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Numbers k such that k*(k+1)/2 + 1 is a square.
(Formerly M1472)
53

%I M1472 #127 Apr 13 2022 13:25:18

%S 0,2,5,15,32,90,189,527,1104,3074,6437,17919,37520,104442,218685,

%T 608735,1274592,3547970,7428869,20679087,43298624,120526554,252362877,

%U 702480239,1470878640,4094354882,8572908965,23863649055,49966575152

%N Numbers k such that k*(k+1)/2 + 1 is a square.

%D A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Problem J/A2, Jul/August 1973 pp. 73-74; solution Jan 1974 (see link).

%D Jeffrey Shallit, personal communication.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Reinhard Zumkeller, <a href="/A006451/b006451.txt">Table of n, a(n) for n = 0..1000</a>

%H A. J. Gottlieb, <a href="/A006451/a006451.pdf">How four dogs meet in a field, etc.</a> (scanned copy)

%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992

%H J. Shallit, <a href="/A006449/a006449.pdf">Letter to N. J. A. Sloane, Oct. 1975</a>

%H Hermann Stamm-Wilbrandt, <a href="/A006451/a006451.svg">4 interlaced bisections</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,6,-6,-1,1).

%F G.f.: x*(-2-3*x+2*x^2+x^3)/(x-1)/(x^2+2*x-1)/(x^2-2*x-1). Conjectured (correctly) by _Simon Plouffe_ in his 1992 dissertation.

%F a(n) = 6*a(n-2) - a(n-4) + 2 with a(0)=0, a(1)=2, a(2)=5, a(3)=15. - _Zak Seidov_, Apr 15 2008

%F a(n) = 3*a(n-2) + 4*sqrt((a(n-2)^2 + a(n-2))/2 + 1) + 1 with a(0) = 0, a(1) = 2. - _Raphie Frank_, Feb 02 2013

%F a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5); a(0)=0, a(1)=2, a(2)=5, a(3)=15, a(4)=32. - _Harvey P. Dale_, Jul 17 2013

%F a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - _Hermann Stamm-Wilbrandt_, Aug 26 2014

%F a(2*n+1) = A098790(2*n+1). - _Hermann Stamm-Wilbrandt_, Aug 26 2014

%F a(2*n) = A098586(2*n-1), for n>0. - _Hermann Stamm-Wilbrandt_, Aug 27 2014

%F a(n) = 8*sqrt(T(a(n-2)) + 1) + a(n-4) where T(a(n)) = A000217(a(n)), and a(-1) = -1, a(0)=0, a(1)=2, a(2)=5. - _Vladimir Pletser_, Apr 29 2017

%p N:= 100: # to get a(0) to a(N)

%p A[0]:= 0: A[1]:= 2: A[2]:= 5: A[3]:= 15:

%p for n from 4 to N do A[n]:= 6*A[n-2] - A[n-4] + 2 od:

%p seq(A[n],n=0..N); # _Robert Israel_, Aug 26 2014

%t LinearRecurrence[{1,6,-6,-1,1},{0,2,5,15,32},30] (* _Harvey P. Dale_, Jul 17 2013 *)

%t Select[Range[10^6], IntegerQ@ Sqrt[# (# + 1)/2 + 1] &] (* _Michael De Vlieger_, Apr 25 2017 *)

%o (PARI) for(n=1,10000,t=n*(n+1)/2+1;if(issquare(t), print1(n,", "))) \\ _Joerg Arndt_, Oct 10 2009

%o (Haskell)

%o a006451 n = a006451_list !! n

%o a006451_list = 0 : 2 : 5 : 15 : map (+ 2)

%o (zipWith (-) (map (* 6) (drop 2 a006451_list)) a006451_list)

%o -- _Reinhard Zumkeller_, Jan 10 2012

%Y Cf. A000124, A006452, A006454, A098586, A098790.

%Y Cf. numbers m such that k*A000217(m)+1 is a square: this sequence for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6. - _Bruno Berselli_, Dec 16 2013

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_ and _Jeffrey Shallit_

%E More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001

%E Edited by _N. J. A. Sloane_, Oct 24 2009, following discussions by several correspondents in the Sequence Fans Mailing List, Oct 10 2009