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 1-3 of 3 results.

A285984 Numbers k such that 27*T(k)+1 is a square, where T(m) is the m-th triangular number A000217(m).

Original entry on oeis.org

0, 110, 374, 107184, 363264, 103968854, 352366190, 100849681680, 341794841520, 97824087261230, 331540643908694, 94889263793711904, 321594082796592144, 92042488055813286134, 311945928772050471470, 89281118524875093838560, 302587229314806160734240, 86602592926640785210117550
Offset: 0

Views

Author

Vladimir Pletser, May 01 2017

Keywords

Comments

Numbers a(n) that make sqrt(27*T(a(n))+1) an integer.
This sequence a(n) gives also the indices of the triangular numbers T(a(n)) such that the 3rd degree Diophantine Bachet-Mordell equation y^2 = x^3+K holds with x = 3*T(a(n)) = A286035(n), y = T(a(n))* sqrt(27*T(a(n))+1) = A286036(n) and K = T(a(n))^2 = A286037(n).

Examples

			k = 110 is a term because 27*(T(110) + 1) = 27 * (110*111/2 + 1) is a square. - _David A. Corneth_, May 02 2017
For n = 2, a(2) = 264*sqrt(27*(a(0)*(a(0)+1)/2)+1)+ a(-2) = 264*sqrt(27*(0*(0+1)/2)+1) + 110 = 374.
For n = 6, a(6) = 264*sqrt(27*(a(4)*(a(4)+1)/2)+1)+ a(2) = 264*sqrt(27*(363264*(363264+1)/2)+1) + 374 = 352366190.

References

  • V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.

Links

Crossrefs

Cf. A286035, A286036, A286037, A285955, A006454, A000217, A006451, A081119, A054504.

Programs

  • Maple
    restart: am2:=110: am1:=0: a0:=0: ap1:=110: print ('0,0','1,110'); for n from 2 to 1000 do a:= 264*sqrt(27* (a0^2+a0)/2+1)+am2; print(n,a); am2:=am1; am1:=a0; a0:=ap1; ap1:=a; end do:
  • Mathematica
    nxt[{a_,b_}]:={b,485*a+242+66*Sqrt[54a^2+54*a+4]}; NestList[nxt,{0,110},20][[All,1]] (* Harvey P. Dale, May 30 2018 *)
  • PARI
    is(n) = issquare(27*binomial(n+1, 2)+1) \\ David A. Corneth, May 02 2017

Formula

a(n) = 264*sqrt(27*T(a(n-2))+1)+ a(n-4) = 264*sqrt(27*(a(n-2)*(a(n-2)+1)/2)+1)+ a(n-4), with a(-2)=110, a(-1)=0, a(0)=0, a(1)=110.
Empirical g.f.: 22*x*(5 + 12*x + 5*x^2) / ((1 - x)*(1 - 970*x^2 + x^4)). - Colin Barker, May 01 2017, verified by Robert Israel, May 03 2017
a(n) = 485*a(n-2)+242+66*sqrt(54*a(n-2)^2+54*a(n-2)+4). - Robert Israel, May 03 2017

A286036 a(n) is the solution y to the Bachet Mordell equation y^2=x^3+K, with x = 3*T(b(n)) and K = (T(b(n)))^2, where T(b(n)) is the triangular number of b(n)= A285984(n).

Original entry on oeis.org

0, 2478630, 96492000, 2262209634604920, 88065491686677120, 2064651070850763887750940, 80374740223699340246041830, 1884345278651963087653858708518360, 73355621393690297028946986338029560, 1719785575058362227821108881720941727234290, 66949481579385248741161156467886515267346140
Offset: 0

Views

Author

Vladimir Pletser, May 01 2017

Keywords

Comments

a(n) is the producs of the triangular number T(b(n)) and the square root of 27 times this triangular number plus one, sqrt(27*T(b(n))+1), where b(n) is the sequence A285984(n) of numbers n such that (27*T(n)+1) is a square.

Examples

			For n = 2, b(n) = 374, a(n)= 96492000.
For n = 3, b(n) = A285984(n) =107184. Therefore, a(n) = T(b(n))* sqrt(27*T(b(n))+1) = A000217(A285984(n))* sqrt(27*A000217(A285984(n))+1) = A000217(107184)* sqrt(27*A000217(107184)+1) =5744258520* sqrt(27*5744258520 +1) = 2262209634604920.

References

  • V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.

Links

Crossrefs

Cf. A285984, A286035, A286037, A285955, A006454, A000217, A006451, A081119, A054504.

Programs

  • Maple
    restart: bm2:=110: bm1:=0: b0:=0: bp1:=110: print ('0,0','1,2478630’); for n from 2 to 1000 do b:= 264*sqrt(27* (b0^2+b0)/2+1)+bm2; T:=b*(b+1)/2; a:= T*sqrt(27*T+1); print(n,a); bm2:=bm1; bm1:=b0; b0:=bp1; bp1:=b; end do:

Formula

Since b(n) = 264*sqrt(27*T(b(n-2))+1)+ b(n-4) = 264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-2)=110, b(-1)=0, b(0)=0, b(1)=110 (see A285984) and a(n) = T(b(n))*sqrt(27*T(b(n))+1) (this sequence), one has :
a(n) = ([264*sqrt(27*T(b(n-2))+1)+ b(n-4)]*[ 264*sqrt(27*T(b(n-2))+1)+ b(n-4)+1]/2) *sqrt(27*([264*sqrt(27*T(b(n-2))+1)+ b(n-4)]*[ 264*sqrt(27*T(b(n-2))+1)+ b(n-4)+1]/2)+1).
Empirical g.f.: 330*x*(1 + x)*(7511 + 284889*x + 108094375*x^2 + 284889*x^3 + 7511*x^4) / ((1 - 912670090*x^2 + x^4)*(1 - 970*x^2 + x^4)). - Colin Barker, May 01 2017

A286037 a(n) = T(A285984(n))^2, where T(m) is the m-th triangular number A000217(m).

Original entry on oeis.org

0, 37271025, 4917515625, 32996505944592590400, 4353432777721630310400, 29211445283110309395256454577225, 3854046352373857001854365165911025, 25860572538708927496411840821477504196161600, 3411945020082158343071838489442339152945921600, 22894081602203374655543296113789919615194083223613314225
Offset: 0

Views

Author

Vladimir Pletser, May 01 2017

Keywords

Comments

a(n) =(T(b(n)))^2, parameters K=a(n) of the Bachet Mordell equation y^2=x^3+K, with x= 3*T(b(n)) and y= T(b(n))*sqrt(27*T(b(n))+1), where T(b(n)) is the triangular number of b(n)= A285984(n).

Examples

			For n = 2, b(n) = 374, a(n)= 4917515625.
For n = 3, b(n) = A285984(n) =107184. Therefore, a(n) = (T(b(n)))^2 = (A000217(A285984(n)))^2 = (A000217(107184))^2 = (5744258520)^2=32996505944592590400.

References

  • V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.

Links

Crossrefs

Cf. A285984, A286035, A286036, A285955, A006454, A000217, A006451, A081119, A054504.

Programs

  • Maple
    restart: bm2:=110: bm1:=0: b0:=0: bp1:=110: print ('0,0','1,4917515625’); for n from 2 to 1000 do b:= 264*sqrt(27*(b0^2+b0)/2+1)+bm2; a:=(b*(b+1)/2)^2; print(n,a); bm2:=bm1; bm1:=b0; b0:=bp1; bp1:=b; end do:

Formula

Since b(n) = 264*sqrt(27*T(b(n-2))+1)+ b(n-4) = 264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-2)=110, b(-1)=0, b(0)=0, b(1)=110 (see A285984) and a(n) = (T(b(n)))^2 (this sequence), one has :
a(n) = ([264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4) ]*[ 264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1]/2)^2.
Empirical g.f.: 27225*x*(1369 + 179256*x + 30879367019*x^2 + 168661970400*x^3 + 30879367019*x^4 + 179256*x^5 + 1369*x^6) / ((1 - x)*(1 - 940898*x + x^2)*(1 - 970*x + x^2)*(1 + 970*x + x^2)*(1 + 940898*x + x^2)). - Colin Barker, May 01 2017
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