A001110 -id:A001110 - cp's OEIS frontend

A081978 Smallest triangular number with exactly n divisors, or 0 if no such number exists.

1, 3, 0, 6, 0, 28, 0, 66, 36, 496, 0, 276, 0, 8128, 1631432881, 120, 0, 300, 0, 528, 0, 38009927549623740385753, 0, 630, 0, 33550336, 2172602007770041, 8256, 0, 209628, 0, 3570, 0, 8589869056, 0, 2016, 0, 137438691328, 0, 3240, 0, 662976, 0, 2096128, 41616

Offset: 1

Amarnath Murthy, Apr 03 2003

a(p)=0 if p is an odd prime. If n is an odd composite number, then a(n) is a square; see A001110 for numbers that are both triangular and square. - Victoria A Sapko (vsapko(AT)frc.mass.edu), Sep 28 2007
From Jon E. Schoenfield, May 25 2014: (Start)
If n is an odd semiprime, then a triangular number t having exactly n divisors must be of the form t = p^(2r) * q^(2s) = (p^r * q^s)^2, where p and q are distinct primes (p < q) and r and s are positive integers such that (2r+1)*(2s+1) = n.
If t is the k-th triangular number k(k+1)/2, it can be factored as t = u * v where
     u = k    and   v = (k+1)/2   if k is odd, or
     u = k/2  and   v = k+1       if k is even.
Since neither p nor q (each of which is greater than 1) can divide both k and (k+1)/2, or both k/2 and k+1, only four cases need to be considered:
   Case 1:  k is even, q^(2s) = k/2, p^(2r) = k+1
   Case 2:  k is even, p^(2r) = k/2, q^(2s) = k+1
   Case 3:  k is odd,  q^(2s) = k,   p^(2r) = (k+1)/2
   Case 4:  k is odd,  p^(2r) = k,   q^(2s) = (k+1)/2
These yield the following equations:
   Case 1:  2 * q^(2s) + 1 = p^(2r)
   Case 2:  2 * p^(2r) + 1 = q^(2s)
   Case 3:  2 * p^(2r) - 1 = q^(2s)
   Case 4:  2 * q^(2s) - 1 = p^(2r)
Case 1 can be ruled out: since q > p, q is odd, so 2 * q^(2s) + 1 == 3 (mod 16), but p^(2r) cannot be 3 (mod 16).
For a Case 2 solution, since q is odd, 2 * p^(2r) + 1 = q^(2s) == 1 (mod 8), so p^(2r) == 0 (mod 4), so p must be even. Therefore, p = 2, and we must satisfy the equation 2 * 2^(2r) + 1 = q^(2s), whose left-hand side, which is divisible by 3 for every nonnegative integer r, is thus the square of a prime iff it is 3^2. So r=1, q=3, and s=1, yielding t = 2^2 * 3^2 = 36, which is the smallest triangular number with exactly 9 divisors, so a(9)=36.
In Case 3, both p and q must be odd, and p^r must be a number w having the property that 2*w^2 - 1 is square (i.e., (q^s)^2); every such number w is in A001653 (1, 5, 29, 169, 985, ...), and the corresponding value of q^s = sqrt(2*w^2 - 1) is in A002315 (1, 7, 41, 239, 1393, ...). (Note that A001653 and A002315 are defined with offsets of 1 and 0, respectively, so A001653(j) corresponds to A002315(j-1).) However, for odd semiprime n > 9, we need r > 1 and/or s > 1. The only nontrivial power (i.e., number of the form x^m, where both x and m are integers greater than 2) in A001653 is A001653(4) = 169 = 13^2 [Pethö], so the only Case 3 solution with r > 1 is 2 * 13^4 - 1 = 239^2, which yields the 15-divisor triangular number 13^4 * 239^2 = 1631432881 = a(15). A Case 3 solution with r = 1 and s > 1 would require 2 * p^2 - 1 = q^(2s), which is impossible since p < q.
Finally, in Case 4, both p and q must be odd, q^s must be in A001653, and p^r must be the corresponding term in A002315. However, using the only nontrivial power in A001653 (i.e., 169 = 13^2) as q^s would not yield a valid solution here because it would mean p = 239 and q = 13 (contradicting p < q). Thus, if a Case 4 solution exists for odd semiprime n > 9, we must have s = 1 and r > 1, so n = (2r+1)*(2s+1) = (2r+1)*3, where 2r+1 is prime. Such a solution requires an index j satisfying two conditions: (1) A001653(j) = q^1 = q is prime, and (2) the corresponding term A002315(j-1) = p^r is a nontrivial prime power. There are no nontrivial powers (whether of primes or composites) among the terms in A002315 below 10^5000. Moreover, the terms in A001653 are the odd-indexed terms from A000129 (Pell numbers), so condition (1) requires that j satisfy A000129(2j-1) = q. A096650 lists the indices of all prime or probable prime Pell numbers up to 100000. A check of the value A002315(j-1) corresponding to each prime or probable prime among the odd-indexed Pell numbers A000129(2j-1) up to j=50000 determined that none were nontrivial powers, so if any Case 4 solution with n > 9 exists, it will yield a triangular number t = p^(2r) * q^2 = (2 * q^2 - 1) * q^2, where q >= A000129(100001) = 3.16...*10^38277, so t > 10^153110. Since there are no nontrivial powers at all in A002315 below 10^5000, and since prime Pell numbers above A000129(50000) seem so scarce, it seems extremely unlikely that any such solution exists.
Thus, a(n) = 0 for every odd semiprime n that is not divisible by 3, and assuming that no Case 4 solution for odd semiprime n > 9 exists, the only nonzero a(n) where n is an odd semiprime greater than 9 is a(15) = 13^4 * 239^2 = 1631432881.
If j is prime and n=2j, then a(n) (if nonzero) must be of the form p^r * q, where p and q are distinct primes, r = j-1, and q is one of 3 functions of p^r:
    q = f1(p^r) = 2*p^r - 1
    q = f2(p^r) = 2*p^r + 1
    q = f3(p^r) = (p^r - 1)/2
Of these, q = f1(p^r) for all but two cases among n < 1000:
    at n=362, q = f2(p^r), with p=3;
    at n=514, q = f3(p^r), with p=331.
Conjecture: a(2j) > 0 for all j > 1. (This conjecture holds at least through n = 2j = 1000. The largest a(n) for even n <= 1000 is a(898) = 20599^448 * (2 * 20599^448 - 1) = 3.21...*10^3865.) (End)
For more known terms, and information about unknown terms, see Links. - Jon E. Schoenfield, May 26 2014
If d(k*(k+1)/2) = 21, note that 2*q^2 = p^6 + 1 = (p^2 + 1)*(p^4 - p^2 + 1) has no prime solutions, so then k = p^2 and k+1 = 2*q^6, where p and q are distinct primes. We can prove 2*x^3 - y^2 = 1 has only one positive solution (1, 1) which shows that p^2 + 1 = 2*q^6 has no prime solutions. In the ring of Gaussian integers, x^3 = (1+y*i)*(1-y*i)/((1+i)*(1-i)) and (1+y*i)/(1+i) is coprime to (1-y*i)/(1-i), thus (1+y*i)/(1+i) = (u+v*i)^3 and (1-y*i)/(1-i) = (u-v*i)^3 for some integers u and v. Note that 1+y*i = (1+i)*(u+v*i)^3 = (u+v)*(u^2+v^2-4*u*v) + (u-v)*(u^2+v^2+4*u*v)*i, we have (u+v)*(u^2+v^2-4*u*v) = 1. Therefore, u = 1 and v = 0 if u > v, which means y = (u-v)*(u^2+v^2+4*u*v) = 1. This implies that a(21) = 0. - Jinyuan Wang, Aug 22 2020
a(n) is a perfect number for all n such that n/2 is in A000043. - J. Lowell, Mar 16 2024

			a(2)=3 because the smallest triangular number with 2 divisors is T(2)=3.

Attila Pethö, The Pell sequence contains only trivial perfect powers, Kossuth Lajos University, Department of Computer Science, H-4010 Debrecen, P.O. Box 12, Hungary, Dedicated to V.T. Sós and A. Hajnal, May 18 2008.
Project Euler, Highly divisible triangular number, Problem 12.
Jon E. Schoenfield, Table of n, a(n) for known terms in n = 1..100, with notes on unknown terms

Greedy inverse of A063440.

Cf. A000129, A001110, A001653, A002315, A081979, A242585, A319037.

More terms from Victoria A Sapko (vsapko(AT)frc.mass.edu), Sep 28 2007
a(14) corrected and a(20) added by Jon E. Schoenfield, May 11 2014
a(21)-a(24) from Jinyuan Wang, Aug 22 2020
a(25)-a(45) from Jon E. Schoenfield, Jan 28 2021

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

A039596 Numbers that are simultaneously triangular and square pyramidal.

Original entry on oeis.org

0, 1, 55, 91, 208335

Offset: 1

Felice Russo

Equivalent to 0^2 + 1^2 + 2^2 + 3^2 + ... + r^2 = 0 + 1 + 2 + 3 + ... + s = n for some r and s.

			1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 2 + 3 + ... + 10 = 55, so 55 is in the sequence.

Joe Roberts, Lure of the Integers, page 245 (entry for 645).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, p. 108.

R. Finkelstein and H. London, On triangular numbers which are sums of consecutive squares, J. Number Theory 4 (1972), 455-462.
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
H. E. Thomas Jr., Problem 5634, Amer. Math. Monthly, 75 (1968), p. 1018.

Intersection of A000217 and A000330.

Cf. A053611, A053612.

q:= n-> issqr(8*n+1):
select(q, [sum(j^2, j=1..n)$n=0..100])[];  # Alois P. Heinz, Oct 17 2024

Additional comments from Jud McCranie, Mar 19 2000

Zero inserted by Daniel Mondot, Sep 07 2023

A046177 Squares (A000290) which are also hexagonal numbers (A000384).

Original entry on oeis.org

1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481

Offset: 1

Eric W. Weisstein

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007
Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011
Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015
Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Colin Barker, Table of n, a(n) for n = 1..327
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (1155,-1155,1).

Cf. A008844, A046176, A253826.
Cf. A001110 (Numbers that are both triangular and square).
Cf. A000290, A000384, A016754, A253826.

LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)

Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015

a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007
a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016
a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016

A164055 Triangular numbers that are one plus a perfect square.

Original entry on oeis.org

1, 10, 325, 11026, 374545, 12723490, 432224101, 14682895930, 498786237505, 16944049179226, 575598885856165, 19553418069930370, 664240615491776401, 22564627508650467250, 766533094678624110085, 26039560591564569275626

Offset: 1

Tanya Khovanova & Alexey Radul, Aug 08 2009

a(n+1) is the second element of the n-th set of three consecutive triangular numbers whose product is a perfect square. - Arkadiusz Wesolowski, Apr 27 2012

The triangular numbers of this sequence satisfy the Diophantine equation T(k) = k*(k+1)/2 = m^2 + 1, which is equivalent to (2k+1)^2 - 2*(2m)^2 = 9. Now, with x=2k+1 and y=2m, the Pell-Fermat equation x^2 - 2*y^2 = 9 appears. The solutions x and y of this equation are respectively in A106329 and A075848. The indices k=(x-1)/2 of the triangular numbers of this sequence are in A072221, while the indices m=y/2 of the corresponding square numbers are in A106328. - Bernard Schott, Mar 09 2019

			10 is in this sequence because it is a triangular number A000217(4) and is equal to a square plus 1: 10 = 3^2 + 1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..500
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001110, A106329, A075848 (solutions of x^2 - 2 * y^2 = 9).

Cf. A072221 (indices of the triangular terms of this sequence), A106328 (indices of the corresponding square numbers).

a164055 n = a164055_list !! (n-1)
a164055_list = 1 : 10 : 325 : zipWith (+) a164055_list
   (map (* 35) $ tail $ zipWith (-) (tail a164055_list) a164055_list)
-- Reinhard Zumkeller, Apr 29 2012

LinearRecurrence[{35,-35,1}, {1,10,325}, 50] (* G. C. Greubel, Sep 09 2017 *)

my(x='x+O('x^50)); Vec(x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1))) \\ G. C. Greubel, Sep 09 2017

A000217 INTERSECT A002522.
a(n) = A000217(A072221(n-1)).
From R. J. Mathar, Sep 22 2009: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
G.f.: x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1)). (End)
A010054(a(n)) * A010052(a(n) - 1) = 1. - Reinhard Zumkeller, Apr 29 2012
a(n) = (A106329(n) - 1)*(A106329(n) + 1)/8 = (A106328(n))^2 + 1. - Bernard Schott, Mar 10 2019
a(n) = (14 + 9*(17+12*sqrt(2))^(1-n) - 9*(-17+12*sqrt(2))*(17+12*sqrt(2))^n) / 32. - Colin Barker, Mar 23 2019
a(n) = 9 * A001110(n) + 1 (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Comment molded into formula by R. J. Mathar, Sep 22 2009

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

A257711 Triangular numbers (A000217) that are the sum of seven consecutive triangular numbers.

Original entry on oeis.org

210, 3486, 51681, 883785, 13125126, 224476266, 3333728685, 57016086141, 846753959226, 14481861401910, 215072171913081, 3678335779997361, 54627484911961710, 934282806257926146, 13875166095466359621, 237304154453733242085, 3524237560763543380386

Offset: 1

Colin Barker, May 05 2015

			210 is in the sequence because T(20) = 210 = 10+15+21+28+36+45+55 = T(4)+ ... +T(10).

Colin Barker, Table of n, a(n) for n = 1..830
Index entries for linear recurrences with constant coefficients, signature (1,254,-254,-1,1).

Cf. A000217, A001110, A129803, A131557, A257712, A257713, A259413, A259414, A259415.

LinearRecurrence[{1, 254, -254, -1, 1}, {210, 3486, 51681, 883785, 13125126}, 30] (* Vincenzo Librandi, Jun 27 2015 *)

Vec(-21*x*(x^4-245*x^2+156*x+10) / ((x-1)*(x^2-16*x+1)*(x^2+16*x+1)) + O(x^100))

G.f.: -21*x*(x^4-245*x^2+156*x+10) / ((x-1)*(x^2-16*x+1)*(x^2+16*x+1)).

16*a(n) = 104 +225*A157456(n+1) +7*(-1)^n*A159678(n+1). - R. J. Mathar, Apr 28 2020

A003556 Numbers that are both square and tetrahedral.

Original entry on oeis.org

0, 1, 4, 19600

Offset: 1

N. J. A. Sloane

A. J. J. Meyl proved in 1878 that only 1, 4 and 19600 are both square and tetrahedral. See link. [Bernard Schott, Dec 23 2012]

			From _Bernard Schott_, Dec 23 2012: (Start)
If S(n) = n^2 and T(m) = m*(m+1)*(m+2)/6, then
-> S(1)= T(1) = 1;
-> S(2)= T(2) = 4;
-> S(140) = T(48) = 19600. (End)

D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 600.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, p. 165 (Rev. ed. 1997).

M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
A. J. J. Meyl, Question 1194, Nouvelles Annales de Mathématiques, 2ème série, tome 17 (1878), p. 464-467.

Intersection of A000290 and A000292.

Select[Rest[FoldList[Plus, 0, Rest[FoldList[Plus, 0, Range[50000]]]]], IntegerQ[Sqrt[ # ]]&]
Intersection[Binomial[# + 2, 3]&/@Range[0, 10000], Range[0,409000]^2] (* Harvey P. Dale, Feb 01 2011 *)

A098602 a(n) = A001652(n) * A046090(n).

Original entry on oeis.org

0, 12, 420, 14280, 485112, 16479540, 559819260, 19017375312, 646030941360, 21946034630940, 745519146510612, 25325704946729880, 860328449042305320, 29225841562491651012, 992818284675673829100, 33726595837410418538400, 1145711440187278556476512

Offset: 0

Charlie Marion, Oct 26 2004

From Ron Knott, Nov 25 2013: (Start)
a(n) = 2*r*(r+1) which is also of form s(s+1) where the s is in A053141.
a(n) is an oblong number (A002378) which is twice another oblong number. (End)
2*a(n)+1 and 4*a(n)+1 are both square. - Paul Cleary, Jun 23 2014

			a(1) = 12 = 2(2*3) = 3*4, a(2) = 420 = 2(14*15) = 20*21.

Reinhard Zumkeller, Table of n, a(n) for n = 0..255
Nikola Adžaga, Andrej Dujella, Dijana Kreso, Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002378, A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(12*x/((1-x)*(x^2-34*x+1)))); // G. C. Greubel, Jul 15 2018

2*Table[ Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Ray Chandler, Nov 10 2004, copied incorrect program from A029549, revised Jul 09 2015 *)
RecurrenceTable[{a[n+3] == 35 a[n+2] - 35 a[n+1] + a[n], a[1] == 0, a[2] == 12, a[3] == 420}, a, {n, 1, 10}] (* Ron Knott, Nov 25 2013 *)
LinearRecurrence[{35, -35, 1}, {0, 12, 420}, 25] (* T. D. Noe, Nov 25 2013 *)
Table[(LucasL[4*n+2, 2] - 6)/16, {n,0,30}] (* G. C. Greubel, Jul 15 2018 *)

concat(0, Vec(12*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016

{a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(4*n+2) - 6)/16), ", ")) \\ G. C. Greubel, Jul 15 2018

a(n) = 2*A029549(n) = 2*A001109(n)*A001109(n+1).
a(n) = (A001653(n)^2 - 1)/2.
a(n) = A053141(n)^2 + A011900(n)^2 - 1.
For n>0, a(n) = A053141(2n) - 2*A001109(n-1)^2.
For n>0, a(n) = 3*(A001542(n)^2 - A001542(n-1)^2).
For n>0, a(n) = A053141(2n-1) + 2*(A001653(2n-1) - A001109(n-1)^2).
a(n+1) + a(n) = 3*A001542(n+1)^2.
a(n+1) - a(n) = A001542(2*n).
a(n+1)*a(n) = 4*(A001109(n)^4 - A001109(n)^2) = 4*A001110(n)*(A001110(n) - 1).
From Ron Knott, Nov 25 2013: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
G.f.: 12*x / ((1-x)*(x^2-34*x+1)). (End)
a(n) = (-6 + (3-2*sqrt(2))*(17+12*sqrt(2))^(-n)+(3+2*sqrt(2))*(17+12*sqrt(2))^n)/16. -  Colin Barker, Mar 02 2016

More terms from Ray Chandler, Nov 10 2004

Corrected by Bill Lam (bill_lam(AT)myrealbox.com), Feb 27 2006

A257712 Triangular numbers (A000217) that are the sum of eight consecutive triangular numbers.

Original entry on oeis.org

120, 276, 1176, 28920, 126756, 306936, 1345620, 33362196, 146264856, 354192420, 1552832856, 38499933816, 168789505620, 408737734296, 1791967758756, 44428890250020, 194782943209176, 471682991173716, 2067929240760120, 51270900848577816, 224779347673872036

Offset: 1

Colin Barker, May 05 2015

			120 is in the sequence because T(15) = 120 = 1+3+6+10+15+21+28+36 = T(1)+ ... +T(8).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1154,-1154,0,0,-1,1).

Cf. A000217, A001110, A129803, A131557, A257711, A257713, A259413, A259414, A259415.

LinearRecurrence[{1, 0, 0, 1154, -1154, 0, 0, -1, 1}, {120, 276, 1176, 28920, 126756, 306936, 1345620, 33362196, 146264856}, 30] (* Vincenzo Librandi, Jun 27 2015 *)
Select[Total/@Partition[Accumulate[Range[5*10^6]],8,1],OddQ[ Sqrt[ 1+8#]]&] (* The program generates the first 16 terms of the sequence *) (* Harvey P. Dale, Feb 27 2022 *)

Vec(-12*x*(3*x^8+7*x^6+13*x^5-3387*x^4+2312*x^3+75*x^2+13*x+10) / ((x-1)*(x^2-6*x+1)*(x^2+6*x+1)*(x^4+34*x^2+1)) + O(x^100))

G.f.: -12*x*(3*x^8+7*x^6+13*x^5-3387*x^4+2312*x^3+75*x^2+13*x+10) / ((x-1)*(x^2-6*x+1)*(x^2+6*x+1)*(x^4+34*x^2+1)).

cp's OEIS Frontend

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