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

A286035 a(n) = 3*T(A285984(n)), where T(m) is the m-th triangular number A000217(m).

Original entry on oeis.org

0, 18315, 210375, 17232775560, 197941645440, 16214284059063255, 186242898311223435, 15255987442587265956120, 175235570535035566127880, 14354328072739259079522561195, 164878797845087651200279041495, 13505958574968967401962031517525680, 155134131134672045268505114018663320
Offset: 0

Views

Author

Vladimir Pletser, May 01 2017

Keywords

Comments

This sequence a(n) gives the solutions x of the 3rd degree Diophantine Bachet-Mordell equation y^2=x^3+K, with y = T(b(n))*sqrt(27*T(b(n))+1) = A286036(n) and K = (T(b(n)))^2 = A286037(n), the square of the triangular number of b(n)= A285984(n).

Examples

			For n = 2, b(n) = 374, a(n)= 210375.
For n = 3, b(n) = A285984(n) =107184. Therefore, a(n) = 3*T(b(n)) = 3*A000217(A285984(n)) = 3*A000217(107184) = 3*5744258520=17232775560.

References

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

Links

Crossrefs

Cf. A285984, A286036, A286037, A285955, A006454, A000217, A006451, A081119, A054504

Programs

  • Maple
    restart: bm2:=110: bm1:=0: b0:=0: bp1:=110: print ('0,0','1,18315'); for n from 2 to 1000 do b:= 264*sqrt(27* (b0^2+b0)/2+1)+bm2; a:=3*b*(b+1)/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) = 3*T(b(n)) (this sequence), one has :
a(n) = 3*[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.
Empirical g.f.: 495*x*(37 + 388*x + 37*x^2) / ((1 - x)*(1 - 970*x + x^2)*(1 + 970*x + x^2)). - 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
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.