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A154145 Indices k such that 15 plus the k-th triangular number is a perfect square. 2
1, 4, 6, 11, 20, 33, 43, 70, 121, 196, 254, 411, 708, 1145, 1483, 2398, 4129, 6676, 8646, 13979, 24068, 38913, 50395, 81478, 140281, 226804, 293726, 474891, 817620, 1321913, 1711963, 2767870, 4765441, 7704676, 9978054, 16132331, 27775028 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

a(1..4)=(1,4,6,11); a(n>4)=6*a(n-2)-a(n-4)+2. [From Ctibor O. Zizka, Nov 13 2009]

LINKS

Table of n, a(n) for n=1..37.

F. T. Adams-Watters, SeqFan Discussion, Oct 2009

FORMULA

{k: 15+k*(k+1)/2 in A000290}.

Conjectures: (Start)

a(n)= +a(n-1) +6*a(n-4) -6*a(n-5) -a(n-8) +a(n-9).

G.f.: x*(-1-3*x-2*x^2-5*x^3-3*x^4+5*x^5+2*x^6+3*x^7+2*x^8)/((x-1) * (x^4+2*x^2-1) * (x^4-2*x^2-1)).

G.f.: ( 4 + (7+4*x+16*x^2+11*x^3)/(x^4-2*x^2-1) + 1/(x-1) + (-4-7*x-3*x^2-2*x^3)/(x^4+2*x^2-1) )/2. (End)

a(1..4) = (1,4,6,11); a(n) = 6*a(n-2) - a(n-4) + 2, for n>4. - Ctibor O. Zizka, Nov 13 2009

EXAMPLE

1*(1+1)/2+15 = 4^2. 4*(4+1)/2+15 = 5^2. 6*(6+1)/2+15 = 6^2. 11*(11+1)/2+15 = 9^2.

MATHEMATICA

Flatten[Position[Accumulate[Range[28000000]], _?(IntegerQ[Sqrt[#+15]]&)]] (* This program will take a long time to run. *) (* Harvey P. Dale, Jun 09 2014 *)

Join[{1, 4, 6}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 15 &]] (* G. C. Greubel, Sep 03 2016 *)

CROSSREFS

Cf. A000217, A000290, A006451.

Sequence in context: A291916 A047811 A244010 * A327553 A302428 A336142

Adjacent sequences:  A154142 A154143 A154144 * A154146 A154147 A154148

KEYWORD

nonn

AUTHOR

R. J. Mathar, Oct 18 2009

EXTENSIONS

a(32)-a(37) from Donovan Johnson, Nov 01 2010

STATUS

approved

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Last modified March 6 20:37 EST 2021. Contains 341850 sequences. (Running on oeis4.)