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.

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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

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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,...}.

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}]

A254668 Number of ways to write n as the sum of a square, a second pentagonal number, and a hexagonal number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 04 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 5, 13, 14, 20, 112, 125.
Compare this conjecture with the conjecture in A160324.
The conjecture that a(n) > 0 for all n = 0,1,2,... appeared in Conjecture 1.2(ii) of the author's JNT paper in the links. - Zhi-Wei Sun, Oct 03 2016

Examples

			a(20) = 1 since 20 = 2^2 + 3*(3*3+1)/2 + 1*(2*1-1).
a(112) = 1 since 112 = 7^2 + 6*(3*6+1)/2 + 2*(2*2-1).
a(125) = 1 since 125 = 5^2 + 8*(3*8+1)/2 + 0*(2*0-1).
		

Crossrefs

Programs

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

A254680 Least positive integer m with A254661(m) = n.

Original entry on oeis.org

1, 3, 7, 17, 21, 51, 66, 72, 157, 147, 121, 136, 246, 297, 332, 367, 402, 506, 547, 577, 677, 796, 892, 731, 926, 1216, 1116, 976, 1181, 1402, 1556, 1416, 1507, 1496, 2287, 1622, 1977, 2112, 1942, 2131, 2017, 2882, 2767, 2501, 3162, 3671, 3097, 3187, 3047, 3762, 3867, 2952, 4356, 4111, 4826, 5112, 5211, 4811, 4686, 5461
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 05 2015

Keywords

Comments

Conjecture: a(n) exists for any n > 0. Moreover, no term is divisible by 5, and no term with n>2 is congruent to 3 modulo 5.

Crossrefs

Programs

  • Maple
    a(3) = 7 since 7 is the first positive integer that can be written as x*(x+1)/2 + (2y)^2 + z*(3*z+1)/2 (with x,y,z nonnegative integers) in exactly 3 ways. In fact, 7 = 0*1/2 + 0^2 +2*(3*2+1)/2 = 1*2/2 + 2^2 +1*(3*1+1)/2 = 2*3/2 + 2^2 + 0*(3*0+1)/2.
  • Mathematica
    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}]
Previous Showing 11-15 of 15 results.