A001110 -id:A001110 - cp's OEIS frontend

A253514 Centered heptagonal numbers (A069099) which are also centered octagonal numbers (A016754).

1, 841, 755161, 678133681, 608963290321, 546848356574521, 491069215240629481, 440979608437728699361, 395999197307865131396641, 355606838202854450265484201, 319334544706965988473273415801, 286762065540017254794549261905041

Offset: 1

Colin Barker, Jan 03 2015

			841 is in the sequence because it is the 16th centered heptagonal number and the 15th centered octagonal number.

Colin Barker, Table of n, a(n) for n = 1..339
Index entries for linear recurrences with constant coefficients, signature (899,-899,1).

Cf. A001110, A016754, A069099, A157877, A157879, A253446, A253447.

Vec(-x*(x^2-58*x+1)/((x-1)*(x^2-898*x+1)) + O(x^100))

a(n) = 899*a(n-1)-899*a(n-2)+a(n-3).
G.f.: -x*(x^2-58*x+1) / ((x-1)*(x^2-898*x+1)).
From Peter Bala, Apr 15 2025; (Start)
a(n) = (1/64)*(-4 + sqrt(14))^2*(15 + 4*sqrt(14) + (449 + 120*sqrt(14))^n)^2 *(449 + 120*sqrt(14))^(-n).
a(-n) = a(n+1).
a(n) = (1/16) * (1 - T(2*n+1, -15)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A001110.
a(n) = A157877(n)^2 = 1 + 7*A157879(n).
a(2) divides a(3*n+2); a(3) divides a(5*n+3); a(4) divides a(7*n+4); a(5) divides a(9*n+5). In general, a(k) divides a((2*k-1)*n + k). (End)

A113449 Sum of the square root of n-th square triangular number and n-th Pell (or lambda) number (A000129).

Original entry on oeis.org

2, 8, 40, 216, 1218, 7000, 40560, 235824, 1373090, 7999592, 46616920, 271683720, 1583441442, 9228858808, 53789455200, 313507253856, 1827252574658, 10650004589000, 62072766255880, 361786571934264, 2108646614622210

Offset: 1

K. B. Subramaniam (subramaniam_kb05(AT)yahoo.co.in), Nov 02 2005

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8,-12,-4,1).

Cf. A000129, A001110, A002203.

Simplify[Table[Sqrt[((17 + 12*Sqrt[2])^n + (17 - 12*Sqrt[2])^n - 2)/32] + ((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2*Sqrt[2]), {n, 1, 25}]] (* Stefan Steinerberger, Jun 17 2007 *)
Table[Sqrt[(LucasL[4*n, 2] - 2)/32] + Fibonacci[n, 2], {n,1,50}] (* G. C. Greubel, Mar 11 2017 *)

x='x+O('x^50); Vec((2*x)*(1-4*x) / ((1-2*x-x^2)*(1-6*x+x^2))) \\ G. C. Greubel, Mar 11 2017

a(n) = sqrt(((17 + 12*sqrt(2))^n + (17 - 12*sqrt(2))^n - 2)/32) + ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)). - Stefan Steinerberger, Jun 17 2007
From G. C. Greubel, Mar 11 2017: (Start)
a(n) = sqrt((Q_{4*n} - 2)/32) + P_{n}, where P_{n} and the Pell numbers and Q_{n} are the Pell-Lucas numbers.
a(n) = 8*a(n-1) - 12*a(n-2) - 4*a(n-3) + a(n-4).
G.f.: (2*x)*(1-4*x) / ((1-2*x-x^2)*(1-6*x+x^2)). (End)

More terms from Stefan Steinerberger, Jun 17 2007

A114620 2*A084158 (twice Pell triangles).

Original entry on oeis.org

0, 2, 10, 60, 348, 2030, 11830, 68952, 401880, 2342330, 13652098, 79570260, 463769460, 2703046502, 15754509550, 91824010800, 535189555248, 3119313320690, 18180690368890, 105964828892652, 617608282987020

Offset: 0

Creighton Dement, Feb 17 2006

Cross-referenced sequences A116484, A001109, A108475, A090390 are also generated by A*B given in the following FAMP code.
Floretion Algebra Multiplication Program, FAMP Code: 1jesleftseq[A*B] with A = - .5'i + .5'j - .5i' + .5j' + 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj' and B = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'
Related to the reciprocals of the differences between successive convergents of the continued fraction of sqrt(2) (i.e., 1, 2, -10, 60, -348, 2030, -11830, 68952, ...). 1/1 + 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 + ... = sqrt(2). 2, 10, 60, ... are products of the denominators of two successive convergents of sqrt(2) (e.g., 11830 = 70*169, cf. A000129 (Pell numbers)). - Gerald McGarvey, Feb 28 2006
a(n) is half of the even leg (b(n)) of the ordered Pythagorean triple (x(n), y(n)=x(n)+1, z(n)). In fact b(n) = x(n) + (1-(-1)^n)/2: x(0)=0, b(0)=0, a(0)=0; x(1)=3, b(1)=4, a(1)=2. - George F. Johnson, Aug 13 2012
Given a square shape composed of A001110(n+1) elements, thinking of it graphically as a sum of layers, each layer having an odd number of elements (all layers together being a sum of consecutive odd numbers), a(n) is the number of last layers that we have to subtract from the square to get a square of squares that is made of A002965(2*(n+1))^4 elements. - Daniel Poveda Parrilla, Jul 17 2016
Also numbers m such that 8*m^2 - 4*m + 1 or 8*m^2 + 4*m + 1 is a perfect square (square roots are then A001653). - Lamine Ngom, Jul 25 2023

Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A116484, A001109, A108475, A090390.

Cf. A000129.

Table[Fibonacci[n, 2] Fibonacci[n + 1, 2], {n, 0, 20}] (* or *)
LinearRecurrence[{5, 5, -1}, {0, 2, 10}, 21] (* or *)
CoefficientList[Series[2 x/((x + 1) (x^2 - 6 x + 1)), {x, 0, 20}], x] (* Michael De Vlieger, Jul 17 2016 *)

G.f.: 2*x/((x+1)*(x^2-6*x+1)).
From George F. Johnson, Aug 13 2012: (Start)
a(n) = ((sqrt(2) + 1)^(2*n+1) - (sqrt(2) - 1)^(2*n+1) - 2*(-1)^n)/8. - corrected by Ilya Gutkovskiy, Jul 18 2016
4*a(n)*(2*a(n) + (-1)^n) + 1 = A000129(2*n+1)^2 is a perfect square.
For n >= 0, a(n+1) = 3*a(n) + (-1)^n + sqrt(4*a(n)*(2*a(n) + (-1)^n) + 1).
For n >  0, a(n-1) = 3*a(n) + (-1)^n - sqrt(4*a(n)*(2*a(n) + (-1)^n) + 1).
a(n+1) = 6*a(n) - a(n-1) + 2*(-1)^n.
a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
For n > 0, a(n+1)*a(n-1) = a(n)*(a(n) + 2*(-1)^n).
a(n) = A046729(n)/2. (End)
a(n) = A000129(n)*A000129(n+1). - Philippe Deléham, Apr 10 2013
a(n) = A002965(2*(n+1))*(A002965(2*(n+1)+1) - A002965(2*(n+1))). - Daniel Poveda Parrilla, Jul 17 2016

A164080 Perfect squares one less than a triangular number.

Original entry on oeis.org

0, 9, 324, 11025, 374544, 12723489, 432224100, 14682895929, 498786237504, 16944049179225, 575598885856164, 19553418069930369, 664240615491776400, 22564627508650467249, 766533094678624110084, 26039560591564569275625

Offset: 1

Tanya Khovanova & Alexey Radul, Aug 09 2009

nonn

			324=18^2 is a perfect square and 325=A000217(25) is a triangular number. Therefore 324 is in this sequence.

Index entries for linear recurrences with constant coefficients, signature (35, -35, 1).

LinearRecurrence[{35,-35,1},{0,9,324},20] (* Harvey P. Dale, Oct 24 2023 *)

a(n) = A164055(n)-1.
a(n) = A072221(n)*(A072221(n)+1)/2 - 1.
a(n) = 35*a(n-1) -35*a(n-2) +a(n-3) = 9*A001110(n-1). G.f.: 9*x^2*(1+x)/((1-x)*(x^2-34*x+1)). [R. J. Mathar, Oct 21 2009]

Comments turned into formulas. - R. J. Mathar, Oct 21 2009

A185097 a(1)=1; thereafter a(n+1) = T(8a(n)), where T(i)=i(i+1)/2 is the i-th triangular number.

Original entry on oeis.org

1, 36, 41616, 55420693056, 98286503002057414584576, 309127573515950117423442905473334441338685531136

Offset: 1

N. J. A. Sloane, Feb 18 2011

nonn

All terms are square and triangular; a subsequence of A001110.

C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 4.

Cf. A000217, A001110, A185096.

T:=n->n*(n+1)/2; t1:=[1]; for n from 1 to 7 do t1:= [op(t1), T(8*t1[nops(t1)])]; od: t1;

NestList[(8#(8#+1))/2&,1,7] (* Harvey P. Dale, Jan 20 2015 *)

a(n) = (1/8)*sinh(2^(n-2)*arccosh(17))^2. [Alexander R. Povolotsky, Aug 14 2011]
a(n+2) = 4*a(n+1)*(a(n+1)/(2*a(n))-1)^2, a(1)=1, a(2)=36. [Alexander R. Povolotsky, Aug 15 2011]
a(n) = A001110(2^(n-1)). - Ivan N. Ianakiev, Mar 11 2014

A229134 Square numbers that are the sum of two non-consecutive triangular numbers.

Original entry on oeis.org

16, 36, 81, 121, 196, 225, 256, 289, 361, 441, 484, 529, 576, 625, 676, 841, 900, 961, 1024, 1156, 1225, 1296, 1444, 1521, 1600, 1681, 1849, 1936, 2116, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3136, 3249, 3481, 3721, 4096, 4356, 4624, 4761, 4900, 5041, 5184

Offset: 1

Jon Perry, Sep 15 2013

nonn

It is well known that tri(n) + tri(n+1) is always a square.

Sequence includes all terms of A001110 > 1. A number m is a term if and only if there exists k > 1 such that m >= tri(k) and 4m - k^2 + 1 is a perfect square. - Chai Wah Wu, Feb 25 2016

			16 = 15+1, 81 = 78+3 = 66+15.

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

Cf. A000217, A001110.

function isSquare(n) {
if (Math.sqrt(n)==Math.floor(Math.sqrt(n))) return true; else return false;
}
a=new Array();
ac=0;
for (i=0;i<100;i++)
for (j=i+2;j<100;j++)
if (isSquare(i*(i+1)/2+j*(j+1)/2)) a[ac++]=i*(i+1)/2+j*(j+1)/2;
a.sort(function(a,b) {return a-b;});
a=trimArray(a);
function trimArray(arr) {
var j,c=new Array(),i;
for (j=0;j

nn = 10000; mx = Floor[Sqrt[1 + 8 nn]/2]; tri = Table[n (n + 1)/2, {n, mx}]; t = {}; Do[s = tri[[i]] + tri[[j]]; If[s <= nn && IntegerQ[Sqrt[s]], AppendTo[t, s]], {i, mx - 2}, {j, i + 2, mx}]; t = Union[t] (* T. D. Noe, Sep 17 2013 *)

from gmpy2 import is_square
A229134_list = []
for i in range(10**3):
    m, m2, j, k = 2, 4, 4*i**2+1, 2*i**2
    while k >= m2 + m:
        if is_square(j-m2):
            A229134_list.append(i**2)
            break
        m2 += 2*m+1
        m += 1 # Chai Wah Wu, Feb 25 2016

Corrected and extended by T. D. Noe, Sep 17 2013

a(2) = 36 reinserted by Chai Wah Wu, Feb 27 2016

A238456 Triangular numbers t such that t+x+y is a square, where x and y are the two squares nearest to t.

Original entry on oeis.org

0, 2211, 5151, 1107816, 20959575, 4237107540, 1564279847151, 61066162885575, 2533192954461975, 2774988107938203, 90728963274006291, 18765679728507154152720

Offset: 1

Alex Ratushnyak, Feb 26 2014

For triangular numbers t such that t*x*y is a square, see A001110 (t is both triangular and square).

a(13) > 5*10^22. - Giovanni Resta, Mar 02 2014

			The two squares nearest to triangular(101)=5151 are 71^2 and 72^2. Because 5151 + 71^2 + 72^2 = 15376 is a perfect square, 5151 is in the sequence.

Cf. A000217, A000290, A001110, A238489.

sqQ[n_]:=Module[{c=Floor[Sqrt[n]]-1,x},x=Total[Take[SortBy[ Range[ c,c+3]^2, Abs[#-n]&],2]];IntegerQ[Sqrt[n+x]]]; Select[ Accumulate[ Range[ 0, 5000000]], sqQ] (* This will generate the first 7 terms of the sequence.  To generate more, increase the second constant within the Range function, but computations will take a long time. *) (* Harvey P. Dale, May 12 2014 *)

def isqrt(a):
    sr = 1 << (int.bit_length(int(a)) >> 1)
    while a < sr*sr:  sr>>=1
    b = sr>>1
    while b:
        s = sr + b
        if a >= s*s:  sr = s
        b>>=1
    return sr
t = i = 0
while 1:
    t += i
    i += 1
    s = isqrt(t)
    if s*s==t:  s-=1
    txy = t + 2*s*(s+1) + 1   # t + s^2 + (s+1)^2
    r = isqrt(txy)
    if r*r==txy:  print(str(t), end=',')

a(12) from Giovanni Resta, Mar 02 2014

A275543 A081585 and A069129 interleaved.

Original entry on oeis.org

1, 1, 9, 17, 33, 49, 73, 97, 129, 161, 201, 241, 289, 337, 393, 449, 513, 577, 649, 721, 801, 881, 969, 1057, 1153, 1249, 1353, 1457, 1569, 1681, 1801, 1921, 2049, 2177, 2313, 2449, 2593, 2737, 2889, 3041, 3201, 3361, 3529, 3697, 3873, 4049, 4233, 4417, 4609

Offset: 0

Daniel Poveda Parrilla, Aug 01 2016

a(A000129(n)) is a square.
(n^2)*a(n) = A275496(n) which is a triangular number.
(A000129(n)^2)*a(A000129(n)) = A275496(A000129(n)) = A001110(n) which is a square triangular number.
a(2n+1)/a(2n) is convergent to 1.

			a(1) = A275496(1) = 1.
a(5) = A275496(5)/25 = 1225/25 = 49.
a(7) = A275496(7)/49 = 4753/49 = 97.
a(12) = A275496(12)/144 = 41616/144 = 289.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..500000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081585(n) = a(2n), A069129(n) = a(2n + 1).

Cf. A000129, A000217, A000290, A001110, A077221, A093954, A275496.

CoefficientList[Series[(1 - x + 7 x^2 + x^3)/((1 - x)^3 (1 + x)), {x, 0, 48}], x] (* or as defined *)
Riffle[LinearRecurrence[{3, -3, 1}, {1, 9, 33}, #], FoldList[#1 + #2 &, 1, 16 Range@ #]] &@ 25 (* Michael De Vlieger, Aug 01 2016, after Vincenzo Librandi at A081585 and Robert G. Wilson v at A069129 *)

a(n)=(-1)^n + 2*n^2 \\ Charles R Greathouse IV, Aug 03 2016

Vec((1-x+7*x^2+x^3)/((1-x)^3*(1+x)) + O(x^100)) \\ Colin Barker, Aug 21 2016

a(0) = 1; a(n) = A275496(n)/(n^2) for n > 0.
From Colin Barker, Aug 01 2016: (Start)
a(n) = (2*n^2 + (-1)^n).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 3.
G.f.: (1 -x +7*x^2 +x^3) / ((1 - x)^3*(1 + x)).
(End)
From Daniel Poveda Parrilla, Aug 18 2016: (Start)
a(2n) = A077221(2n) + 1.
a(2n + 1) = A077221(2n + 1). (End)
Sum_{n>=0} 1/a(n) = (1 + (tan(c) + coth(c))*c)/2, where c = Pi/(2*sqrt(2)) is A093954. - Amiram Eldar, Aug 21 2022

A328791 Triangular numbers of the form k^2 + 3.

Original entry on oeis.org

3, 28, 903, 30628, 1040403, 35343028, 1200622503, 40785822028, 1385517326403, 47066803275628, 1598885794044903, 54315050194251028, 1845112820810490003, 62679520857362409028, 2129258596329511416903, 72332112754346025765628, 2457162575051435364614403

Offset: 1

Jon E. Schoenfield, Oct 27 2019

nonn

There exist triangular numbers of the form k^2 + j for j=0 (A001110), j=1 (A164055), j=2 (A214838), and j=3 (this sequence), but not for j=4,7,8,13,16,18,... (A328792).

Cf. A001110, A164055, A214838, A328792.
Intersection of A000217 and A117950.
Cf. A276598 (the k's).

limit = 10**7 # rough limit for k
A000217 = set(k*(k+1)//2 for k in range(14*limit//10))
A117950 = set(k**2 + 3 for k in range(limit))
print(sorted(A000217 & A117950)) # Michael S. Branicky, Mar 28 2021

a(1) = 3, a(2) = 28; for n > 2, a(n) = 34*a(n-1) - a(n-2) - 46.

A328792 Numbers that are not the difference between any triangular number and the largest square that does not exceed it.

Original entry on oeis.org

4, 7, 8, 13, 16, 18, 22, 23, 25, 26, 31, 33, 34, 37, 38, 40, 43, 47, 48, 49, 52, 58, 59, 60, 61, 63, 64, 67, 68, 70, 73, 76, 79, 81, 83, 85, 86, 88, 92, 93, 94, 97, 98, 99, 102, 103, 106, 108, 112, 113, 114, 115, 118, 121, 123, 124, 125, 130, 133, 134, 138

Offset: 1

Jon E. Schoenfield, Oct 27 2019

nonn

			For any triangular number t, let f(t) = t - floor(sqrt(t))^2.
0 is not a term: for each term t in A001110, f(t) = 0.
1 is not a term: for each term t > 1 in A164055, f(t) = 1.
2 is not a term: for each term t in A214838, f(t) = 2.
3 is not a term: for each term t > 3 in A328791, f(t) = 3.
4 is a term, however: there exists no triangular number t such that f(t) = 4.

The complement of A230044.

Cf. A001110, A064784, A164055, A214838, A328791.

cp's OEIS Frontend

A253514 Centered heptagonal numbers (A069099) which are also centered octagonal numbers (A016754).

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A113449 Sum of the square root of n-th square triangular number and n-th Pell (or lambda) number (A000129).

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A114620 2*A084158 (twice Pell triangles).

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A164080 Perfect squares one less than a triangular number.

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A185097 a(1)=1; thereafter a(n+1) = T(8*a(n)), where T(i)=i*(i+1)/2 is the i-th triangular number.

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A229134 Square numbers that are the sum of two non-consecutive triangular numbers.

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A238456 Triangular numbers t such that t+x+y is a square, where x and y are the two squares nearest to t.

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A275543 A081585 and A069129 interleaved.

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A185097 a(1)=1; thereafter a(n+1) = T(8a(n)), where T(i)=i(i+1)/2 is the i-th triangular number.