A129445 -id:A129445 - cp's OEIS frontend

A129446 Primes p such that p^2 is a centered triangular number; or primes in A129445.

2, 19, 79, 1861, 7741, 1647522841025041

Offset: 1

Alexander Adamchuk, Apr 15 2007, Apr 26 2007

nonn

The next terms are too large to include, see the b-file.

Alexander Adamchuk, Apr 26 2007, Table of n, a(n) for n = 1..9

Cf. A005448 (centered triangular numbers).
Cf. A129444 (numbers k such that centered triangular number A005448(k) = 3*k*(k-1)/2 + 1 is a perfect square).
Cf. A129445 (numbers k > 0 such that k^2 is a centered triangular number).

Do[ f = 3n(n-1)/2 + 1; If[ PrimeQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,150000} ]

A129444 Numbers k such that the centered triangular number A005448(k) = 3k(k-1)/2 + 1 is a perfect square.

Original entry on oeis.org

0, 1, 2, 7, 16, 65, 154, 639, 1520, 6321, 15042, 62567, 148896, 619345, 1473914, 6130879, 14590240, 60689441, 144428482, 600763527, 1429694576, 5946945825, 14152517274, 58868694719, 140095478160, 582740001361, 1386802264322

Offset: 1

Alexander Adamchuk, Apr 15 2007

Corresponding numbers m > 0 such that m^2 is a centered triangular number are listed in A129445 = {1, 2, 8, 19, 79, 188, 782, 1861, 7741, 18422, 76628, 182359, ...}.

			G.f. = x^2 + 2*x^3 + 7*x^4 + 16*x^5 + 65*x^6 + 154*x^7 + 639*x^8 + 1520*x^9 + ...

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Cf. A005448 (centered triangular numbers).
Cf. A129445 (numbers k > 0 such that k^2 is a centered triangular number).
Cf. A000290, A004189, A249483.

I:=[0,1,2,7,16,65]; [n le 6 select I[n] else 11*Self(n-2) -11*Self(n-4) +Self(n-6): n in [1..40]]; // G. C. Greubel, Feb 07 2024

Do[ f = 3n(n-1)/2 + 1; If[ IntegerQ[ Sqrt[f] ], Print[ n ] ], {n,1,150000} ]
a[1]=0;a[2]=1;a[3]=2;a[4]=7;a[5]=16;a[6]=65;a[n_]:=a[n]=11(a[n-2]-a[n-4])+a[n-6];Table[a[n], {n, 100}] (* Zak Seidov, Apr 17 2007 *)
LinearRecurrence[{1,10,-10,-1,1},{0,1,2,7,16},30] (* Harvey P. Dale, Dec 06 2012 *)

{a(n) = my(m); m = if( n<1, 2-n, n-1); (n<1) + (-1)^(n<1) * polcoeff( (x + x^2 - 5*x^3 - x^4) / ((1 - x) * (1 - 10*x^2 + x^4)) + x * O(x^m), m)}; /* Michael Somos, Apr 05 2008 */

def A129444_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(1+x-5*x^2-x^3)/((1-x)*(1-10*x^2+x^4)) ).list()
a=A129444_list(40); a[1:] # G. C. Greubel, Feb 07 2024

a(n) = 1/2 + sqrt(1/4 + (2/3)*(A129445(n)^2 - 1)).
a(n) = 11*(a(n-2) - a(n-4)) + a(n-6); a(1)=0; a(2)=1; a(3)=2; a(4)=7; a(5)=16; a(6)=65. - Zak Seidov, Apr 17 2007
a(n) = 1 - a(-n+3) for all n in Z. - Michael Somos, Apr 05 2008
G.f.: x^2*(1 + x - 5*x^2 - x^3) / ((1 - x) * (1 - 10*x^2 + x^4)). - Michael Somos, Apr 05 2008
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5); a(1)=0, a(2)=1, a(3)=2, a(4)=7, a(5)=16. - Harvey P. Dale, Dec 06 2012
a(n) = (1/2)*(2*[n=0] + 1 + ((1+(-1)^n)/2)*(31*b(n/2) - 3*b(n/2 + 1)) + ((1-(-1)^n)/2)*(13*b((n-1)/2) - b((n+1)/2))), where b(n)=A004189(n). - G. C. Greubel, Feb 07 2024

More terms from Zak Seidov, Apr 17 2007

A165517 Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290).

Original entry on oeis.org

0, 5, 14, 63, 152, 637, 1518, 6319, 15040, 62565, 148894, 619343, 1473912, 6130877, 14590238, 60689439, 144428480, 600763525, 1429694574, 5946945823, 14152517272, 58868694717, 140095478158, 582740001359, 1386802264320

Offset: 1

Ant King, Sep 25 2009, Oct 01 2009

Those perfect squares that can be expressed as the sum of three consecutive triangular numbers correspond to integer solutions of the equation T(k)+T(k+1)+T(k+2)=s^2, or equivalently to 3k^2 + 9k + 8 = 2s^2. Hence solutions occur whenever (3k^2 + 9k + 8)/2 is a perfect square, or equivalently when s>=2 and sqrt(24s^2 - 15) is congruent to 3 mod 6. This sequence returns the index of the smallest of the 3 triangular numbers, the values of s^2 are given in A165516 and, with the exception of the first term, the values of s are in A129445.

			The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 = T(63) + T(64) + T(65). Hence a(4)=63.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Cf. A000290, A129445, A000217, A165516.

I:=[0, 5, 14, 63, 152]; [n le 5 select I[n] else Self(n-1) + 10*Self(n-2) - 10*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // G. C. Greubel, Oct 21 2018

TriangularNumber[ n_ ]:=1/2 n (n+1);Select[ Range[ 0,10^7 ], IntegerQ[ Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1 ]+TriangularNumber[ #+2 ] ] ] & ]
CoefficientList[Series[x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)), {x,0,50}], x] (* or *) LinearRecurrence[{1,10,-10,-1,1}, {0, 5, 14, 63, 152}, 50] (* G. C. Greubel, Feb 17 2017 *)

x='x+O('x^50); concat([0], Vec(x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)))) \\ G. C. Greubel, Feb 17 2017

a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5).
G.f.: x(x^3 + x^2 - 9x - 5)/((x-1)(x^4 - 10x^2 + 1)).
a(n) = 10*a(n-2) - a(n-4) + 12. - Zak Seidov, Sep 25 2009

a(1) = 0 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander R. Povolotsky

More terms from Zak Seidov, Sep 25 2009

A085376 Ratio-dependent insertion sequence I(0.36704) (see the link below).

Original entry on oeis.org

1, 3, 11, 30, 109, 297, 1079, 2940, 10681, 29103, 105731, 288090, 1046629, 2851797, 10360559, 28229880, 102558961, 279447003, 1015229051, 2766240150, 10049731549, 27382954497, 99482086439, 271063304820, 984771132841

Offset: 1

John W. Layman, Jun 26 2003

nonn

This sequence is the ratio-determined insertion sequence (RDIS) "twin" of I(0.37802)=A080874 and "child" of I(0.33344)=A001835 and I(0.38208)=A001906 in the RDIS recurrence tree (see the link for an explanation of terms). See A082630, A082981, A085348 and A085349 for recent examples of RDIS sequences.

Conjecture: partial sums of A129445. - Sean A. Irvine, Jul 14 2022

John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A001835, A001906, A080874, A082630, A082981, A085348, A085349.

It is conjectured that a(n) = 10*a(n-2) - a(n-4).

Apparently a(n)*a(n+3) = -3 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004

A237610 Positive integers k such that x^2 - 10xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

8, 15, 20, 23, 24, 32, 47, 60, 71, 72, 80, 87, 92, 95, 96, 116, 128, 135, 152, 159, 167, 180, 188, 191, 200, 207, 212, 215, 216, 239, 240, 263, 276, 284, 288, 303, 311, 320, 335, 344, 348, 359, 368, 375, 380, 383, 384, 392, 404, 423, 431, 447, 456, 464, 479

Offset: 1

Colin Barker, Feb 10 2014

nonn

			15 is in the sequence because x^2 - 10xy + y^2 + 15 = 0 has integer solutions, for example (x, y) = (2, 19).

Cf. A072256 (k = 8), A129445 (k = 15), A080806 (k = 20), A074061 (k = 23), A001079 (k = 24).

Cf. A031363, A084917, A237351, A237599, A237606, A237609.

is(n)=m=bnfisintnorm(bnfinit(x^2-10*x+1),-n);#m>0&&denominator(polcoeff(m[1],1))==1 \\ Ralf Stephan, Feb 11 2014

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Original entry on oeis.org

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055

Offset: 1

Bruno Berselli, Dec 10 2013

For n>1, partial sums of A080872 starting from A080872(1).

			153 is in the sequence because 3*153*154/2+1 = 188^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Sequence A129444 gives n+1.
Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.
a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.

A165516 Perfect squares (A000290) that can be expressed as the sum of three consecutive triangular numbers (A000217).

Original entry on oeis.org

4, 64, 361, 6241, 35344, 611524, 3463321, 59923081, 339370084, 5871850384, 33254804881, 575381414521, 3258631508224, 56381506772644, 319312633001041, 5524812282304561, 31289379402593764, 541375222159074304, 3066039868821187801, 53049246959306977201, 300440617765073810704, 5198284826789924691364

Offset: 1

Ant King, Sep 25 2009, Oct 01 2009

Those perfect squares that can be expressed as the sum of three consecutive triangular numbers correspond to integer solutions of the equation T(k)+T(k+1)+T(k+2)=s^2, or equivalently to 3k^2+9k+8=2s^2. Hence solutions occur whenever 1/2 (3k^2+9k+8) is a perfect square, or equivalently when s>=2 and sqrt(24s^2-15) is congruent to 3 mod 6. Furthermore, with the exception of the first term, the members of this sequence are precisely those perfect squares that are also centered triangular numbers (A005448). For s>=2, the values of s are in A129445, and the corresponding indices of the smallest of the 3 triangular numbers are given in A165517.

			The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 (=T63+T64+T65), and hence a(4)=6241.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000290, A129445, A005448, A000290, A000217, A165517.

I:=[4, 64, 361, 6241, 35344]; [n le 5 select I[n] else Self(n-1) + 98*Self(n-2) - 98*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // G. C. Greubel, Oct 21 2018

Select[Range[2,1.8 10^7],Mod[Sqrt[24#^2-15],6]==3 &]^2
CoefficientList[Series[(4 + 60 x - 95 x^2 + x^4)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 14 2014 *)
LinearRecurrence[ {1,98,-98,-1,1}, {4, 64, 361, 6241, 35344}, 50] (* G. C. Greubel, Oct 21 2018 *)

Vec(O(x^66)+x*(4+60*x-95*x^2+x^4)/((1-x)*(1-10*x+x^2)*(1+10*x+x^2))) \\ Joerg Arndt, Mar 13 2014

a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
a(n) = 98*a(n-2) - a(n-4) - 30. - Ant King, Dec 09 2010
a(n) = (1/32)*(10 -3*(sqrt(6)-3) * (5-2*sqrt(6))^n + (2+ sqrt(6)) * (-5-2*sqrt(6))^n -(sqrt(6)-2) *(2*sqrt(6)-5)^n + 3*(3+sqrt(6)) *(5+2*sqrt(6))^n).
G.f.: x*(4+60*x-95*x^2+x^4)/((1-x)*(1-10*x+x^2)*(1+10*x+x^2)).
16*a(n) = 5 +9*A072256(n+1) +2*(-1)^n*A054320(n). - R. J. Mathar, Apr 28 2020

a(1) = 4 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander R. Povolotsky

a(16)-a(21) added by Alex Ratushnyak, Mar 12 2014

A249483 Squares (A000290) which are also centered triangular numbers (A005448).

Original entry on oeis.org

1, 4, 64, 361, 6241, 35344, 611524, 3463321, 59923081, 339370084, 5871850384, 33254804881, 575381414521, 3258631508224, 56381506772644, 319312633001041, 5524812282304561, 31289379402593764, 541375222159074304, 3066039868821187801, 53049246959306977201

Offset: 1

Colin Barker, Jan 13 2015

Apart from the first term the same as A165516. - R. J. Mathar, Jan 20 2015

			64 is in the sequence because the 8th square is 64, which is also the 7th centered triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000290, A005448, A129444, A129445.

I:=[1,4,64,361,6241]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Jan 20 2015

CoefficientList[Series[(x^2 - 5 x + 1) (x^2 + 8 x + 1) / ((1 - x) (x^2 - 10 x + 1) (x^2 + 10 x + 1)), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 20 2015 *)

Vec(-x*(x^2-5*x+1)*(x^2+8*x+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))

a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).

G.f.: -x*(x^2-5*x+1)*(x^2+8*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).

A257274 Numbers whose square can be written as sum of at least 3 consecutive triangular numbers (A000217).

Original entry on oeis.org

8, 10, 19, 26, 44, 58, 79, 84, 91, 105, 111, 121, 135, 140, 154, 172, 184, 188, 195, 203, 208, 212, 217, 222, 230, 240, 246, 265, 276, 286, 292, 316, 322, 329, 338, 364, 390, 426, 429, 462, 490, 498, 506, 548, 550, 605, 704, 714, 715, 763, 770, 780, 782, 807

Offset: 1

M. F. Hasler, May 02 2015

nonn

Any square can trivially be written as sum of two consecutive triangular numbers T = A000217, since T(n-1) + T(n) = n(n-1)/2 + n(n+1)/2 = n*2n/2 = n^2. But it seems nontrivial to determine the squares that can be written as sum of more than 2 consecutive triangular numbers.
Some of these have two different decompositions of this form, e.g., 286^2 = T(13)+...+T(78) = T(75)+...+T(96), 826^2 = T(13)+...+T(159) = T(43)+...+T(160). What is the sequence of these numbers?
The terms > 2 of A129445, numbers k > 0 such that k^2 is a centered triangular number, form a subsequence: they correspond to k^2 = T(n-2) + T(n-1) + T(n), with n in sequence A129444: numbers n such that centered triangular number A005448(n) = 3n(n-1)/2 + 1 is a perfect square.
Terms > 2 of sequence A075870 also form a subsequence, namely the numbers whose square is the sum of four triangular numbers T(n-3)+...+T(n), with n given by twice the terms > 1 of A046090 or A182435.

			8^2 = T(5)+T(6)+T(7), 10^2 = T(5)+T(6)+T(7)+T(8), 19^2 = T(14)+T(15)+T(16), 26^2 = T(3)+...+T(15), 44^2 = T(13)+...+T(23), ...

Chai Wah Wu, Table of n, a(n) for n = 1..10000

{a=[];(S(n)=binomial(n+2,3)); for(n=1,999,for(k=1,n-3,issquare(S(n)-S(k))&&a=concat(a,sqrtint(S(n)-S(k))))); Set(a)[1..50]}

a(14), a(43)-a(54) from Chai Wah Wu, Jan 20 2016

cp's OEIS Frontend

A129446 Primes p such that p^2 is a centered triangular number; or primes in A129445.

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A129444 Numbers k such that the centered triangular number A005448(k) = 3*k*(k-1)/2 + 1 is a perfect square.

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A165517 Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290).

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A085376 Ratio-dependent insertion sequence I(0.36704) (see the link below).

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A237610 Positive integers k such that x^2 - 10xy + y^2 + k = 0 has integer solutions.

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PARI

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

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Programs

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Formula

A165516 Perfect squares (A000290) that can be expressed as the sum of three consecutive triangular numbers (A000217).

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Magma

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A129444 Numbers k such that the centered triangular number A005448(k) = 3k(k-1)/2 + 1 is a perfect square.