A165517 -id:A165517 - cp's OEIS frontend

A116476 Numbers n such that T(n) + T(n+1) + ... + T(n+10) is a square, where T(m) = A000217(m) is the m-th triangular number.

13, 46, 229, 1608, 7335, 20304, 92391, 635710, 2892133, 8001886, 36403981, 250470288, 1139495223, 3152724936, 14343078279, 98684659918, 448958227885, 1242165625054, 5651136440101, 38881505539560, 176888402293623, 489410103548496

Offset: 1

Edward Fedorovich (chipramy(AT)012.net.il), Mar 29 2006

Positive integers n such that 11*n^2 + 121*n + 440 = 2*m^2 for some integer m. - Max Alekseyev, Jan 20 2010

			13 belongs to this sequence since T(13) + T(14) + ... + T(23) = 91 + 105 + 120 + 136 + 153 + 171 + 190 + 210 + 231 + 253 + 276 = 1936 = 44^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,394,-394,0,0,-1,1).

Cf. A176541, A176542, A165517, A202391

For[n = 1, n < 100000, n++, If[IntegerQ[Sqrt[Sum[i*(i+1)/2, {i, n, n + 10}]]], Print[n]]] (* Stefan Steinerberger, Mar 30 2006 *)
LinearRecurrence[{1,0,0,394,-394,0,0,-1,1},{13,46,229,1608,7335,20304,92391,635710,2892133},30] (* Harvey P. Dale, Sep 01 2017 *)

For n>8, a(n) = 394*a(n-4) - a(n-8) + 2156. - Max Alekseyev, Jan 20 2010

G.f.: x*(2*x^8+7*x^7+15*x^6+33*x^5-605*x^4-1379*x^3-183*x^2-33*x-13)/((x-1)*(x^8-394*x^4+1)). - Colin Barker, Nov 22 2012

Extended by Max Alekseyev, Jan 20 2010

A176541 Numbers n such that there exist n consecutive triangular numbers which sum to a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 11, 13, 22, 23, 25, 27, 32, 37, 39, 46, 47, 48, 49, 50, 52, 59, 66, 71, 73, 83, 94, 98, 100, 104, 107, 109, 111, 118, 121, 128, 143, 146, 147, 148, 157, 167, 176, 179, 181, 183, 191, 192, 193, 194, 200, 214, 219, 227, 239, 241, 242, 243, 244, 253, 263

Offset: 1

Andrew Weimholt, Apr 20 2010

nonn

Numbers n such that there exists some x >= 0 such that A000292(x+n) - A000292(x) is a square. Terms of this sequence, for which only a finite number of solutions x exist, are given in A176542.
Integer n is in the sequence if there exist non-degenerate solutions to the Diophantine equation: 8x^2 - n*y^2 - A077415(n) = 0. A degenerate solution is one involving triangular numbers with negative indexes.
The sum of n consecutive triangular numbers starting at the j-th is Sum_{k=j..j+n-1} A000217(k) = n*(n^2 + 3*j*n + 3*j^2 - 1)/6, see A143037. - R. J. Mathar, May 06 2015

			0 is in the sequence because the sum of 0 consecutive triangular numbers is 0 (a square).
1 is in the sequence because there exist triangular numbers which are squares (cf. A001110).
2 is in the sequence because ANY 2 consecutive triangular numbers sum to a square.
3 is in the sequence because there are infinitely many solutions (cf. A165517).
4 is in the sequence because there infinitely many solutions (cf. A202391).
5 is NOT in the sequence because no 5 consecutive triangular numbers sum to a square.
For n=8, solutions to the Diophantine equation exist, but start at A000217(-2) and A000217(-6): 1 + 0 + 0 + 1 + 3 + 6 + 10 + 15 = 36 and 15 + 10 + 6 + 3 + 1 + 0 + 0 + 1 = 36. There are no non-degenerate solutions for n=8. Hence, 8 is not included in the sequence.
For n=11, there exist infinitely many solutions (cf. A116476), so 11 is in the sequence.

Cf. A176542, A000217, A000292, A001110, A077415.

More terms from Max Alekseyev, May 10 2010

A176542 Numbers n such that there are only a finite nonzero number of sets of n consecutive triangular numbers that sum to a square.

Original entry on oeis.org

32, 50, 98, 128, 200, 242, 338, 392, 512, 578, 722, 800, 968, 1058, 1250, 1352, 1568, 1682, 1922, 2048, 2312, 2450, 2738, 2888, 3200, 3362, 3698, 3872, 4232, 4418, 4802, 5000, 5408, 5618, 6050, 6272, 6728, 6962, 7442, 7688, 8192, 8450, 8978, 9248, 9800

Offset: 1

Andrew Weimholt, Apr 20 2010

nonn

Members of A176541, for which there are only a finite number of solutions.
Integer n is in this sequence if n=2*m^2 and the equation (2*x-m*y)*(2*x+m*y)=A077415(n)/2 has integer solutions with y>=n. - Max Alekseyev, May 10 2010
It seems that a(n) = 2*A001651(n+2)^2. - Colin Barker, Sep 25 2015

			32 is in this sequence because there is only one set of 32 consecutive triangular numbers that sum to a square (namely, A000217(26) thru A000217(57), which sum to 29584 = 172^2).
3 is NOT in this sequence, because there are infinitely many sets of 3 consecutive triangular numbers that sum to a square (cf. A165517).
4 is NOT in this sequence, because there are infinitely many sets of 4 consecutive triangular numbers that sum to a square (cf. A202391).
5 is NOT in this sequence, because there are NO sets of 5 consecutive triangular numbers that sum to a square.
11 is NOT in this sequence, since there are infinitely many sets of 11 consecutive triangular numbers that sum to a square (cf. A116476).

Cf. A176541, A000217, A000292, A077415.

Conjectures from Colin Barker, Sep 24 2015: (Start)
a(n) = (9*n^2+24*n+16)/2 for n even.
a(n) = (9*n^2+30*n+25)/2 for n odd.
a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>5.
G.f.: -2*x*(4*x^4-3*x^3-8*x^2+9*x+16) / ((x-1)^3*(x+1)^2).
(End)

Terms a(6) onward from Max Alekseyev, May 10 2010

A257293 Numbers n such that T(n) + T(n+1) + ... + T(n+12) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

3, 29, 75, 432, 998, 3624, 8310, 44717, 102443, 370269, 848195, 4561352, 10448838, 37764464, 86508230, 465213837, 1065679683, 3851605709, 8822991915, 47447250672, 108688879478, 392826018504, 899858667750, 4839154355357, 11085200027723, 40064402282349

Offset: 1

M. F. Hasler, May 04 2015

It is well known that T(n)+T(n+1) is always a square. T(n)+T(n+1)+T(n+2) is a square for n in A165517. T(n)+T(n+1)+T(n+2)+T(n+3) is a square for n in A202391. There is no sequence of 5, 6, 7, 8, 9 or 10 consecutive T(i)'s which sum to a square, cf. A176541. The next possible length is 11, see A116476. Then comes this sequence, corresponding to length 13.

Positive integers y in the solutions to 2*x^2-13*y^2-169*y-728 = 0. - Colin Barker, May 04 2015

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,102,-102,0,0,-1,1).

Cf. A176541, A176542, A000217, A000292, A001110, A077415.

Cf. A116476 (length 11).

I:=[3,29,75,432,998,3624,8310,44717,102443]; [n le 9 select I[n] else Self(n-1)+102*Self(n-4)-102*Self(n-5)-Self(n-8)+Self(n-9): n in [1..40]]; // Vincenzo Librandi, May 05 2015

Select[Range[10^5],IntegerQ[Sqrt[(#^2+13*#+56)*13/2]]&] (* Ivan N. Ianakiev, May 04 2015 *)
LinearRecurrence[{1, 0, 0, 102, -102, 0, 0, -1, 1}, {3, 29, 75, 432, 998, 3624, 8310, 44717, 102443}, 50] (* Vincenzo Librandi, May 05 2015 *)

for(n=0,10^8,issquare(binomial(n+14,3)-binomial(n+1,3))&&print1(n","))

Vec(x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)) + O(x^100)) \\ Colin Barker, May 04 2015

G.f.: x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)). - Colin Barker, May 04 2015

A202391 Indices of the smallest of four consecutive triangular numbers summing up to a square.

Original entry on oeis.org

5, 39, 237, 1391, 8117, 47319, 275805, 1607519, 9369317, 54608391, 318281037, 1855077839, 10812186005, 63018038199, 367296043197, 2140758220991, 12477253282757, 72722761475559, 423859315570605, 2470433131948079

Offset: 1

Max Alekseyev, Dec 18 2011

nonn

Positive integers n such that A000217(n) + A000217(n + 1) + A000217(n + 2) + A000217(n + 3) is a square (=A075870(n+1)^2).

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A176541, A176542, A165517, A116476, A075870

a(n) = A002315(n) - 2.
G.f.: x*(1+x)*(x-5) / ( (x-1)*(1-6*x+x^2) ). - R. J. Mathar, Dec 19 2011
a(n+2) = 6*a(n+1) - a(n) + 8; a(n+3) = 7*a(n+2) - 7*a(n+1) + a(n); a(n+1) = (-4 + (7 + 5*r)*(3 + 2*r)^n + (7 - 5*r)*(3 - 2*r)^n)/2 where r = sqrt(2). - Paul Weisenhorn, Jan 13 2013

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

A262489 The index of the first of two consecutive positive triangular numbers (A000217) the sum of which is equal to the sum of three consecutive positive triangular numbers.

Original entry on oeis.org

7, 18, 78, 187, 781, 1860, 7740, 18421, 76627, 182358, 758538, 1805167, 7508761, 17869320, 74329080, 176888041, 735782047, 1751011098, 7283491398, 17333222947, 72099131941, 171581218380, 713707828020, 1698478960861, 7064979148267, 16813208390238

Offset: 1

Colin Barker, Sep 24 2015

For the index of the first of the corresponding three consecutive triangular numbers, see A165517.

			7 is in the sequence because T(7)+T(8) = 28+36 = 64 = 15+21+28 = T(5)+T(6)+T(7), where T(k) is the k-th triangular number.

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

Cf. A000217, A165517, A262490, A262491, A262492.

Vec(-x*(x^4-x^3-10*x^2+11*x+7)/((x-1)*(x^4-10*x^2+1)) + O(x^30))

a(n) = a(n-1)+10*a(n-2)-10*a(n-3)-a(n-4)+a(n-5) for n>5.

G.f.: -x*(x^4-x^3-10*x^2+11*x+7) / ((x-1)*(x^4-10*x^2+1)).

A319588 First of three consecutive triangular numbers that add up to a perfect square.

Original entry on oeis.org

0, 15, 105, 2016, 11628, 203203, 1152921, 19968040, 113108320, 1957220895, 11084786065, 191793185496, 1086209028828, 18793829460003, 106437529743441, 1841604033412080, 10429792989769440, 180458406785594575, 1022013288177368025, 17683082313822046576

Offset: 1

Harvey P. Dale, Sep 23 2018

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,108,-108,-982,982,108,-108,-1,1).

Cf. A000217, A165517.

Select[Partition[Accumulate[Range[2500000]],3,1],IntegerQ[Sqrt[ Total[ #]]]&] [[All,1]]
(#(#+1))/2&/@LinearRecurrence[{1,10,-10,-1,1},{0,5,14,63,152},20]
LinearRecurrence[{1, 108, -108, -982, 982, 108, -108, -1, 1},{0, 15, 105, 2016, 11628, 203203, 1152921, 19968040, 113108320},20] (* Ray Chandler, Mar 01 2024 *)

concat(0, Vec(x^2*(15 + 90*x + 291*x^2 - 108*x^3 - 83*x^4 + 2*x^5 + x^6) / ((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)*(1 - 10*x^2 + x^4)) + O(x^40))) \\ Colin Barker, Sep 24 2018

a(n) = A000217(A165517(n)). - Alois P. Heinz, Sep 24 2018
From Colin Barker, Sep 24 2018: (Start) Heinz's formula implies a g.f. and a recurrence:
G.f.: x^2*(15 + 90*x + 291*x^2 - 108*x^3 - 83*x^4 + 2*x^5 + x^6) / ((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) + 108*a(n-2) - 108*a(n-3) - 982*a(n-4) + 982*a(n-5) + 108*a(n-6) - 108*a(n-7) - a(n-8) + a(n-9) for n>9.
(End)

cp's OEIS Frontend

A116476 Numbers n such that T(n) + T(n+1) + ... + T(n+10) is a square, where T(m) = A000217(m) is the m-th triangular number.

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A176541 Numbers n such that there exist n consecutive triangular numbers which sum to a square.

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A176542 Numbers n such that there are only a finite nonzero number of sets of n consecutive triangular numbers that sum to a square.

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A257293 Numbers n such that T(n) + T(n+1) + ... + T(n+12) is a square, where T = A000217 (triangular numbers).

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A202391 Indices of the smallest of four consecutive triangular numbers summing up to a square.

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

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A262489 The index of the first of two consecutive positive triangular numbers (A000217) the sum of which is equal to the sum of three consecutive positive triangular numbers.

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A319588 First of three consecutive triangular numbers that add up to a perfect square.

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