A006451 Numbers k such that k*(k+1)/2 + 1 is a square.
0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074, 6437, 17919, 37520, 104442, 218685, 608735, 1274592, 3547970, 7428869, 20679087, 43298624, 120526554, 252362877, 702480239, 1470878640, 4094354882, 8572908965, 23863649055, 49966575152
Offset: 0
References
- 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).
- Jeffrey Shallit, personal communication.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
- A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
- Hermann Stamm-Wilbrandt, 4 interlaced bisections
- Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).
Crossrefs
Programs
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Haskell
a006451 n = a006451_list !! n a006451_list = 0 : 2 : 5 : 15 : map (+ 2) (zipWith (-) (map (* 6) (drop 2 a006451_list)) a006451_list) -- Reinhard Zumkeller, Jan 10 2012
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Maple
N:= 100: # to get a(0) to a(N) A[0]:= 0: A[1]:= 2: A[2]:= 5: A[3]:= 15: for n from 4 to N do A[n]:= 6*A[n-2] - A[n-4] + 2 od: seq(A[n],n=0..N); # Robert Israel, Aug 26 2014
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Mathematica
LinearRecurrence[{1,6,-6,-1,1},{0,2,5,15,32},30] (* Harvey P. Dale, Jul 17 2013 *) Select[Range[10^6], IntegerQ@ Sqrt[# (# + 1)/2 + 1] &] (* Michael De Vlieger, Apr 25 2017 *) -
PARI
for(n=1,10000,t=n*(n+1)/2+1;if(issquare(t), print1(n,", "))) \\ Joerg Arndt, Oct 10 2009
Formula
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.
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
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
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
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n+1) = A098790(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n) = A098586(2*n-1), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
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
Extensions
More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001
Edited by N. J. A. Sloane, Oct 24 2009, following discussions by several correspondents in the Sequence Fans Mailing List, Oct 10 2009