A001110 -id:A001110 - cp's OEIS frontend

A257713 Triangular numbers (A000217) that are the sum of ten consecutive triangular numbers.

1485, 7260, 28920, 142845, 2112540, 10440165, 41673885, 205953660, 3046252485, 15054681960, 60093684540, 296985006165, 4392693942120, 21708840917445, 86655051404085, 428252172907560, 6334261618255845, 31304133548245020, 124956524030977320, 617539336347666645

Offset: 1

Colin Barker, May 05 2015

nonn

			1485 is in the sequence because T(54) = 1485 = 78+91+105+120+136+153+171+190+210+231 = T(12)+ ... +T(21).

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, A001110, A129803, A131557, A257711, A257712, A259413, A259414, A259415.

 LinearRecurrence[{1, 0, 0, 1442, -1442, 0, 0, -1, 1}, {1485, 7260, 28920, 142845, 2112540, 10440165, 41673885, 205953660, 3046252485}, 30] (* Vincenzo Librandi, Jun 27 2015 *)

Vec(-15*x*(8*x^8-5*x^7+5*x^5-11445*x^4+7595*x^3+1444*x^2+385*x+99) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)*(x^4+38*x^2+1)) + O(x^100))

G.f.: -15*x*(8*x^8-5*x^7+5*x^5-11445*x^4+7595*x^3+1444*x^2+385*x+99) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)*(x^4+38*x^2+1)).

A259413 Triangular numbers (A000217) that are the sum of eleven consecutive triangular numbers.

Original entry on oeis.org

2145, 3916, 7381, 13530, 843051, 1547920, 2926990, 5374281, 335521560, 616057651, 1164924046, 2138939715, 133536727236, 245189386585, 463636832725, 851292621696, 53147281907775, 97584759792586, 184526294489911, 338812324484700, 21152484662556621

Offset: 1

Colin Barker, Jun 26 2015

			2145 is in the sequence because T(65) = 2145 = 105 + 120 + 136 + 153 + 171 + 190 + 210 + 231 + 253 + 276 + 300 = T(14) + ... + T(24).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,398,-398,0,0,-1,1).

Cf. A000217, A001110, A129803, A131557, A257711, A257712, A257713, A259414, A259415.

LinearRecurrence[{1, 0, 0, 398, -398, 0, 0, -1, 1}, {2145, 3916, 7381, 13530, 843051, 1547920, 2926990, 5374281, 335521560}, 30] (* Vincenzo Librandi, Jun 27 2015 *)

Vec(-11*x*(6*x^8-x^7+x^5-2199*x^4+559*x^3+315*x^2+161*x+195)/((x-1)*(x^4-20*x^2+1)*(x^4+20*x^2+1)) + O(x^30))

G.f.: -11*x*(6*x^8-x^7+x^5-2199*x^4+559*x^3+315*x^2+161*x+195) / ((x-1)*(x^4-20*x^2+1)*(x^4+20*x^2+1)).

A259414 Triangular numbers (A000217) that are the sum of thirteen consecutive triangular numbers.

Original entry on oeis.org

2080, 414505, 28815436, 49317346, 3428789455, 698283666730, 48548229019381, 83089887991201, 5776831256176630, 1176469718198438755, 81794153348207147926, 139990009467226925656, 9732816854065394603605, 1982118534159467652450580, 137806953149317550935817071

Offset: 1

Colin Barker, Jun 26 2015

			2080 is in the sequence because T(64) = 2080 = 66 + 78 + 91 + 105 + 120 + 136 + 153 + 171 + 190 + 210 + 231 + 253 + 276 = T(11) + ... + T(23).

Colin Barker, Table of n, a(n) for n = 1..641
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1684802,-1684802,0,0,-1,1).

Cf. A000217, A001110, A129803, A131557, A257711, A257712, A257713, A259413, A259415.

 LinearRecurrence[{1, 0, 0, 1684802, -1684802, 0, 0, -1, 1}, {2080, 414505, 28815436, 49317346, 3428789455, 698283666730, 48548229019381, 83089887991201, 5776831256176630}, 30] (* Vincenzo Librandi, Jun 27 2015 *)

Vec(-13*x*(7*x^8 +153*x^6 +31725*x^5 -9608927*x^4 +1577070*x^3 +2184687*x^2 +31725*x +160) / ((x -1)*(x^2 -36*x -1)*(x^2 +36*x -1)*(x^4 +1298*x^2 +1)) + O(x^20))

G.f.: -13*x*(7*x^8 +153*x^6 +31725*x^5 -9608927*x^4 +1577070*x^3 +2184687*x^2 +31725*x +160) / ((x -1)*(x^2 -36*x -1)*(x^2 +36*x -1)*(x^4 +1298*x^2 +1)).

A259415 Triangular numbers (A000217) that are the sum of seventeen consecutive triangular numbers.

Original entry on oeis.org

1326, 9180, 24531, 1979055, 5325216, 39529386, 106368405, 8616365901, 23185550130, 172110498456, 463127571831, 37515654714891, 100949879501796, 749369070309030, 2016457340944761, 163343152011830505, 439535752164830646, 3262752760014579156

Offset: 1

Colin Barker, Jun 26 2015

			1326 is in the sequence because T(51) = 1326 = 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 + 91 + 105 + 120 + 136 + 153 + 171 + 190 = T(3) + ... + T(19).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,4354,-4354,0,0,-1,1).

Cf. A000217, A001110, A129803, A131557, A257711, A257712, A257713, A259413, A259414.

 LinearRecurrence[{1, 0, 0, 4354, -4354, 0, 0, -1, 1}, {1326, 9180, 24531, 1979055, 5325216, 39529386, 106368405, 8616365901, 23185550130}, 30] (* Vincenzo Librandi, Jun 27 2015 *)
Module[{nn=10^6},Select[Total/@Partition[Accumulate[Range[nn]],17,1],OddQ[ Sqrt[8#+1]]&]] (* Harvey P. Dale, Mar 19 2023 *)

Vec(-51*x*(11*x^8 +15*x^6 +154*x^5 -47593*x^4 +38324*x^3 +301*x^2 +154*x +26) / ((x -1)*(x^2 -8*x -1)*(x^2 +8*x -1)*(x^4 +66*x^2 +1)) + O(x^30))

G.f.: -51*x*(11*x^8 +15*x^6 +154*x^5 -47593*x^4 +38324*x^3 +301*x^2 +154*x +26) / ((x -1)*(x^2 -8*x -1)*(x^2 +8*x -1)*(x^4 +66*x^2 +1)).

A054686 Multiset consisting of squares and triangular numbers.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 6, 9, 10, 15, 16, 21, 25, 28, 36, 36, 45, 49, 55, 64, 66, 78, 81, 91, 100, 105, 120, 121, 136, 144, 153, 169, 171, 190, 196, 210, 225, 231, 253, 256, 276, 289, 300, 324, 325, 351, 361, 378, 400, 406, 435, 441, 465, 484, 496, 528

Offset: 1

Michael Somos, Apr 19 2000

Terms of A001110 occur twice. [Reinhard Zumkeller, Aug 03 2011]

Hofstadter, D. R., Fluid Concepts and Creative Analogies: Computer Models of the Fundamental Mechanisms of Thought, (together with the Fluid Analogies Research Group), NY: Basic Books, 1995. p. 15.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A005214, A000217, A000290.

a054686_list = merge a000290_list a000217_list where
   merge xs'@(x:xs) ys'@(y:ys)
     | x <= y    = x : merge xs ys'
     | otherwise = y : merge xs' ys
-- Reinhard Zumkeller, Aug 03 2011

stnos[max_]:=Module[{sqmax=Floor[Sqrt[max]],trmax=Floor[(Sqrt[ 8max+1]- 1)/2]}, Sort[Join[Range[0,sqmax]^2,Accumulate[Range[0,trmax]]]]]; stnos[ 528] (* Harvey P. Dale, Feb 06 2012 *)

upTo(lim)=vecsort(concat(vector(sqrtint(lim\1)+1,n,(n-1)^2),vector(floor(sqrt(2+2*lim)+1/2),n,n*(n-1)/2))) \\ Charles R Greathouse IV, Aug 04 2011

Offset fixed by Reinhard Zumkeller, Aug 04 2011

A076708 Numbers n such that triangular numbers T(n) and T(n+1) sum to another triangular number (that is also a perfect square).

Original entry on oeis.org

0, 5, 34, 203, 1188, 6929, 40390, 235415, 1372104, 7997213, 46611178, 271669859, 1583407980, 9228778025, 53789260174, 313506783023, 1827251437968, 10650001844789, 62072759630770, 361786555939835, 2108646576008244, 12290092900109633, 71631910824649558

Offset: 1

Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002

From T(k)+T(k+1) = (k*(k+1)+(k+1)*(k+2))/2 = (k+1)^2 any two consecutive triangular numbers sum to a square, the above sequence gives the sums that are also triangular. The units digit cycles through 0, 5, 4, 3, 8, 9, 0, 5, ...

Let P(b,e) be the polynomial 1+4*b+4*b^2+4*e+4*e^2. It appears that sequences A076708 and A076049 are special cases of the sequence of integers b such that P(b,b+n) is a perfect square. A076708 and A076049 for example are respectively the sequences of b's such that P(b,b+1) and P(b,b+2) are perfect squares. In fact it appears to be true that the sequence of integers b such that P(b,b+n) is a perfect square has the property that t(b)+t(b+n) is a triangular number. I have not had time to prove this but I do have evidence produced by Mathematica to support the assertion. - Robert Phillips (bobanne(AT)bellsouth.net), Sep 04 2009; corrected Sep 08 2009

			a(1) = (sqrt(2)*((3+2*sqrt(2))^2-(3-2*sqrt(2))^2)-8)/8 = (sqrt(2)*(9+12*sqrt(2)+8-9+12*sqrt(2)-8)-8)/8 = (sqrt(2)*24*sqrt(2)-8)/8 = (48-8)/8 = 40/8 = 5.
T(5) + T(6) = 15 + 21 = 36 = T(8).

Colin Barker, Table of n, a(n) for n = 1..1000
B. Polster, M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658, 2015
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001108, A001109, A001110.

Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/(4 Sqrt[2]) - 1, {n, 1, 20}] (* Zerinvary Lajos, Jul 14 2009 *)

concat(0, Vec(x^2*(x-5)/((x-1)*(x^2-6*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Recursion: a(n+2) = 6*a(n+1)-a(n)+4, with a(0)=0 and a(1)=5.
G.f.: (5*x^2-x^3)/((1-x)*(1-6*x+x^2)).
Closed form: a(n)= ( sqrt(2)*( (3+2*sqrt(2))^(n+1) - (3-2*sqrt(2))^(n+1) )-8 )/8.
Also, if the entries in A001109 are denoted by b(n) then a(n) = b(n+1)-1.
a(n) = sqrt(A001110(n)) - 1. - Ivan N. Ianakiev, May 03 2014

A229131 Numbers k such that the distance between the k-th triangular number and the nearest square is exactly 1.

Original entry on oeis.org

1, 2, 4, 5, 15, 25, 32, 90, 148, 189, 527, 865, 1104, 3074, 5044, 6437, 17919, 29401, 37520, 104442, 171364, 218685, 608735, 998785, 1274592, 3547970, 5821348, 7428869, 20679087, 33929305, 43298624

Offset: 1

Ralf Stephan, Sep 15 2013

nonn

The k-th triangular number (A000217(k)) is a square plus or minus one.

Union of A006451 (k-th triangular number is a square minus one) and A072221 (k-th triangular number is a square plus one).

			A000217(4)=10 and 10 - 3^2 = 1 so 4 is in the sequence.
A000217(5)=15 and 4^2 - 15 = 1 so 5 is in the sequence.

Cf. A000217, A001110, A006451, A072221, A229083, A229118.

for(n=1,10^8,for(i=-1,1,f=0;if(i&&issquare(n*(n+1)/2+i),f=1;break));if(f,print1(n,",")))

G.f.: (-x^7 + 2*x^6 - 2*x^5 + 4*x^4 - 5*x^3 + 2*x^2 + x + 1)/((1-6*x^3+x^6)*(1-x)) (conjectured).

A342709 12-gonal (dodecagonal) square numbers.

Original entry on oeis.org

1, 64, 3025, 142129, 6677056, 313679521, 14736260449, 692290561600, 32522920134769, 1527884955772561, 71778070001175616, 3372041405099481409, 158414167969674450625, 7442093853169599697984, 349619996931001511354641, 16424697761903901433970161

Offset: 1

Bernard Schott, Mar 19 2021

The 12-gonal square numbers k correspond to the nonnegative integer solutions of the Diophantine equation k = d*(5*d-4) = c^2, equivalent to (5*d-2)^2 - 5*c^2 = 4. Substituting x = 5*d-2 and y = c gives the Pell-Fermat's equation x^2 - 5*y^2 = 4.
The solutions x are in A342710, while corresponding solutions y that are also the indices c of the squares which are 12-gonal are in A033890.
The indices d of the corresponding 12-gonal which are squares are in A081068.

			142129 = 169*(5*169-4) = 377^2, so 142129 is the 169th 12-gonal number and the 377th square, hence 142129 is a term.

Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Intersection of A000290 (squares) and A051624 (12-gonal numbers).

Similar for n-gonal squares: A001110 (triangular), A036353 (pentagonal), A046177 (hexagonal), A036354 (heptagonal), A036428 (octagonal), A036411 (9-gonal), A188896 (there are no 10-gonal squares > 1), A333641 (11-gonal), this sequence (12-gonal).

with(combinat):
seq(fibonacci(4*n-2)^2, n=1..16);

Table[Fibonacci[4*n - 2]^2, {n, 1, 16}] (* Amiram Eldar, Mar 19 2021 *)

a(n) = fibonacci(4*n-2)^2; \\ Michel Marcus, Mar 21 2021

G.f.: x*(1 + 16*x + x^2)/((1 - x)*(1 - 47*x + x^2)). - Stefano Spezia, Mar 20 2021
a(n) = 48*a(n-1) - 48*a(n-2) + a(n-3). - Kevin Ryde, Mar 20 2021
a(n) = 9*A161582(n) + 1. - Hugo Pfoertner, Mar 19 2021
a(n) = A033890(n-1)^2.

A181412 Squares whose reverse is a triangular number; trailing zeros are permitted.

Original entry on oeis.org

1, 100, 10000, 1000000, 1306449, 100000000, 130644900, 1044000721, 10000000000, 12957041241, 13064490000, 104400072100, 1000000000000, 1019072079081, 1174279984164, 1295704124100, 1306449000000, 6454272356676, 10440007210000

Offset: 1

Harvey P. Dale, Jan 30 2011

Suggested by T. D. Noe.

			1306449 is 1143 squared, and its reverse, 9446031, is a triangular number.

Giovanni Resta, Table of n, a(n) for n = 1..73 (terms < 10^24)

Cf. A066702, A001110, A066703, A179889, A066528, A069673.

trnos = Accumulate[Range[300000]]; Select[Range[210000]^2, MemberQ[trnos, FromDigits[Reverse[IntegerDigits[#]]]] &]

a(12)-a(19) from Donovan Johnson, Feb 12 2011

A182334 Triangular numbers that differ from a square by 1.

Original entry on oeis.org

0, 1, 3, 10, 15, 120, 325, 528, 4095, 11026, 17955, 139128, 374545, 609960, 4726275, 12723490, 20720703, 160554240, 432224101, 703893960, 5454117903, 14682895930, 23911673955, 185279454480, 498786237505, 812293020528, 6294047334435, 16944049179226

Offset: 1

Arkadiusz Wesolowski, Apr 25 2012

From Robert G. Wilson v, Jun 20 2015: (Start)
Actually this sequence is the union of two subsequences; the triangular numbers that are less than a square by 1 and those that are greater than a square by 1.
The first sequence by index of the triangular numbers is A072221: b(n) = 6b(n-1) - b(n-2) + 2, with b(0)=1, b(1)=4.
And obviously the second sequence by index of the triangular numbers is A006451: c(n) = 6c(n-2) - c(n-4) + 2 with c(0)=0, c(1)=2, c(2)=5, c(3)=15.
(End)

			T(2) = 3 = 2^2 - 1, T(4) = 10 = 3^2 + 1,  T(5) = 15 = 4^2 - 1, and T(15) = 120 = 11^2 - 1.

Edward J. Barbeau, Pell's Equation (Springer 2003) at 17.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,35,0,0,-35,0,0,1).

Subsequence of A000217 and of A087279.

Cf. A001110, A006451, A072221, A229131, A299921.

I:=[0,1,3,10,15,120,325,528,4095,11026,17955]; [n le 11 select I[n] else 35*Self(n-3)-35*Self(n-6)+Self(n-9): n in [1..30]]; // Vincenzo Librandi, Jun 21 2015

lst = {}; Do[t = n*(n + 1)/2; If[IntegerQ[(t - 1)^(1/2)] || IntegerQ[(t + 1)^(1/2)], AppendTo[lst, t]], {n, 0, 10^4}]; lst (* Arkadiusz Wesolowski, Aug 06 2012 *)
b[n_] := b[n] = 6 b[n - 1] - b[n - 2] + 2; b[0] = 1; b[1] = 4; c[n_] := c[n] = 6 c[n - 2] - c[n - 4] + 2; c[0] = 0; c[1] = 2; c[2] = 5; c[3] = 15; #(# + 1)/2 & /@ Union@ Join[ Array[b, 9, 0], Array[c, 18, 0]] (* or *)
#(# + 1)/2 & /@ Join[{0, 1}, LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {2, 4, 5, 15, 25, 32, 90}, 35]] (* or *)
#(# + 1)/2 & /@ CoefficientList[ Series[x + x^2 (1 + x) (2 + x^2 - 3 x^3 + x^4)/((1 - x) (1 - 6 x^3 + x^6)), {x, 0, 36}], x] (* Robert G. Wilson v, Jun 20 2015 *)
a[n_] := a[n] = 35 a[n - 3] - 35 a[n - 6] + a[n - 9]; a[1] = 0; a[2] = 1; a[3] = 3; a[4] = 10; a[5] = 15; a[6] = 120; a[7] = 325; a[8] = 528; a[9] = 4095; a[10] = 11026; a[11] = 17955; Array[a, 36] (* Robert G. Wilson v after Charles R Greathouse IV, Apr 25 2012 *)
Select[Accumulate[Range[0,6*10^6]],AnyTrue[Sqrt[#+{1,-1}],IntegerQ]&] (* or *) LinearRecurrence[{0,0,35,0,0,-35,0,0,1},{0,1,3,10,15,120,325,528,4095,11026,17955},40] (* The first program uses the AnyTrue function from Mathematica version 10 *) (* Harvey P. Dale, Dec 24 2015 *)

concat(0, Vec(x^2*(1+3*x+10*x^2-20*x^3+15*x^4-25*x^5+38*x^6+x^8-x^9)/((1-x)*(1+x+x^2)*(1-34*x^3+x^6)) + O(x^30))) \\ Colin Barker, Sep 17 2016

a(n) = 35*a(n-3) - 35*a(n-6) + a(n-9). - Charles R Greathouse IV, Apr 25 2012

G.f.: x^2*(1+3*x+10*x^2-20*x^3+15*x^4-25*x^5+38*x^6+x^8-x^9) / ((1-x)*(1+x+x^2)*(1-34*x^3+x^6)). - Colin Barker, Sep 17 2016

cp's OEIS Frontend

A257713 Triangular numbers (A000217) that are the sum of ten consecutive triangular numbers.

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A259413 Triangular numbers (A000217) that are the sum of eleven consecutive triangular numbers.

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A259414 Triangular numbers (A000217) that are the sum of thirteen consecutive triangular numbers.

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A259415 Triangular numbers (A000217) that are the sum of seventeen consecutive triangular numbers.

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A054686 Multiset consisting of squares and triangular numbers.

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A076708 Numbers n such that triangular numbers T(n) and T(n+1) sum to another triangular number (that is also a perfect square).

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A229131 Numbers k such that the distance between the k-th triangular number and the nearest square is exactly 1.

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