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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A230596 Number of ways to write n = x + y + z with 0 < x <= y <= z such that x*y*z is a triangular number, and that x is a triangular number of the form (p^2 - 1)/8 with p an odd prime.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 2, 1, 1, 2, 2, 1, 1, 2, 2, 4, 1, 1, 2, 3, 2, 5, 1, 3, 3, 3, 3, 2, 8, 1, 4, 2, 2, 3, 5, 1, 3, 6, 3, 5, 3, 1, 6, 4, 5, 3, 3, 1, 6, 6, 3, 4, 2, 4, 3, 8, 3, 3, 8, 5, 2, 4, 4, 6, 6, 3, 6, 2, 3, 12, 7, 1, 10, 7, 3, 4, 5, 3, 7, 8, 2, 5, 4, 6, 4, 2, 5, 6, 6, 4, 4, 13, 6, 9, 6, 4, 10, 7, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 24 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7.
(ii) For any integer n > 7, there are positive integers x, y, z with x + y + z = n such that x*y*z is a triangular number and x is among 1, 2, 3, 4, 5, 6.
Note that a(3k) and a(3k+2) are positive for every k = 1, 2, 3, .... In fact, 3k = 1 + k + (2k-1) with 1*k*(2k-1) = 2k*(2k-1)/2 a triangular number, and 3k+2 = 1 + k + (2k+1) with 1*k*(2k+1) = 2k(2k+1)/2 a triangular number.

Examples

			a(10) = 1 since 10 = 3 + 3 + 4, and 3 = (5^2-1)/8 with 5 an odd prime, and 3*3*4 = 8*9/2 is a triangular number.
a(31) = 1 since 31 = 3 + 11 + 17, and 3 = (5^2-1)/8 with 5 an odd prime, and 3*11*17 = 33*34/2 is a triangular number.

Links

Crossrefs

Cf. A000217, A000040, A132399, A229166, A230121, A230451.

Programs

  • Mathematica
    TQ[n_]:=IntegerQ[Sqrt[8n+1]]
a[n_]:=Sum[If[TQ[(Prime[i]^2-1)/8*y*(n-(Prime[i]^2-1)/8-y)],1,0],{i,2,PrimePi[Sqrt[8n/3+1]]},{y,(Prime[i]^2-1)/8,(n-(Prime[i]^2-1)/8)/2}]
Table[a[n],{n,1,100}]

A227877 Number of ways to write n = x + y + z (x, y, z > 0) such that x*y and x*z are triangular numbers, and 6*y-1 and 6*z+1 are both prime.

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 2, 3, 3, 7, 3, 6, 3, 3, 2, 3, 7, 6, 7, 5, 4, 5, 10, 2, 10, 4, 5, 2, 2, 9, 5, 9, 2, 4, 3, 4, 5, 7, 5, 11, 12, 5, 8, 11, 12, 5, 11, 3, 7, 11, 4, 10, 6, 2, 9, 11, 8, 7, 9, 8, 9, 4, 3, 4, 10, 6, 9, 15, 9, 17, 3, 3, 8, 12, 10, 5, 1, 7, 9, 16, 8, 17, 6, 8, 16, 6, 8, 8, 10, 1, 6, 4, 8, 5, 23, 11, 2, 9, 6, 14
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 25 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 4.
For n = 4*k - 1, we have n = (2k-1) + k + k with (2k-1)*k = 2k*(2k-1)/2 a triangular number. For n = 4*k + 1, we have n = (2k+1) + k + k with (2k+1)*k = 2k*(2k+1)/2 a triangular number. For n = 4*k + 2, we have n = (2k+1) + k + (k+1), and (2k+1)*k = 2k*(2k+1)/2 and (2k+1)*(k+1) = (2k+1)(2k+2)/2 are both triangular numbers.
For n = 5*k, we have n = k + (2k-1) + (2k+1), and k*(2k-1) = 2k*(2k-1)/2 and k*(2k+1) = 2k*(2k+1)/2 are both triangular numbers. For n = 5*k - 2, we have n = k + (2k-1) + (2k-1) with k*(2k-1) = 2k*(2k-1)/2 a triangular number. For n = 5*k + 2, we have n = k + (2k+1) + (2k+1) with k*(2k+1) = 2k*(2k+1)/2 a triangular number.

Examples

			a(77) = 1 since 77 = 1 + 10 + 66, and 1*10 = 4*5/2 and 1*66 = 11*12/2 are triangular numbers, and 6*10 - 1 = 59 and 6*66 + 1 = 397 are both prime.
a(90) = 1 since 90 = 45 + 22 + 23, and 45*22 = 44*45/2 and 45*23 = 45*46/2 are triangular numbers, and 6*22 - 1 = 131 and 6*23 + 1 = 139 are both prime.

Links

Crossrefs

Cf. A000040, A000217, A132399, A229166, A230121, A230451, A230596.

Programs

  • Mathematica
    TQ[n_]:=IntegerQ[Sqrt[8n+1]]
a[n_]:=Sum[If[PrimeQ[6j-1]&&PrimeQ[6(n-i-j)+1]&&TQ[i*j]&&TQ[i(n-i-j)],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]
Showing 1-2 of 2 results.

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