1,2

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

Recursion: a(n+2)=6*a(n+1)-a(n)+4 with a(0)=0 and a(1)=5; g.f.: A(x)=(5x-x^2)/((1-x)*(1-6x+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(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).

Expand[Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/(4 Sqrt[2]) - 1, {n, 1, 20}]] [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 14 2009]

Cf. A001108, A001109, A001110.

Sequence in context: A034224 A167023 A121831 this_sequence A127816 A024063 A015545

Adjacent sequences: A076705 A076706 A076707 this_sequence A076709 A076710 A076711

easy,nonn

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