n) exists for any n > 0. Moreover, no term a(n) is congruent to 3 modulo 5.

			a(3) = 5 since 5 is the least positive integer that can be written as x(x+1)/2 + (2y)^2 + z(3z-1)/2 (with x,y,z nonnegative integers) in exactly 3 ways. In fact, 5 = 0*1/2 + 0^2 + 2*(3*2-1)/2 = 0*1/2 + 2^2 + 1*(3*1-1)/2 = 1*2/2 + 2^2 + 0*(3*0-1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..200
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Cf. A000217, A000290, A000326, A160325.

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[Do[m=0;Label[aa];m=m+1;r=0;Do[If[TQ[m-4y^2-z(3z-1)/2],r=r+1;If[r>n,Goto[aa]]],{y,0,Sqrt[m/4]},{z,0,(Sqrt[24(m-4y^2)+1]+1)/6}];
If[r==n,Print[n," ", m];Goto[bb],Goto[aa]]];Label[bb];Continue,{n,1,60}]

A262815 Number of ordered ways to write n as x^3 + y(y+1)/2 + z(3*z+1)/2, where x, y and z are nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 5, 12, 19, 20, 75, 83, 97, 117.
Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 20 2023
See also A262813 and A262816 for similar conjectures.
By Theorem 1.7(i) in the linked paper, each natural number can be written as the sum of a triangular number, an even square and a generalized pentagonal number.

			a(0) = 1 since 0 = 0^3 + 0*1/2 + 0*(3*0+1)/2.
a(5) = 1 since 5 = 0^3 + 2*3/2 + 1*(3*1+1)/2.
a(12) = 1 since 12 = 0^3 + 4*5/2 + 1*(3*1+1)/2.
a(19) = 1 since 19 = 1^3 + 2*3/2 + 3*(3*3+1)/2.
a(20) = 1 since 20 = 2^3 + 4*5/2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 2^3 + 4*5/2 + 6*(3*6+1)/2.
a(83) = 1 since 83 = 0^3 + 3*4/2 + 7*(3*7+1)/2.
a(97) = 1 since 97 = 3^3 + 10*11/2 + 3*(3*3+1)/2.
a(117) = 1 since 117 = 0^3 + 13*14/2 + 4*(3*4+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. A000217, A000578, A005449, A160325, A262813, A262816.

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-z(3z+1)/2],r=r+1],{x,0,n^(1/3)},{z,0,(Sqrt[24(n-x^3)+1]-1)/6}];Print[n," ",r];Continue,{n,0,100}]

A262816 Number of ordered ways to write n as x^3 + y^2 + z(3z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 216.
(ii) Any positive integer can be written as x^3 + y*(3*y-1)/2 + z*(3*z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.
Conjectures (i) and (ii) verified for n up to 10^9. - Mauro Fiorentini, Jul 21 2023
See also A262813 and A262815 for similar conjectures.
By Theorem 1.7(ii) in the linked paper, any nonnegative integer can be written as x^2 + y^2 + z*(3*z-1)/2, where x, y and z are integers.

			a(1) = 1 since 1 = 0^3 + 0^2 + 1*(3*1-1)/2.
a(4) = 1 since 4 = 1^3 + 1^2 + (-1)*(3*(-1)-1)/2.
a(8) = 2 since 8 = 0^3 + 1^2 + (-2)*(3*(-2)-1)/2 = 1^3 + 0^2 + (-2)*(3*(-2)-1)/2.
a(216) = 1 since 216 = 2^3 + 14^2 + 3*(3*3-1)/2.

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. A000290, A000326, A000578, A001318, A160325, A160326, A262813, A262815.

PenQ[n_]:=n>0&&IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[PenQ[n-x^3-y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Continue,{n,1,100}]

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.

Original entry on oeis.org

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}]

A160326 Number of ways to express n=0,1,2,... as the sum of two squares and a pentagonal number.

Original entry on oeis.org

1, 3, 3, 1, 2, 5, 4, 1, 1, 5, 6, 2, 1, 5, 5, 2, 4, 6, 5, 1, 3, 6, 5, 3, 1, 8, 8, 4, 2, 4, 8, 4, 5, 1, 4, 5, 4, 10, 6, 6, 5, 8, 6, 1, 3, 6, 6, 4, 6, 4, 7, 8, 8, 8, 5, 7, 4, 4, 6, 5, 6, 8, 7, 4, 8, 8, 6, 5, 4, 7, 7, 8, 7, 7, 8, 8, 8, 7, 3, 4, 12, 4, 4, 7, 3, 13, 12, 12, 5, 2, 12, 4, 5, 6, 6, 8, 10, 8, 3, 5, 11

Offset: 0

Zhi-Wei Sun, May 08 2009

nonn

In April 2009, Zhi-Wei Sun conjectured that a(n)>0 for every n=0,1,2,3,... Note that pentagonal numbers are more sparse than squares. The Gauss-Legendre theorem asserts that n is the sum of three squares if and only if it is not of the form 4^a(8b+7) (a,b=0,1,2,...).

			For n=5 the a(5)=5 solutions are 0+0+5, 1+4+0, 4+1+0, 0+4+1, 4+0+1.

Zhi-Wei Sun, Table of n, a(n) for n = 0..50000
B. K. Oh and Z. W. Sun, Mixed sums of squares and triangular numbers (III), J. Number Theory 129(2009), 964-969.
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, Various new conjectures involving polygonal numbers and primes (a message to Number Theory List), 2009.
Zhi-Wei Sun, Mixed Sums of Primes and Other Terms (a webpage).
Z. W. Sun, On universal sums of polygonal numbers, preprint, arXiv:0905.0635 [math.NT], 2009-2015.

Cf. A000290, A000326, A160324, A160325.

SQ[x_]:=x>-1&&IntegerPart[Sqrt[x]]^2==x RN[n_]:=Sum[If[SQ[n-y^2-(3z^2-z)/2],1,0], {y,0,Sqrt[n]},{z,0,Sqrt[n-y^2]}] Do[Print[n," ", RN[n]],{n,0,50000}]

a(n) = |{: x,y=0,1,2,... & x^2+y^2+(3z^2-z)/2=n}|.

A240088 The number of ways of writing n as an ordered sum of a triangular number (A000217), a square (A000290) and a pentagonal number (A000326).

Original entry on oeis.org

1, 3, 3, 2, 3, 4, 4, 4, 3, 3, 5, 5, 5, 3, 3, 7, 7, 5, 2, 6, 5, 4, 8, 5, 6, 4, 8, 7, 5, 7, 4, 9, 6, 5, 4, 3, 9, 12, 9, 4, 7, 9, 8, 4, 6, 8, 7, 8, 4, 8, 9, 10, 9, 6, 10, 6, 7, 10, 9, 8, 7, 11, 7, 4, 10, 8, 10, 10, 7, 5, 10, 14, 11, 7, 6, 11, 10, 10, 4, 11, 10, 10, 13, 8, 7, 7, 13, 12, 8, 8, 6, 10, 17, 8, 10, 7, 16, 10, 3, 12, 9

Offset: 0

Robert G. Wilson v, Mar 31 2014

nonn

0 and 1 are triangular numbers, square numbers and pentagonal numbers.
It is conjectured that a(n) is always positive - this is one of the conjectures in Conjecture 1.1 of Sun (2009). - N. J. A. Sloane, Apr 01 2014
Note that both the conjecture in A160325 and the conjecture in A160324 imply that a(n) is always positive. - Zhi-Wei Sun, Apr 01 2014
a(n) > 0 for all n < 10^10. - Robert G. Wilson v, Aug 20 2016
Least number to be represented k ways, k >= 1: 0, 3, 1, 5, 10, 19, 15, 22, 31, 51, 61, 37, 82, 71, 126, 96, 92, 136, 162, 187, 206, 276, 191, 261, 236, 247, 317, 302, 401, 292, 422, 547, 456, 544, 551, 612, 591, 577, 521, 666, 742, 726, 682, 877, 796, 1052, 961, 1046, 1171, 1027, ..., . A275999.
Greatest number (conjectured) to be represented k ways, k >= 1: 0, 18, 168, 78, 243, 130, 553, 455, 515, 658, 865, 945, 633, 1918, 2258, 1385, 1583, 2828, 2135, 2335, 2785, 4533, 3168, 3478, 2790, 3868, 4193, 7328, 4953, 5278, 6390, 8148, 8015, 4585, 9160, 10485, 7613, 12333, 12025, 10178, 9923, 9720, 12558, 11340, 17420, 11753, 14893, 16155, 16415, 14343, ..., .
Conjectured lists of numbers that are represented in k >= 1 ways:
1: 0;
2: 3, 18;
3: 1, 2, 4, 8, 9, 13, 14, 35, 98, 168;
4: 5, 6, 7, 21, 25, 30, 34, 39, 43, 48, 63, 78;
5: 10, 11, 12, 17, 20, 23, 28, 33, 69, 193, 203, 230, 243;
6: 19, 24, 32, 44, 53, 55, 74, 90, 111, 130;
7: 15, 16, 27, 29, 40, 46, 56, 60, 62, 68, 73, 84, 85, 95, 108, 113, 123, 135, 139, 163, 165, 273, 553;
8: 22, 26, 42, 45, 47, 49, 59, 65, 83, 88, 89, 93, 112, 119, 125, 134, 140, 144, 186, 205, 233, 244, 320, 405, 455;
9: 31, 36, 38, 41, 50, 52, 58, 100, 109, 124, 160, 214, 249, 308, 358, 515; ..., .

Robert G. Wilson v and Robert Israel, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Science China Mathematics, Vol. 58, No. 7 (2015), 1367-1396; arXiv:0905.0635 [math.NT], 2009-2015 [Edited, _Felix Fröhlich_, Aug 24 2016].
Zhi-Wei Sun, Various new conjectures involving polygonal numbers and primes, a message to the Number Theory List, May 8 2009.

Cf. A000925, A008443, A101428, A115171, A115172, A115173, A115174, A115175, A115176, A115177, A144642.

Cf. A160324, A160325, A160326. - Zhi-Wei Sun, Apr 01 2014

# requires Maple 17 and up
with(SignalProcessing):
N:= 10000;  # to get terms up to a(N)
A:= Array(0..N,datatype=float);
B:= Array(0..N,datatype=float);
C:= Array(0..N,datatype=float);
for i from 0 to floor(sqrt(N)) do A[i^2]:= 1 od:
for i from 0 to floor((1+sqrt(1+8*N))/2) do B[i*(i-1)/2]:= 1 od:
for i from 0 to floor((1+sqrt(1+24*N))/6) do C[i*(3*i-1)/2]:= 1 od:
R:= Convolution(Convolution(A,B),C);
R:= evalhf(map(round,R));
# Note that a(i) = R[i+1] for i from 0 to N
# Robert Israel, Apr 01 2014

p = Table[n (3n - 1)/2, {n, 0, 26}]; s = Table[n^2, {n, 0, 32}]; t = Table[n (n + 1)/2, {n, 0, 45}]; a = Sort@ Flatten@ Table[ p[[i]] + s[[j]] + t[[k]], {i, 26}, {j, 32}, {k, 45}]; Table[ Count[a, n], {n, 0, 105}]

A287616 Number of ways to write n as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, May 27 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 4, 7, 9, 22.
It was proved in arXiv:1502.03056 that each n = 0,1,2,... can be written as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z integers. The author would like to offer 135 US dollars as the prize for the first proof of the conjecture that a(n) is always positive.
See over 400 similar conjectures in the linked a-file.

			a(4) = 1 since 4 = 1*(1+1)/2 + 0*(3*0+1)/2 + 1*(5*1+1)/2.
a(7) = 1 since 7 = 0*(0+1)/2 + 2*(3*2+1)/2 + 0*(5*0+1)/2.
a(9) = 1 since 9 = 3*(3+1)/2 + 0*(3*0+1)/2 + 1*(5*1+1)/2.
a(22) = 1 since 22 = 5*(5+1)/2 + 2*(3*2+1)/2 + 0*(5*0+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, List of conjectural tuples (a,b,c,d,e,f) with {x*(ax+b)/2 + y*(cy+d)/2 + z*(ez+f)/2: x,y,z = 0,1,2,...} = {0,1,2,...}
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 universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.

Cf. A000217, A000290, A005449, A160324, A160325, A160326, A254668, A286944.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
Do[r=0;Do[If[TQ[n-x(3x+1)/2-y(5y+1)/2],r=r+1],{x,0,(Sqrt[24n+1]-1)/6},{y,0,(Sqrt[40(n-x(3x+1)/2)+1]-1)/10}];Print[n," ",r],{n,0,80}]

A085263 Number of ways to write n as the sum of a squarefree number (A005117) and a positive square (A000290).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 0, 3, 3, 2, 2, 3, 3, 2, 2, 3, 4, 2, 1, 4, 4, 2, 1, 5, 4, 3, 2, 2, 5, 2, 3, 6, 6, 3, 2, 6, 4, 3, 2, 5, 6, 3, 2, 5, 6, 3, 2, 4, 6, 4, 3, 4, 6, 4, 1, 7, 5, 3, 3, 7, 6, 4, 4, 6, 8, 3, 3, 6, 7, 2, 4, 8, 5, 4, 3, 7, 9, 4, 2, 8, 9, 4, 3, 6, 6, 5, 4, 7, 9, 5, 3, 8, 4, 3, 5, 9

Offset: 1

Reinhard Zumkeller, Jun 23 2003

a(A085265(n))>0; a(A085266(n))=1; a(A085267(n))>1.
a(A085264(n))=n and a(i)<>n for i < A085264(n).
First occurrence of k: 2, 6, 11, 23, 30, 38, 62, 71, 83, 110, 138, 155, 182, 203, 227, 263, 302, 327, 383, 435, 447, 503, 542, 602, 635, ..., . Conjecture: For each k above, there is a finite number of terms; for example, only the two numbers 1 and 13 cannot be represented as the sum of a squarefree number and a square. The number of k terms beginning with 0: 2, 9, 19, 27, 38, 36, 57, 63, 62, 74, 94, ..., . - Robert G. Wilson v, May 16 2014

			a(11)=3:
11 = 1 + 10 = A000290(1) + A005117(7)
   = 4 + 7  = A000290(2) + A005117(6)
   = 9 + 2  = A000290(3) + A005117(2).

T. D. Noe, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Square Numbers.
Eric Weisstein's World of Mathematics, Squarefree
Robert G. Wilson v, Plot of first 100000 terms

Cf. A000290, A002636, A005117, A115288, A160324, A160325, A160326, A240088, A242442, A242443.

f[n_] := Count[ SquareFreeQ@# & /@ (n - Range[1, Floor[ Sqrt[ n]]]^2), True]; Array[f, 105] (* Robert G. Wilson v, May 16 2014 *)

a(n) = sum(k=1, n-1, issquare(k) * issquarefree(n-k)); \\ Michel Marcus, Oct 30 2020

a(n+1) = Sum_{k=1..n} A008966(k)*A010052(n-k+1). - Reinhard Zumkeller, Nov 04 2009
a(n) < sqrt(n). - Robert G. Wilson v, May 17 2014
G.f.: (Sum_{i>=1} x^(i^2))*(Sum_{j>=1} mu(j)^2*x^j). - Ilya Gutkovskiy, Feb 06 2017

A271325 Number of ordered ways to write n as x^3 + y^2 + z*(3z+1), where x, y and z are integers with x positive and y nonnegative.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 04 2016

nonn

We guess that a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 6, 7, 8, 13, 18, 20, 23, 25, 44, 49, 55, 59, 121, 238.
Based on our computation, we propose the following general conjecture (which extends the conjectures in A262813 and A270469).
Conjecture: Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Every natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers, where P(x,y,z) is any of the following cubic polynomials: x^3+T(y)+z^2, a*x^3+T(y)+pen(z) (a = 1,2,3,4), x^3+T(y)+z*(5z+1)/2, x^3+T(y)+z*(3z+r) (r = 1,2), x^3+T(y)+z*(7z+3)/2, x^3+T(y)+z*(9z+j)/2 (j = 5,7), x^3+T(y)+z*(5z+r) (r = 2,3), x^3+T(y)+2z*(3z+r) (r = 1,2), x^3+T(y)+z*(6z+5), x^3+T(y)+z*(13z+j)/2 (j = 3,7,9), x^3+T(y)+z*(7z+k) (k = 2,6), a*x^3+y^2+pen(z) (a = 1,2,3,4), x^3+y^2+z*(5z+3)/2, x^3+y^2+2*pen(z), x^3+2*T(y)+pen(z), x^3+2*T(y)+z(5z+j)/2 (j = 1,3), a*x^3+2*T(y)+z*(3z+2) (a = 1,2,3), x^3+2*T(y)+z*(7z+3)/2, x^3+4*T(y)+pen(z), x^3+2y^2+pen(z), x^3+pen(y)+c*pen(z) (c = 1,2,3,4), x^3+b*pen(y)+z*(5z+j)/2 (b = 1,2; j = 1,3), x^3+pen(y)+z*(7z+k)/2 (k = 1,3,5), x^3+pen(y)+z*(4z+j) (j = 1,3), x^3+pen(y)+z*(9z+5)/2, a*x^3+pen(y)+z*(9z+r)/2 (a = 1,2; r = 1,7), x^3+pen(y)+z*(5z+r) (r = 1,2,3,4), a*x^3+pen(y)+z*(11z+9)/2 (a = 1,2), x^3+pen(y)+2z*(3z+2),x^3+pen(y)+z*(13z+11)/2, x^3+pen(y)+z*(7z+k) (k = 4,5,6), x^3+pen(y)+3z*(5z+3)/2, x^3+pen(y)+z*(15z+11)/2, x^3+pen(y)+z*(8z+7), x^3+pen(y)+z*(11z+7), x^3+2*pen(y)+z*(7z+j)/2 (j = 1,5), x^3+2*pen(y)+3*pen(z), x^3+2*pen(y)+z*(4z+1), x^3+2*pen(y)+z*(7z+2), x^3+y*(5y+j)/2+z*(7z+k)/2 (j = 1,3; k = 3,5), x^3+y*(5y+3)/2+z*(9z+7)/2, x^3+y*(3y+2)+z*(4z+1), x^3+y*(3y+2)+z*(5z+1)/2, x^3+y*(7y+3)/2+z*(7z+5)/2, 2x^3+T(y)+z*(5z+3)/2, 2x^3+T(y)+z*(3z+r) (r = 1,2), 2x^3+T(y)+z*(5z+4), 2x^3+2*T(y)+z*(5z+3)/2, 2x^3+3*T(y)+pen(z), 2x^3+y^2+2*pen(z), 2x^3+pen(y)+pen(z), a*x^3+pen(y)+3*pen(z) (a = 2,3,4), a*x^3+pen(y)+z*(7z+5)/2 (a = 2,3,4), 2x^3+pen(y)+z*(5z+k) (k = 1,3), 2x^3+y*(5y+3)/2+z*(7z+5)/2, 2x^3+2*pen(y)+z*(3z+2), 2x^3+2*pen(y)+z*(7z+5)/2, 2x^3+y*(3y+2)+z*(4z+3), 3x^3+pen(y)+z*(7z+3)/2, 4x^3+y^2+z*(5z+1)/2, 4x^3+pen(y)+z*(4z+3).
The listed ternary polynomials in the conjecture should exhaust all those P(x,y,z) = a*x^3+y*(s*y+t)/2+z*(u*z+v)/2 with a,s,u > 0, 0 <= t <= s, 0 <= v <= u, s == t (mod 2), u == v (mod 2), and (s-2t)*(u-2v) nonzero, such that any natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers. Note that those numbers y*(2y+1) with y integral are just triangular numbers.
Conjecture verified for all polynomials up to 10^11. - Mauro Fiorentini, Aug 03 2023
See also A271106 for another general conjecture on universal sums.

			a(13) = 1 since 13 = 2^3 + 1^2 + 1*(3*1+1).
a(18) = 1 since 18 = 2^3 + 0^2 + (-2)*(3*(-2)+1).
a(20) = 1 since 20 = 1^3 + 3^2 + (-2)*(3*(-2)+1).
a(23) = 1 since 23 = 2^3 + 1^2 + 2*(3*2+1).
a(25) = 1 since 25 = 1^3 + 0^2 + (-3)*(3*(-3)+1).
a(44) = 1 since 44 = 2^3 + 6^2 + 0*(3*0+1).
a(49) = 1 since 49 = 1^3 + 2^2 + (-4)*(3*(-4)+1).
a(55) = 1 since 55 = 3^3+ 2^2 + (-3)*(3*(-3)+1).
a(59) = 1 since 59 = 2^3 + 7^2 + (-1)*(3*(-1)+1).
a(121) = 1 since 121 = 3^3 + 8^2 + 3*(3*3+1).
a(238) = 1 since 238 = 4^3 + 12^2 + 3*(3*3+1).

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), 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, A000326, A000578, A001318, A160325, A160326, A262813, A262815, A262816, A270488, A270469, A270516, A270533, A270559, A270566, A271106.

pQ[n_]:=pQ[n]=IntegerQ[Sqrt[12n+1]]
Do[r=0;Do[If[pQ[n-x^3-y^2],r=r+1],{x,1,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Label[aa];Continue,{n,1,70}]

A254661 Number of ways to write n as the sum of a triangular number, an even square and a second pentagonal number.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Feb 04 2015

nonn

Conjecture: (i) a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 2, 4, 9, 11, 13, 20, 35, 69, 75, 188.
(ii) For each a = 2,3, any nonnegative integer n can be written as x(x+1)/2 + a*y^2 + z*(3*z+1)/2 with x,y,z nonnegative integers.
Compare part (i) of this conjecture with the conjecture in A160325.

			a(20) = 1 since 20 = 1*2/2 + 2^2 + 3*(3*3+1)/2.
a(35) = 1 since 35 = 7*8/2 + 0^2 + 2*(3*2+1)/2.
a(69) = 1 since 69 = 2*3/2 + 8^2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 9*10/2 + 2^2 + 4*(3*4+1)/2.
a(188) = 1 since 188 = 1*2/2 + 0^2 + 11*(3*11+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Cf. A000217, A000290, A016742, A005449, A160325.

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-4y^2-z(3z+1)/2],r=r+1],{y,0,Sqrt[n/4]},{z,0,(Sqrt[24(n-4y^2)+1]-1)/6}];
Print[n," ",r];Label[aa];Continue,{n,0,100}]

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A254677 Least positive integer m with A160325(m) = n.

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A262815 Number of ordered ways to write n as x^3 + y*(y+1)/2 + z*(3*z+1)/2, where x, y and z are nonnegative integers.

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A262816 Number of ordered ways to write n as x^3 + y^2 + z*(3*z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

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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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A160326 Number of ways to express n=0,1,2,... as the sum of two squares and a pentagonal number.

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A240088 The number of ways of writing n as an ordered sum of a triangular number (A000217), a square (A000290) and a pentagonal number (A000326).

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A287616 Number of ways to write n as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z nonnegative integers.

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A085263 Number of ways to write n as the sum of a squarefree number (A005117) and a positive square (A000290).

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A262815 Number of ordered ways to write n as x^3 + y(y+1)/2 + z(3*z+1)/2, where x, y and z are nonnegative integers.

A262816 Number of ordered ways to write n as x^3 + y^2 + z(3z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.