A076049 -id:A076049 - cp's OEIS frontend

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

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

A239969 Least positive k such that triangular(n) + triangular(n+k) is a triangular number (A000217), or -1 if no such k exists.

Original entry on oeis.org

2, 5, 1, 3, 20, 2, 4, 16, 3, 5, 31, 4, 6, 119, 5, 7, 16, 6, 8, 103, 7, 9, 2, 8, 10, 26, 9, 11, 464, 10, 12, 1, 11, 13, 313, 12, 5, 58, 13, 15, 37, 14, 3, 493, 15, 17, 31, 16, 18, 47, 17, 2, 79, 9, 20, 796, 19, 21, 883, 20, 22, 89, 4, 23, 58, 22, 24, 100, 23, 25, 1276

Offset: 3

Alex Ratushnyak, Mar 30 2014

nonn

In other words, smallest solution k>0 to 4*k^2 + 8*(k + 1)*n + 8*n^2 + 4*k + 1 = m^2. - Ralf Stephan, Apr 01 2014

			a(3) = 2 because triangular(3)+triangular(3+2)=21 is a triangular number.
a(5) = 1 because triangular(5)+triangular(5+1)=36 is a triangular number.
In other words, k=a(3)=2 is the smallest positive solution to 4*k^2 + 28*k + 97 = m^2, and k=a(5)=1 is the smallest positive solution to 4*k^2 + 44*k + 241 = m^2.

Reinhard Zumkeller, Table of n, a(n) for n = 3..1000

Cf. A000217, A082183, A076708, A076049, A239970.

Cf. A010054.

a239969 n = head [k | k <- [1..],
                      a010054 (a000217 n + a000217 (n + k)) == 1]
-- Reinhard Zumkeller, Apr 03 2014

triangular(n) = n*(n+1)/2;
is_triangular(n) = issquare(8*n+1);
s=[]; for(n=3, 100, k=1; while(!is_triangular(triangular(n)+triangular(n+k)), k++); s=concat(s, k)); s \\ Colin Barker, Mar 31 2014

First PROG corrected by Colin Barker, Apr 04 2014

cp's OEIS Frontend

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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