A124174 -id:A124174 - cp's OEIS frontend

A217278 Sequences A124174 and A006454 interlaced.

0, 0, 1, 3, 10, 15, 45, 120, 351, 528, 1540, 4095, 11935, 17955, 52326, 139128, 405450, 609960, 1777555, 4726275, 13773376, 20720703, 60384555, 160554240, 467889345, 703893960, 2051297326, 5454117903, 15894464365, 23911673955, 69683724540, 185279454480

Offset: 0

Raphie Frank, Sep 29 2012

a(2n) and 2*a(2n) + 1 are triangular.

a(2n + 1) is triangular and a(2n + 1)/2 is the harmonic mean of consecutive triangular numbers (therefore, a(2n + 1) + 1 is square).

			a(18) = 35*(52326 - 1540) + 45 = 1777555,
a(19) = 35*(139128 - 4095) + 120 = 4726275.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,-34,0,-1,0,1).

Cf. (sqrt(8a(2n) + 1) - 1)/2 = A216134(n) = A216162(2n + 1).
Cf. sqrt(a(2n+1) + 1) = A006452(n + 1) = A216162(2n + 2).
Cf. (sqrt(8a(2n+1) + 1) - 1)/2 = A006451(n).

LinearRecurrence[{0,1,0,34,0,-34,0,-1,0,1},{0,0,1,3,10,15,45,120,351,528},40] (* Harvey P. Dale, Aug 04 2019 *)

concat([0,0], Vec(-x^2*(3*x^5+x^4+12*x^3+9*x^2+3*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)) + O(x^100))) \\ Colin Barker, Jun 23 2015

a(n) = 35*(a(n-4) - a(n-8)) + a(n-12).
lim n --> infinity a(2n)/a(2n - 1) = (3 + sqrt(8))/2.
From Raphie Frank, Dec 21 2015: (Start)
a(2*n + 1) = 1/64*(((4 + sqrt(2)) * (1 - (-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2)) + (4 - sqrt(2)) * (1+(-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2))))^2 - 1.
a(2*n + 2) = 1/2*(3*(a(2*n + 1)) + sqrt((a(2*n + 1)) + 1) * sqrt(8*(a(2*n + 1)) + 1) + 1).
(End)

A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

Original entry on oeis.org

0, 3, 15, 120, 528, 4095, 17955, 139128, 609960, 4726275, 20720703, 160554240, 703893960, 5454117903, 23911673955, 185279454480, 812293020528, 6294047334435, 27594051024015, 213812329916328, 937385441796000, 7263325169820735, 31843510970040003, 246739243443988680

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Alternative definition: a(n) is triangular and a(n)/2 is the harmonic average of consecutive triangular numbers. See comments and formula section of A005563, of which this sequence is a subsequence. - Raphie Frank, Sep 28 2012
As with the Sophie Germain triangular numbers (A124174), 35 = (a(n) - a(n-6))/(a(n-2) - a(n-4)). - Raphie Frank, Sep 28 2012
Sophie Germain triangular numbers of the second kind as defined in A217278. - Raphie Frank, Feb 02 2013
Triangular numbers m such that m+1 is a square. - Bruno Berselli, Jul 15 2014
From Vladimir Pletser, Apr 30 2017: (Start)
Numbers a(n) which are the triangular number T(b(n)), where b(n) is the sequence A006451(n) of numbers n such that T(n)+1 is a square.
a(n) also gives the x solutions of the 3rd-degree Diophantine Bachet-Mordell equation y^2 = x^3 + K, with y = T(b(n))*sqrt(T(b(n))+1) = A285955(n) and K = T(b(n))^2 = A285985(n), the square of the triangular number of b(n) = A006451(n).
Also: This sequence is a subsequence of A000217(n), namely A000217(A006451(n)). (End)

			From _Raphie Frank_, Sep 28 2012: (Start)
35*(528 - 15) + 0 = 17955 = a(6),
35*(4095 - 120) + 3 = 139128 = a(7),
35*(17955 - 528) + 15 = 609960 = a(8),
35*(139128 - 4095) + 120 = 4726275 = a(9). (End)
From _Raphie Frank_, Feb 02 2013: (Start)
a(7) = 139128 and a(9) = 4726275.
a(9) = (2*(sqrt(8*a(7) + 1) - 1)/2 + 3*sqrt(a(7) + 1) + 1)^2 - 1 = (2*(sqrt(8*139128 + 1) - 1)/2 + 3*sqrt(139128 + 1) + 1)^2 - 1 = 4726275.
a(9) = 1/2*((3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)^2 + (3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)) = 1/2*((3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)^2 + (3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)) = 4726275. (End)
From _Vladimir Pletser_, Apr 30 2017: (Start)
For n=2, b(n)=5, a(n)=15
For n=5, b(n)=90, a(n)= 4095
For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. (End)

Edward J. Barbeau, Pell's Equation, New York: Springer-Verlag, 2003, p. 17, Exercise 1.2.
Allan J. Gottlieb, How four dogs meet in a field, and other problems, Technology Review, Jul/August 1973, pp. 73-74.
Vladimir Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vladimir Pletser, Table of n, a(n) for n = 0..1000 (first 60 terms from Vincenzo Librandi)
M.A. Bennett and A. Ghadermarzi, Data on Mordell's curve.
Michael A. Bennett and Amir Ghadermarzi, Mordell's equation : a classical approach, arXiv:1311.7077 [math.NT], 2013.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Jeffrey Shallit, Letter to N. J. A. Sloane, Oct. 1975.
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Eric Weisstein's World of Mathematics, Mordell Curve.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. sqrt(a(n) + 1) = A006452(n + 1) = A216162(2n + 2) and (sqrt(8a(n) + 1) - 1)/2 = A006451.
Cf. A217278, A124174, A216134. - Raphie Frank, Feb 02 2013
Subsequence of A182334.
Cf. A001109, A245031.
Cf. A285955, A285985, A000217, A006451, A081119, A054504. - Vladimir Pletser, Apr 30 2017

I:=[0,3,15,120,528,4095]; [n le 6 select I[n] else 35*(Self(n-2) - Self(n-4)) + Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 21 2015

A006454:=-3*z*(1+4*z+z**2)/(z-1)/(z**2-6*z+1)/(z**2+6*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print ('0,0','1,3','2,15'); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=b*(b+1)/2; print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do: # Vladimir Pletser, Apr 30 2017

Clear[a]; a[0] = a[1] = 1; a[2] = 2; a[3] = 4; a[n_] := 6a[n - 2] - a[n - 4]; Array[a, 40]^2 - 1 (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)
LinearRecurrence[{1,34,-34,-1,1},{0,3,15,120,528},30] (* Harvey P. Dale, Feb 18 2023 *)

concat(0, Vec(3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30))) \\ Colin Barker, Apr 30 2017

a(n) = A006451(n)*(A006451(n)+1)/2.
a(n) = A006452(n)^2 - 1. - Joerg Arndt, Mar 04 2011
a(n) = 35*(a(n-2) - a(n-4)) + a(n-6). - Raphie Frank, Sep 28 2012
From Raphie Frank, Feb 01 2013: (Start)
a(0) = 0, a(1) = 3, and a(n+2) = (2x + 3y + 1)^2 - 1  = 1/2*((3x + 4y + 1)^2 + (3x + 4y + 1)) where x = (sqrt(8*a(n) + 1) - 1)/2 = A006451(n) =  1/2*(A216134(n + 1) + A216134(n - 1)) and y = sqrt(a(n) + 1) = A006452(n + 1) = 1/2*(A216134(n + 1) - A216134(n - 1)).
Note that A216134(n + 1) = x + y, and A216134(n + 3) = (2x + 3y + 1) + (3x + 4y + 1) = (5x + 7y + 2), where A216134 gives the indices of the Sophie Germain triangular numbers. (End)
a(n) = (1/64)*(((4 + sqrt(2))*(1 -(-1)^(n+1)*sqrt(2))^(2* floor((n+1)/2)) + (4 - sqrt(2))*(1 + (-1)^(n+1)*sqrt(2))^(2*floor((n+1)/2))))^2 - 1. - Raphie Frank, Dec 20 2015
From Vladimir Pletser, Apr 30 2017: (Start)
Since b(n) = 8*sqrt(T(b(n-2))+1)+ b(n-4) = 8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-1)=-1, b(0)=0, b(1)=2, b(2)=5 (see A006451) and a(n) = T(b(n)) (this sequence), we have:
a(n) = ((8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4))*(8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1)/2). (End)
From Colin Barker, Apr 30 2017: (Start)
G.f.: 3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) for n > 4.
(End)
a(n)  = (A001109(n/2+1) - 2*A001109(n/2))^2 - 1 if n is even, and (A001109((n+3)/2) - 4*A001109((n+1)/2))^2 - 1 if n is odd (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Better description from Harvey P. Dale, Jan 28 2001
More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001
Minor edits by N. J. A. Sloane, Oct 24 2009

A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

Original entry on oeis.org

0, 1, 4, 9, 26, 55, 154, 323, 900, 1885, 5248, 10989, 30590, 64051, 178294, 373319, 1039176, 2175865, 6056764, 12681873, 35301410, 73915375, 205751698, 430810379, 1199208780, 2510946901, 6989500984, 14634871029, 40737797126, 85298279275, 237437281774

Offset: 0

Raphie Frank, Sep 01 2012

Numbers n such that 2*triangular(n) + 1 is a triangular number. Equivalently, numbers n such that n^2 + n + 1 is a triangular number. - Alex Ratushnyak, Apr 18 2013

For n > 0, a(n) is the n-th almost cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Colin Barker, Table of n, a(n) for n = 0..1000
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Wikipedia, Pell numbers
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A124174, A000129, A001333, A006451, A006452, A124124, A079496.

Cf. A000217, A069017 (triangular numbers of the form k^2 + k + 1).

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 4, 9, 26}, 40] (* T. D. Noe, Sep 03 2012 *)

Vec( x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 13 2014

isok(n) = ispolygonal(n*(n+1) + 1, 3); \\ Michel Marcus, Aug 13 2014

G.f.: x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)). - R. J. Mathar, Sep 08 2012
sqrt(2) = lim_{k->infinity} ((a(2k+1) + a(2k) + 1)/2)/(a(2k+1) - a(2k)) = lim_{k->infinity} A001333(2k + 1)/A000129(2k + 1).
1 + (sqrt 2) = lim_{k->infinity} (a(2k + 1) - a(2k))/(a(2k + 1) - 2*a(2k) +  a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A000129(2k).
1 + 1/(sqrt 2) = lim_{k->infinity} (a(2k+1) - a(2k))/(a(2k) - a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A001333(2k).
a(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013
From Raphie Frank, Jan 04 2013: (Start)
A124174(n) = a(n)*(a(n) + 1)/2.
A079496(n) = a(n + 1) - a(n).
A000129(2n) = a(2n) - 2*a(2n - 1) +  a(2n - 2).
A000129(2n) = a(2n + 1) - 2*a(2n) +  a(2n - 1).
A000129(2n + 1) = a(2n + 1) - a(2n).
A001333(2n) = a(2n) - a(2n - 1).
A001333(2n + 1) = (a(2n + 1) + a(2n) + 1)/2.
A006451(n + 1) = (a(n + 2) + a(n))/2.
A006452(n + 2) = (a(n + 2) - a(n))/2.
A124124(n + 2) = (a(n + 2) + a(n))/2 + (a(n + 2) - a(n)).
(End)
a(n + 2) = sqrt(8*a(n)^2 + 8*a(n) + 9) + 3*a(n) + 1; a(0) = 0, a(1) = 1. - Raphie Frank, Feb 02 2013
a(n) = (3/8 + sqrt(2)/4)*(1 + sqrt(2))^n + (-1/8 - sqrt(2)/8)*(-1 + sqrt(2))^n + (3/8 - sqrt(2)/4)*(1 - sqrt(2))^n + (-1/8 + sqrt(2)/8)*(-1 - sqrt(2))^n - 1/2. - Robert Israel, Aug 13 2014
E.g.f.: (1/4)*(-2*cosh(x) - 2*sinh(x) + 2*cosh(sqrt(2)*x)*(cosh(x) + 2*sinh(x)) + sqrt(2)*(cosh(x) + 3*sinh(x))*sinh(sqrt(2)*x)). - Stefano Spezia, Dec 10 2019

A069017 Triangular numbers of the form k^2 + k + 1.

Original entry on oeis.org

1, 3, 21, 91, 703, 3081, 23871, 104653, 810901, 3555111, 27546753, 120769111, 935778691, 4102594653, 31788928731, 139367449081, 1079887798153, 4734390674091, 36684396208461, 160829915470003, 1246189583289511, 5463482735306001, 42333761435634903

Offset: 1

Amarnath Murthy, Apr 02 2002

G. C. Greubel, Table of n, a(n) for n = 1..1001
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A124174.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (x^4 +2*x^3 -16*x^2 +2*x +1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)) )); // G. C. Greubel, Dec 01 2018

Do[a = n(n + 1) + 1; b = Floor[Sqrt[2a]]; If[b(b + 1) == 2a, Print[a]], {n, 1, 106}] (* Robert G. Wilson v *)
Select[Table[n^2+n+1,{n,0,206*10^6}],OddQ[Sqrt[8#+1]]&] (* The program takes a long time to run. *) (* Harvey P. Dale, Sep 22 2017 *)
CoefficientList[Series[(x^4 +2*x^3 -16*x^2 +2*x +1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)), {x,0,50}], x] (* G. C. Greubel, Dec 01 2018 *)

Vec((x^4+2*x^3-16*x^2+2*x+1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)) +O(x^66)) /* Joerg Arndt, Mar 25 2013 */

s=((x^4 +2*x^3 -16*x^2 +2*x +1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2))).series(x, 50); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 01 2018

G.f.: (x^4 + 2*x^3 - 16*x^2 + 2*x + 1)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
From Zak Seidov, Sep 25 2010: (Start)
a(n) = 34*a(n-2) - a(n-4) - 11.
a(n) = 2*A124174(n) + 1. (End)
a(n) = (A077443(n)^2 - 1)/2. - Amiram Eldar, Dec 01 2018

Program and terms from Robert G. Wilson v

a(18)-a(22) from Alex Ratushnyak, Mar 23 2013

A274579 Values of k such that 2k+1 and 5k+1 are both triangular numbers.

Original entry on oeis.org

0, 1, 7, 27, 540, 2002, 10660, 39501, 779247, 2887450, 15372280, 56960982, 1123674201, 4163701465, 22166817667, 82137697110, 1620337419162, 6004054625647, 31964535704101, 118442502272205, 2336525434757970, 8657842606482076, 46092838318496542

Offset: 1

Colin Barker, Jun 29 2016

Intersection of A074377 and A085787.

			7 is in the sequence because 2*7+1 = 15, 5*7+1 = 36, and 15 and 36 are both triangular numbers.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1442,-1442,0,0,-1,1).

Cf. A000217, A074377, A085787, A124174.

concat(0, Vec(x^2*(1+6*x+20*x^2+513*x^3+20*x^4+6*x^5+x^6)/((1-x)*(1+6*x-x^2)*(1-6*x-x^2)*(1+38*x^2+x^4)) + O(x^30)))

isok(n) = ispolygonal(2*n+1, 3) && ispolygonal(5*n+1, 3); \\ Michel Marcus, Jun 29 2016

G.f.: x^2*(1+6*x+20*x^2+513*x^3+20*x^4+6*x^5+x^6) / ((1-x)*(1+6*x-x^2)*(1-6*x-x^2)*(1+38*x^2+x^4)).

A274680 Values of n such that 2n+1 and 4n+1 are both triangular numbers.

Original entry on oeis.org

0, 16065, 545751, 21394547226, 726784809030, 28491418065071115, 967869505172593485, 37942420317086720855700, 1288925370210688376036076, 50528452330120333959563160501, 1716479960463788790499334882595, 67289447366315927998308608003134830

Offset: 1

Colin Barker, Jul 02 2016

			16065 is in the sequence because 2*16065+1 = 32131, 4*16065+1 = 64261, and 32131 and 64261 are both triangular numbers.

Colin Barker, Table of n, a(n) for n = 1..325
Index entries for linear recurrences with constant coefficients, signature (1,1331714,-1331714,-1,1).

Cf. A124174 (2*n+1 and 9*n+1), A274579 (2*n+1 and 5*n+1), A274603 (2*n+1 and 3*n+1).

Rest@ CoefficientList[Series[459 x^2 (35 + 1154 x + 35 x^2)/((1 - x) (1 - 1154 x + x^2) (1 + 1154 x + x^2)), {x, 0, 12}], x] (* Michael De Vlieger, Jul 02 2016 *)

isok(n) = ispolygonal(2*n+1, 3) && ispolygonal(4*n+1, 3)

concat(0, Vec(459*x^2*(35+1154*x+35*x^2)/((1-x)*(1-1154*x+x^2)*(1+1154*x+x^2)) + O(x^20)))

Intersection of A074377 and A274681.

G.f.: 459*x^2*(35+1154*x+35*x^2) / ((1-x)*(1-1154*x+x^2)*(1+1154*x+x^2)).

A274756 Values of n such that 2n+1 and 6n+1 are both triangular numbers.

Original entry on oeis.org

0, 945, 13167, 35578242, 495540990, 1338951572595, 18649189618605, 50390103447476100, 701843601611053692, 1896381151803363988917, 26413182084381205040235, 71368408216577696911440390, 994033693861758668873164410, 2685878672926303893761783662455

Offset: 1

Colin Barker, Jul 04 2016

Intersection of A074377 and A274757.

			945 is in the sequence because 2*945+1 = 1891, 6*945+1 = 5671, and 1891 and 5671 are both triangular numbers.

Colin Barker, Table of n, a(n) for n = 1..400
Index entries for linear recurrences with constant coefficients, signature (1,37634,-37634,-1,1).

Cf. A124174 (2*n+1 and 9*n+1), A274579 (2*n+1 and 5*n+1), A274603 (2*n+1 and 3*n+1), A274680 (2*n+1 and 4*n+1).

isok(n) = ispolygonal(2*n+1, 3) && ispolygonal(6*n+1, 3)

concat(0, Vec(63*x^2*(15+194*x+15*x^2)/((1-x)*(1-194*x+x^2)*(1+194*x+x^2)) + O(x^20)))

G.f.: 63*x^2*(15+194*x+15*x^2) / ((1-x)*(1-194*x+x^2)*(1+194*x+x^2)).

a(n) = a(n-1)+37634*a(n-2)-37634*a(n-3)-a(n-4)+a(n-5). - Wesley Ivan Hurt, Apr 24 2021

A174685 Indices i such that 2*A000326(i)+1 is also a pentagonal number.

Original entry on oeis.org

0, 75, 244, 86359, 281384, 99658019, 324716700, 115005267375, 374722790224, 132715978892539, 432429775201604, 153154124636722439, 499023585859860600, 176739727114798801875, 575872785652503930604, 203957491936353180641119

Offset: 1

Jonathan Vos Post, Mar 27 2010

			Using P(n) = A000326(n) we have:
a(1) = 0 because 2*P(0)+1 = 2*0+1 = 1 = P(1).
a(2) = 75 because 2*P(75)+1 = 2*8400+1 = 16801 = P(106).
a(3) = 244 because 2*P(244)+1 = 2*89182+1 = 178365 = P(345).

Ray Chandler, Table of n, a(n) for n = 1..653 (terms to 1000 digits)
Eric Weisstein's World of Mathematics, Pentagonal Number.
Eric Weisstein's World of Mathematics, Sophie Germain prime.
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Cf. A000326, A005384, A124174, A260937.

LinearRecurrence[{1,1154,-1154,-1,1},{0,75,244,86359,281384},30] (* Harvey P. Dale, Jul 17 2014 *)

/* Jack Brennen, who extended this sequence, found that, other than the trivial pair (0,1), you need to solve: u^2 - 2*v^2 = 23. And take only those values where u and v are both congruent to 5 mod 6. Then you have P((u+1)/6) = 2*P((v+1)/6)+1. The following PARI/GP will give the nontrivial answers: */
{forstep(i=7,47,8,
  print(Vec(lift(Mod((x+1),x^2-2)^i*(4*x-3)+(x+1))/6));
  print(Vec(lift(Mod((x+1),x^2-2)^i*(4*x+3)+(x+1))/6)))}

G.f.: x^2*(-75-169*x+435*x^2+x^3) / ( (x-1)*(x^2-34*x+1)*(x^2+34*x+1) ) with a(n) = 1/6 -7*A029547(n) +239*A029547(n-1) +35*A029547(n)*(-1)^n/6 +1183*A029547(n-1)*(-1)^(n-1)/6. - R. J. Mathar, Oct 25 2011

a(n)= a(n-1)+ 1154*a(n-2)-1154*a(n-3)-a(n-4)+a(n-5). - Harvey P. Dale, Jul 17 2014

A274832 Values of n such that 2n+1 and 7n+1 are both triangular numbers (A000217).

Original entry on oeis.org

0, 27, 297, 24570, 267030, 22064157, 239792967, 19813588740, 215333817660, 17792580624687, 193369528466037, 15977717587380510, 173645621228683890, 14347972600887073617, 155933574493829667507, 12884463417879004727880, 140028176249837812737720

Offset: 1

Colin Barker, Jul 08 2016

Intersection of A074377 and A274830.

			27 is in the sequence because 2*27+1 = 55, 7*27+1 = 190, and 55 and 190 are both triangular numbers.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (1,898,-898,-1,1).

Cf. A000217, A074377, A274830.

Cf. A124174 (2*n+1 and 9*n+1), A274579 (2*n+1 and 5*n+1), A274603 (2*n+1 and 3*n+1), A274680 (2*n+1 and 4*n+1), A274756 (2*n+1 and 7*n+1).

LinearRecurrence[{1, 898, -898, -1, 1}, {0, 27, 297, 24570, 267030}, 20] (* Paolo Xausa, Oct 21 2024 *)

isok(n) = ispolygonal(2*n+1, 3) && ispolygonal(7*n+1, 3)

concat(0, Vec(27*x^2*(1+10*x+x^2)/((1-x)*(1-30*x+x^2)*(1+30*x+x^2)) + O(x^20)))

G.f.: 27*x^2*(1+10*x+x^2) / ((1-x)*(1-30*x+x^2)*(1+30*x+x^2)).

A217000 Triangular numbers of the form 2p-1 where p is prime.

Original entry on oeis.org

3, 21, 45, 105, 253, 325, 465, 561, 861, 1081, 1225, 1485, 1653, 1953, 3741, 4005, 4753, 6441, 7021, 7381, 8001, 9045, 10153, 13041, 15753, 19701, 20301, 21945, 23005, 23653, 24753, 25425, 28441, 32385, 35245, 37401, 38781, 41041, 43365, 45753, 46665, 48205

Offset: 1

César Eliud Lozada, Sep 22 2012

nonn

Indexes n in A000217(n): A217001.
The only triangular odd number with the form 2p+1 and p prime is 15=2*7+1?
The only triangular even numbers with the form 2p and p prime are {6,10}?
From Daniel Starodubtsev, Mar 13 2020: (Start)
Proof that 15 is the only triangular number of the form 2p + 1 where p is prime: we can express T(n)=n*(n+1)/2 and p=(T(n)-1)/2=(n*(n+1)/2-1)/2=(n+2)*(n-1)/4, which can be prime only if n+2=4 or n-1=4, from which we get the only possible value n=5 (T(n)=15).
It can also be easily seen that {6,10} are the only possible values of T(n) such that T(n)/2 is prime. (End)

			For A000217 = {0, 1, 3, 6, 10, 15, 21, 28,...}, A000217(6) = 21 = 2*(11)-1. As 11 is prime then A000217(6) is in the sequence. A000217(5) = 15 = 2*(8)-1. As 8 is not prime then A000217(5) is not in the sequence.

T. D. Noe, Table of n, a(n) for n = 1..10000

Subsequence of A000217.

Cf. A124174 (2*tr+1 is also a triangular number), A217001.

tn := unapply(n*(n+1)/2,n):
f := unapply((t+1)/2,t):
T := []: N := []: P := []:
for k from 0 to 5000 do
  t:=tn(k):
  p := f(k):
  if p = floor(p) then
    p = floor(p):
    if isprime(p) then
      T := [op(T), t]:
      N := [op(N), k]:
      P := [op(P), p]:
    end if:
  end if:
  if nops(T) = 50 then
    break:
  end if:
end do:
T := T;

tri = 0; t = {}; Do[tri = tri + n; If[PrimeQ[(tri + 1)/2], AppendTo[t, tri]], {n, 500}]; t (* T. D. Noe, Sep 24 2012 *)

cp's OEIS Frontend

A217278 Sequences A124174 and A006454 interlaced.

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A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

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A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

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A069017 Triangular numbers of the form k^2 + k + 1.

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A274579 Values of k such that 2*k+1 and 5*k+1 are both triangular numbers.

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A274680 Values of n such that 2*n+1 and 4*n+1 are both triangular numbers.

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A274756 Values of n such that 2*n+1 and 6*n+1 are both triangular numbers.

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A274579 Values of k such that 2k+1 and 5k+1 are both triangular numbers.

A274680 Values of n such that 2n+1 and 4n+1 are both triangular numbers.

A274756 Values of n such that 2n+1 and 6n+1 are both triangular numbers.

A274832 Values of n such that 2n+1 and 7n+1 are both triangular numbers (A000217).