cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

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.

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

Views

Author

Zhi-Wei Sun, Oct 03 2015

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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}]

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

Views

Author

Robert G. Wilson v, Mar 31 2014

Keywords

Comments

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; ..., .

Crossrefs

Programs

  • Maple
    # 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
  • Mathematica
    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

Views

Author

Zhi-Wei Sun, May 27 2017

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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

Views

Author

Reinhard Zumkeller, Jun 23 2003

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Mathematica
    f[n_] := Count[ SquareFreeQ@# & /@ (n - Range[1, Floor[ Sqrt[ n]]]^2), True]; Array[f, 105] (* Robert G. Wilson v, May 16 2014 *)
  • PARI
    a(n) = sum(k=1, n-1, issquare(k) * issquarefree(n-k)); \\ Michel Marcus, Oct 30 2020

Formula

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

Views

Author

Zhi-Wei Sun, Apr 04 2016

Keywords

Comments

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.

Examples

			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).
		

Crossrefs

Programs

  • Mathematica
    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}]

A242442 Number of ways of writing n, a positive integer, as an unordered sum of a triangular number (A000217), an odd square (A016754) and a pentagonal number (A000326).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, May 14 2014

Keywords

Comments

It is conjectured that only 18 cannot be so represented. See Sun, p. 4, Remark 1.2 (b).

Crossrefs

Programs

  • Mathematica
    planeFigurative[n_, r_] := (n - 2) Binomial[r, 2] + r; s = Sort@ Flatten@ Table[ planeFigurative[3, i] + planeFigurative[4, j] + planeFigurative[5, k], {i, 0, 20}, {j, 1, 11, 2}, {k, 0, 8}]; Table[ Count[s, n], {n, 0, 104}]

A242443 Number of ways of writing n, a positive integer, as an unordered sum of a triangular number (A000217), an even square (A016742) and a generalized pentagonal number (A001318).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 3, 4, 1, 4, 3, 4, 2, 2, 5, 3, 5, 3, 5, 4, 5, 7, 3, 4, 4, 6, 6, 4, 6, 3, 5, 7, 6, 4, 1, 7, 7, 6, 5, 6, 9, 5, 7, 7, 8, 6, 8, 4, 6, 6, 7, 9, 4, 10, 3, 6, 9, 7, 8, 5, 9, 7, 6, 7, 5, 11, 9, 7, 3, 7, 12, 13, 7, 7, 6, 9, 11, 6, 11, 8, 7, 10, 10, 8, 8, 8, 11, 5, 8, 5, 8, 11, 10, 10, 6, 14, 10, 6, 7, 7
Offset: 1

Views

Author

Robert G. Wilson v, May 14 2014

Keywords

Comments

It is conjectured (1.1) and then proved by theorem 1.2 that all positive integers can be so represented [Sun, pp. 4-5].

Crossrefs

Programs

  • Mathematica
    planeFigurative[n_, r_] := pf[n, r] = (n - 2) Binomial[r, 2] + r; s = Sort@ Table[ planeFigurative[3, i] + planeFigurative[3, j] + planeFigurative[3, k], {i, 0, 14}, {j, 0, 10, 2}, {k, -8, 8}]; Table[ Count[s, n], {n, 0, 50}]

A270966 Number of ways to write n as x^2 + y^2 + z*(3z+1)/2, where x, y and z are integers with 0 <= x <= y such that x or y has the form p-1 with p prime.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 3, 3, 5, 3, 2, 4, 2, 3, 3, 2, 4, 3, 5, 4, 2, 4, 4, 5, 2, 3, 2, 4, 5, 4, 5, 3, 6, 6, 4, 4, 4, 3, 4, 5, 1, 3, 5, 8, 5, 3, 6, 3, 4, 4, 4, 4, 4, 5, 3, 3, 6, 5, 8, 4, 2, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 27 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 49, 608.
(ii) Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Any positive integer can be written as (p-1)^2+P(x,y) with p prime and x and y integral, where the polynomial P(x,y) is either of the following ones: T(x)+2*pen(y), 2*T(x)+pen(y), T(x)+y*(5y+1)/2, T(x)+y*(9y+5)/2, pen(x)+y*(5y+j)/2 (j = 1,3), pen(x)+y*(7y+k)/2 (k = 3,5), pen(x)+y*(4y+j) (j = 1,3), pen(x)+y*(5y+r) (r = 1,2,3,4), pen(x)+2y*(3y+i) (i = 1,2), pen(x)+6*pen(y), x*(5x+1)/2+y*(3y+2), x*(5x+1)/2+y*(9y+7)/2, x*(5x+3)/2+y*(3y+i) (i = 1,2), x*(5x+3)/2+y*(9y+5)/2.
See also A270928 for a similar conjecture involving T(p-1) = p*(p-1)/2 with p prime.

Examples

			a(1) = 1 since 1 = 0^2 + (2-1)^2 + 0*(3*0+1)/2 with 2 prime.
a(12) = 2 since 12 = (2-1)^2 + 2^2 + 2*(2*3+1)/2 = (2-1)^2 + 3^2 + 1*(3*1+1)/2 with 2 prime.
a(49) = 1 since 49 = (2-1)^2 + 6^2 + (-3)*(3*(-3)+1)/2 with 2 prime.
a(608) = 1 since 608 = (7-1)^2 + 14^2 + (-16)*(3*(-16)+1)/2 with 7 prime.
		

Crossrefs

Programs

  • Mathematica
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
    Do[r=0;Do[If[(PrimeQ[x+1]||PrimeQ[y+1])&&pQ[n-x^2-y^2],r=r+1],{x,0,Sqrt[n/2]},{y,x,Sqrt[n-x^2]}];Print[n," ",r];Continue,{n,1,70}]

A165141 The least positive integer that can be written in exactly n ways as the sum of a square, a pentagonal number and a hexagonal number.

Original entry on oeis.org

3, 9, 1, 6, 16, 36, 50, 37, 66, 82, 167, 121, 162, 236, 226, 276, 302, 446, 478, 532, 457, 586, 677, 521, 666, 852, 976, 877, 1006, 1046, 1277, 1381, 1857, 1556, 1507, 1657, 1832, 1732, 2336, 2299, 2007, 2677, 2326, 2117, 2591, 2502, 2516, 2592, 3106, 3557
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 05 2009

Keywords

Comments

On Sep 04 2009, Zhi-Wei Sun conjectured that the sequence A160324 contains every positive integer, i.e., for each positive integer n there exists a positive integer s which can be written in exactly n ways as the sum of a square, a pentagonal number and a hexagonal number. Based on this conjecture we create the current sequence. It seems that 0.9 < a(n)/n^2 < 1.6 for n > 33. Zhi-Wei Sun conjectured that a(n)/n^2 has a limit c with 1.1 < c < 1.2. On Sun's request, his friend Qing-Hu Hou produced a list of a(n) for n = 1..913 (see the b-file).

Examples

			For n=5 the a(5)=16 solutions are 0^2+1+15 = 1^2+0+15 = 2^2+12+0 = 3^2+1+6 = 4^2+0+0 = 16.
		

Crossrefs

Programs

Formula

a(n) = min{m>0: m=x^2+(3y^2-y)/2+(2z^2-z) has exactly n solutions with x,y,z=0,1,2,...}.
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