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.

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.

Crossrefs

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

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.

Crossrefs

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.

Crossrefs

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.