A001110 -id:A001110 - cp's OEIS frontend

A214838 Triangular numbers of the form k^2 + 2.

3, 6, 66, 171, 2211, 5778, 75078, 196251, 2550411, 6666726, 86638866, 226472403, 2943171003, 7693394946, 99981175206, 261348955731, 3396416785971, 8878171099878, 115378189547778, 301596468440091, 3919462027838451, 10245401755863186, 133146330756959526, 348042063230908203

Offset: 1

Alex Ratushnyak, Mar 07 2013

Corresponding k values are in A077241.

Except 3, all terms are in A089982: in fact, a(2) = 3+3 and a(n) = (k-2)*(k-1)/2+(k+1)*(k+2)/2, where k = sqrt(a(n)-2) > 2 for n > 2. [Bruno Berselli, Mar 08 2013]

			2211 is in the sequence because 2211 = 47^2 + 2.

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A000217, A001110, A001219, A006454, A029549, A069017, A077241, A089982, A164055, A165892.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)))); // Bruno Berselli, Mar 08 2013

LinearRecurrence[{1, 34, -34, -1, 1}, {3, 6, 66, 171, 2211}, 25] (* Bruno Berselli, Mar 08 2013 *)

t[n]:=((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4$
makelist(expand(t[n]*(t[n]+1)/2), n, 1, 25); /* Bruno Berselli, Mar 08 2013 */

for(n=1, 10^9, t=n*(n+1)/2; if(issquare(t-2), print1(t,", "))); \\ Joerg Arndt, Mar 08 2013

import math
for i in range(2, 1<<32):
      t = i*(i+1)//2 - 2
      sr = int(math.sqrt(t))
      if sr*sr == t:
          print(f'{sr:10} {i:10} {t+2}')

G.f.: -3*x*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)). - Joerg Arndt, Mar 08 2013

a(n) = A000217(t), where t = ((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4. - Bruno Berselli, Mar 08 2013

A242585 Number of divisors of the n-th positive number that is both triangular and square.

Original entry on oeis.org

1, 9, 9, 45, 9, 405, 15, 189, 81, 729, 27, 6075, 27, 1215, 2025, 729, 81, 45927, 27, 32805, 2025, 6561, 81, 229635, 243, 2187, 2187, 18225, 9, 7381125, 243, 24057, 2187, 19683, 3645, 6200145, 729, 19683, 19683, 1240029, 81, 22143375, 243, 295245, 492075, 19683

Offset: 1

Jon E. Schoenfield, May 25 2014

nonn

Number of divisors of A001110(n).
Since each term in A001110 is a square, each term in this sequence is an odd number.
Adding terms to this sequence requires obtaining the prime factorization of terms from A001110. This is facilitated by the observation that A001110(n) = (A000129(n)*A001333(n))^2, but after a few hundred terms, it becomes difficult to obtain the prime factorization of some of the values of A000129(n) and A001333(n).
Obviously, a(1) is the only term whose value is 1. It can be shown (see A081978) that 15 appears only at a(7).
Conjectures:
(1) There exist only 5 triangular numbers with exactly 9 divisors: the 2nd, 3rd, 5th, 29th, and 59th terms of A001110.
(2) A001110(4)=41616 is the only triangular number with exactly 45 divisors.
(3) A001110(6)=48024900 is the only triangular number with exactly 405 divisors.
(4) A001110(8)=55420693056 is the only triangular number with exactly 189 divisors.

			a(2) = 9 because A001110(2) = 36 = 2^2 * 3^2 has (2+1)*(2+1) = 9 divisors.
a(4) = 45 because A001110(4) = 41616 = 2^4 * 3^2 * 17^2 has (4+1)*(2+1)*(2+1) divisors.
a(6) = 405 because A001110(6) = 48024900 = 2^2 * 3^4 * 5^2 * 7^2 * 11^2 has (2+1)*(4+1)*(2+1)*(2+1)*(2+1) = 405 divisors.

Jon E. Schoenfield, Table of n, a(n) for n = 1..300
Jon E. Schoenfield, Table of values of n <= 300 such that n-th square triangular number has exactly k divisors

Cf. A001110, A000217, A081978.

a:=0; b:=1; NumberOfDivisors(b); for n in [2..46 by 2] do a:=34*b-a+2; NumberOfDivisors(a); b:=34*a-b+2; NumberOfDivisors(b); end for;

Conjecture: a(n) == 0 mod 9 for n different from 1 and 7 [tested up to n = 300]. - Ivan N. Ianakiev, May 29 2014

A257707 Numbers n such that T(n) + T(n+1) + ... + T(n+22) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

56, 470, 1094, 7856, 128534, 201539, 3293081, 23435699, 53805155, 382911281, 6256309475, 9809462822, 160274811896, 1140616029542, 2618697452438, 18636292598096, 304494582579398, 477426555904883, 7800575092244921, 55513782134933123, 127452004956911987

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-23*y^2-529*y-4048 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,48670,-48670,0,0,0,0,-1,1).

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

Cf. A116476 (length 11), A257293 (length 13), A257708 (length 25), A257709 (length 27), A257710 (length 37).

LinearRecurrence[{1, 0, 0, 0, 0, 48670, -48670, 0, 0, 0, 0, -1, 1}, {56, 470, 1094, 7856, 128534, 201539, 3293081, 23435699, 53805155, 382911281, 6256309475, 9809462822, 160274811896}, 50] (* Vincenzo Librandi, May 05 2015 *)

Vec(x*(10*x^12 +3*x^11 +66*x^10 +414*x^9 +624*x^8 +6762*x^7 -366022*x^6 -73005*x^5 -120678*x^4 -6762*x^3 -624*x^2 -414*x -56) / ((x -1)*(x^12 -48670*x^6 +1)) + O(x^100))

G.f.: x*(10*x^12 +3*x^11 +66*x^10 +414*x^9 +624*x^8 +6762*x^7 -366022*x^6 -73005*x^5 -120678*x^4 -6762*x^3 -624*x^2 -414*x -56) / ((x -1)*(x^12 -48670*x^6 +1)).

A257708 Numbers n such that T(n) + T(n+1) + ... + T(n+24) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

25, 55, 208, 382, 1273, 2287, 7480, 13390, 43657, 78103, 254512, 455278, 1483465, 2653615, 8646328, 15466462, 50394553, 90145207, 293721040, 525404830, 1711931737, 3062283823, 9977869432, 17848298158, 58155284905, 104027505175, 338953840048, 606316732942

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-25*y^2-625*y-5200 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

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

Cf. A116476 (length 11), A257293 (length 13), A257707 (length 23), A257709 (length 27), A257710 (length 37).

LinearRecurrence[{1, 6, -6, -1, 1}, {25, 55, 208, 382, 1273}, 50] (* Vincenzo Librandi, May 05 2015 *)

Vec(x*(x^2+4*x+5)*(2*x^2-2*x-5)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^100))

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

A257709 Numbers n such that T(n) + T(n+1) + ... + T(n+26) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

8, 14, 39, 53, 103, 112, 206, 264, 509, 647, 1141, 1230, 2160, 2734, 5159, 6525, 11415, 12296, 21502, 27184, 51189, 64711, 113117, 121838, 212968, 269214, 506839, 640693, 1119863, 1206192, 2108286, 2665064, 5017309, 6342327, 11085621, 11940190, 20870000

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-27*y^2-729*y-6552 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,10,-10,0,0,0,0,-1,1).

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

Cf. A116476 (length 11), A257293 (length 13), A257707 (length 23), A257708 (length 25), A257710 (length 37).

I:=[8,14,39,53,103,112,206,264,509,647,1141,1230, 2160]; [n le 13 select I[n] else Self(n-1)+10*Self(n-6)-10*Self(n-7)-Self(n-12)+Self(n-13): n in [1..40]]; // Vincenzo Librandi, May 05 2015

LinearRecurrence[{1, 0, 0, 0, 0, 10, -10, 0, 0, 0, 0, -1, 1}, {8, 14, 39, 53, 103, 112, 206, 264, 509, 647, 1141, 1230, 2160}, 50] (* Vincenzo Librandi, May 05 2015 *)
Position[Total/@Partition[Accumulate[Range[70000]],27,1],?(IntegerQ[ Sqrt[ #]]&)]//Flatten (* The program generates the first 22 terms of the sequence. To generate more, increase the Range constant but the program may take a long time to run. *) (* _Harvey P. Dale, Jul 27 2021 *)

Vec(x*(2*x^12+x^11+6*x^10+2*x^9+5*x^8+2*x^7-14*x^6-9*x^5-50*x^4-14*x^3-25*x^2-6*x-8) / ((x-1)*(x^12-10*x^6+1)) + O(x^100))

G.f.: x*(2*x^12+x^11+6*x^10+2*x^9+5*x^8+2*x^7-14*x^6-9*x^5-50*x^4-14*x^3-25*x^2-6*x-8) / ((x-1)*(x^12-10*x^6+1)).

A257710 Numbers n such that T(n) + T(n+1) + ... + T(n+36) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

5, 32, 291, 661, 4102, 8515, 13685, 113558, 182368, 377701, 2290342, 5027232, 30483491, 63130838, 101378488, 840238915, 1349295285, 2794368792, 16944086651, 37191598501, 225516999142, 467042067835, 749998177365, 6216087516438, 9982086472888, 20672740082341

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-37*y^2-1369*y-16872 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,7398,-7398,0,0,0,0,0,0,-1,1).

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

Cf. A116476 (length 11), A257293 (length 13), A257707 (length 23), A257708 (length 25), A257709 (length 27).

LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 7398, -7398, 0, 0, 0, 0, 0, 0, -1, 1}, {5, 32, 291, 661, 4102, 8515, 13685, 113558, 182368, 377701, 2290342, 5027232, 30483491, 63130838, 101378488, 840238915, 1349295285}, 50] (* Vincenzo Librandi, May 05 2015 *)

Vec(x*(5*x^16 +27*x^15 +10*x^14 +27*x^13 +259*x^12 +370*x^11 +3441*x^10 +4413*x^9 -31820*x^8 -99873*x^7 -5170*x^6 -4413*x^5 -3441*x^4 -370*x^3 -259*x^2 -27*x -5) / ((x -1)*(x^8 -86*x^4 -1)*(x^8 +86*x^4 -1)) + O(x^100))

G.f.: x*(5*x^16 +27*x^15 +10*x^14 +27*x^13 +259*x^12 +370*x^11 +3441*x^10 +4413*x^9 -31820*x^8 -99873*x^7 -5170*x^6 -4413*x^5 -3441*x^4 -370*x^3 -259*x^2 -27*x -5) / ((x -1)*(x^8 -86*x^4 -1)*(x^8 +86*x^4 -1)).

A275496 a(n) = n^2(2n^2 + (-1)^n).

Original entry on oeis.org

0, 1, 36, 153, 528, 1225, 2628, 4753, 8256, 13041, 20100, 29161, 41616, 56953, 77028, 101025, 131328, 166753, 210276, 260281, 320400, 388521, 468996, 559153, 664128, 780625, 914628, 1062153, 1230096, 1413721, 1620900, 1846081, 2098176, 2370753, 2673828

Offset: 0

Daniel Poveda Parrilla, Jul 30 2016

All terms of this sequence are triangular numbers. Graphically, for each term of the sequence, one corner of the square of squares (4th power) will be part of the corresponding triangle's hypotenuse if the term is an odd number. Otherwise, it will not be part of it.
a(A000129(n)) is a square triangular number.
a(2^((A000043(n) - 1)/2)) - 2^A000043(n) is a perfect number.

			a(5) = 5^4 + Sum_{k=0..(5^2 - (5 mod 2))} 2k = 625 + Sum_{k=0..(25 - 1)} 2k = 625 + 600 = 1225.
a(12) = 12^4 + Sum_{k=0..(12^2 - (12 mod 2))} 2k = 20736 + Sum_{k=0..(144 - 0)} 2k = 20736 + 20880 = 41616.

Colin Barker and Daniel Poveda Parrilla, Table of n, a(n) for n = 0..46340 [n = 1 through 1000 by Colin Barker, Aug 02 2016; and n=1001 to 46340 by Daniel Poveda Parrilla, Aug 04 2016]
Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Cf. A000040, A000129, A000217, A001110, A077221, A093954, A275543.

Table[n^2 ((-1)^n + 2 n^2), {n, 0, 34}] (* or *)
CoefficientList[Series[x (1 + 34 x + 79 x^2 + 156 x^3 + 79 x^4 + 34 x^5 +
x^6)/((1 - x)^5 (1 + x)^3), {x, 0, 34}], x] (* Michael De Vlieger, Aug 01 2016 *)
LinearRecurrence[{2,2,-6,0,6,-2,-2,1},{0,1,36,153,528,1225,2628,4753},40] (* Harvey P. Dale, Sep 10 2016 *)

a(n)=n=n^2; if(n%2,2*n-1,2*n+1)*n \\ Charles R Greathouse IV, Jul 30 2016

concat(0, Vec(x*(1+34*x+79*x^2+156*x^3+79*x^4+34*x^5+x^6)/((1-x)^5*(1+x)^3) + O(x^100))) \\ Colin Barker, Aug 01 2016

a(n) = n^4 + Sum_{k=0..(n^2 - (n mod 2))} 2k.
a(n) = A275543(n)*(n^2).
From Colin Barker, Aug 01 2016 and Aug 04 2016: (Start)
a(n) = n^2*(2*n^2 + (-1)^n).
a(n) = 2*n^4 + n^2 for n even.
a(n) = 2*n^4 - n^2 for n odd.
G.f.: x*(1 +34*x +79*x^2 +156*x^3 +79*x^4 +34*x^5 +x^6) / ((1-x)^5*(1+x)^3).
(End)
a(n) = n^2*A000217(2n-1) + 2n*A000217(n-(n mod 2)) for n > 0.
E.g.f.: x*(2*(1 + 7*x + 6*x^2 + x^3)*exp(x) - exp(-x)). - G. C. Greubel, Aug 05 2016
a(n) = A000217(A077221(n)).
a(n) = (A001844(A077221(n)) - 1)/4.
Sum_{n>=1} 1/a(n) = 1 - Pi^2/12 + (tan(c) - coth(c))*c, where c = Pi/(2*sqrt(2)) is A093954. - Amiram Eldar, Aug 21 2022

New name from Colin Barker, Aug 04 2016

A299921 Squares that differ from a triangular number by 1.

Original entry on oeis.org

0, 1, 4, 9, 16, 121, 324, 529, 4096, 11025, 17956, 139129, 374544, 609961, 4726276, 12723489, 20720704, 160554241, 432224100, 703893961, 5454117904, 14682895929, 23911673956, 185279454481, 498786237504, 812293020529, 6294047334436, 16944049179225, 27594051024016

Offset: 1

N. J. A. Sloane, Mar 17 2018

nonn

Squares k such that 8*k-7 or 8*k+9 is a square. - Robert Israel, Mar 18 2018

Robert Israel, Table of n, a(n) for n = 1..1959
Index entries for linear recurrences with constant coefficients, signature (0,0,35,0,0,-35,0,0,1).

Cf. A001110, A164080, A182334, A229131.

f:= gfun:-rectoproc({a(n+9) = 35*a(n+6) - 35*a(n+3) + a(n), seq(a(i)=[0, 1, 4, 9, 16, 121, 324, 529, 4096][i],i=1..9)}, a(n), remember):
map(f, [$1..50]); # Robert Israel, Mar 18 2018

LinearRecurrence[{0, 0, 35, 0, 0, -35, 0, 0, 1}, {0, 1, 4, 9, 16, 121, 324, 529, 4096}, 50] (* Jean-François Alcover, Sep 17 2022 *)

isok(n) = issquare(n) && (ispolygonal(n+1, 3) || ispolygonal(n-1, 3)); \\ Michel Marcus, Mar 17 2018

From Robert Israel, Mar 18 2018: (Start)
G.f.: x^2*(1+4*x+9*x^2-19*x^3-19*x^4+9*x^5+4*x^6+x^7)/(1-35*x^3+35*x^6-x^9).
a(n) = 35*a(n-3) - 35*a(n-6) + a(n-9). (End)

More terms from Altug Alkan, Mar 17 2018

A112352 Triangular numbers that are the sum of two distinct positive triangular numbers.

Original entry on oeis.org

21, 36, 55, 66, 91, 120, 136, 171, 231, 276, 351, 378, 406, 496, 561, 666, 703, 741, 820, 861, 946, 990, 1035, 1081, 1176, 1225, 1326, 1378, 1431, 1485, 1540, 1596, 1653, 1711, 1770, 1891, 1953, 2016, 2080, 2211, 2278, 2346, 2556, 2701, 2775, 2850, 2926

Offset: 1

Rick L. Shepherd, Sep 05 2005

nonn

Subsequence of A089982: it doesn't require the two positive triangular numbers to be distinct.
Subsequence of squares: 36, 1225, 41616, 1413721,... is also in A001110. - Zak Seidov, May 07 2015
First term with 2 representations is 231: 21+210=78+153, first term with 3 representations is 276: 45+211=66+120=105+171; apparently the number of representations is unbounded. - Zak Seidov, May 11 2015

			36 is a term because 36 = 15 + 21 and these three numbers are distinct triangular numbers (A000217(8) = A000217(5) + A000217(6)).

Zak Seidov, Table of n, a(n) for n = 1..1000

Cf. A000217 (triangular numbers), A112353 (triangular numbers that are the sum of three distinct positive triangular numbers), A089982.

Cf. A001110. - Zak Seidov, May 07 2015

N:= 10^5: # to get all terms <= N
S:= {}:
for a from 1 to floor(sqrt(1+8*N)/2) do
  for b from 1 to a-1 do
    y:= a*(a+1)/2 + b*(b+1)/2;
      if y > N then break fi;
      if issqr(8*y+1) then S:= S union {y} fi
  od
od:
sort(convert(S,list)); # Robert Israel, May 13 2015

Select[Union[Total/@Subsets[Accumulate[Range[100]],{2}]],OddQ[ Sqrt[ 1+8#]]&] (* Harvey P. Dale, Feb 28 2016 *)

Offset corrected by Arkadiusz Wesolowski, Aug 06 2012

A214937 Square numbers that can be expressed as sums of a positive square number and a positive triangular number.

Original entry on oeis.org

4, 16, 25, 49, 64, 81, 100, 121, 169, 196, 256, 289, 361, 400, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1156, 1225, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2401, 2500, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481

Offset: 1

Ivan N. Ianakiev, Jul 30 2012

nonn

Theorem (I. N. Ianakiev): There are infinitely many such numbers. Proof: There are infinitely many square triangular numbers (A001110) and every (2t+1)-th of them is odd because A001110(0)=0, A001110(1)=1 and A001110(n)=34*a(n-1)-a(n-2)+2, for n>=2. Any sqrt(A001110(2t+1)) is odd (i. e. is in A005408) and can be written as p^2-q^2 because A005408(n)=A000290(n+1)-A000290(n). The unique values of p and q (p>q>0) for each sqrt(A001110(2t+1)) generate (when t>0) a unique Pythagorean triple with a unique hypotenuse (a=p^2-q^2, b=2pq, c=p^2+q^2). Therefore, there are infinitely many such hypotenuses squared.

			4 and 49 are in the sequence because 2^2=1^2+2*3/2 and 7^2=2^2+9*10/2

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000217, A000290, A182427.

cp's OEIS Frontend

A214838 Triangular numbers of the form k^2 + 2.

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A242585 Number of divisors of the n-th positive number that is both triangular and square.

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A257707 Numbers n such that T(n) + T(n+1) + ... + T(n+22) is a square, where T = A000217 (triangular numbers).

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A257708 Numbers n such that T(n) + T(n+1) + ... + T(n+24) is a square, where T = A000217 (triangular numbers).

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A257709 Numbers n such that T(n) + T(n+1) + ... + T(n+26) is a square, where T = A000217 (triangular numbers).

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A257710 Numbers n such that T(n) + T(n+1) + ... + T(n+36) is a square, where T = A000217 (triangular numbers).

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A275496 a(n) = n^2*(2*n^2 + (-1)^n).

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A275496 a(n) = n^2(2n^2 + (-1)^n).