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A006451 Numbers n such that n*(n+1)/2+1 is a square.
(Formerly M1472)
52
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 1. - Ctibor O. Zizka, Nov 10 2009

For n>0, with T(n) the n-th triangular number,

  a(2n-1)=A002315(n-1)+A001108(n) and T(a(2n-1)+1=(A001653(n)+A001109(n))^2;

  e.g., a(5)=41+49=90 and T(90)+1=(29+35)^2;

  a(2n)=A002315(n)-A001108(n)-1 and T(a(2n))+1=(A001653(n+1)-A001109(n))^2;

  e.g., a(6)=239-49-1=189 and T(189)+1=(169-35)^2.

  In general, if b(n)=the numbers such that T(b(n))+T(k) is a square, then

  b(2n-1)=k*A002315(n-1)+A001108(n) and T(b(2n-1)+T(k))=(k*A001653(n)+A001109(n))^2;

  e.g., for k=3, b(5)=3*41+49=172 and T(172)+6=(3*29+35)^2;

  b(2n)=k*A002315(n)-A001108(n)-1 and T(b(2n))+1=(k*A001653(n+1)-A001109(n))^2;

    e.g., for k=4, b(6)=4*239-49-1=906 and T(906)+10=(4*169-35)^2.

  See A154138, A154140, A154143, A154145, A154148, A154153.

- Charlie Marion, Dec 05 2010

a(2n-1)=A001652(n)-A001653(n); a(2n)=A001652(n-1)+A001653(n+1).

  In general, indices k such that A001109(2j) plus the k-th triangular

  number is a perfect square may be found as follows:

  b(2n-1)=A001652(n+j-1)-A001653(n-j);

  b(2n)=A001652(n-j-1)+A001653(n+j);

  indices k such that A001109(2j-1) plus the k-th triangular number is

  a perfect square may be found as follows:

  b(2n-1)=A001652(n+j-1)-A001653(n-j+1);

  b(2n)=A001652(n-j)+A001653(n+j). See also A154140.

  - Charlie Marion, Mar 11 2011

(((a(n)+ sqrt ((a(n)(a(n) + 1)/2) + 1)) = a(n) + A006452(n + 1) = A216134(n + 1) gives the indices of the Sophie Germain triangular numbers  > 0. (((a(n)+ 2*sqrt ((a(n)(a(n) + 1)/2) + 1)) = a(n) + 2*A006452(n + 1) = A124124(n + 1) gives the indices of the triangular numbers, tr,  such that (tr - 1)/2 is a Sophie Germain triangular number > 0. For example, 32 = a(4) and (32+ sqrt ((32(32 + 1)/2) + 1)) = 32 + 23 = 55 = a(4) + A006452(5) = A216134(5) and (32+ 2*sqrt ((32(32 + 1)/2) + 1)) = 32 + 23 + 23 = 78 = a(4) + 2*A006452(5) = A124124(5). Additionally, a(n+1) = (A216134(n) + A216134(n + 2))/2. For example, 32 = (A216134(3) + A216134(5))/2 = (9 + 55)/2. - Raphie Frank, Jan 03 2013

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.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

Hermann Stamm-Wilbrandt, 4 interlaced bisections

Index entries for linear recurrences with constant coefficients, signature (1, 6, -6, -1, 1).

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) = 6a(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(0)=0, a(1)=2, a(2)=5, a(3)=15, a(4)=32, a(n)=a(n-1)+6*a(n-2)- 6*a(n-3)- a(n-4)+a(n-5). - 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

Starting from n=7

n=odd: a(n)= Int(a(n-1)*a(n-2)/a(n-3))-4 = Int(2.093836*a(n-1))-4

n=even: a(n)= Int(a(n-1)*a(n-2)/a(n-3))-2 = Int(2.783612*a(n-1))-2

or

n=2m+1: a(2m+1)= Int(a(2m)*a(2m-1)/a(2m-2))-4 = Int(2.093836*a(2m))-4

n=2m: a(2m)= Int(a(2m-1)*a(2m-2)/a(2m-3))-2 = Int(2.783612*a(2m-1))-2. - Vladimir Pletser, Apr 25 2017

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

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

MATHEMATICA

a[0] = a[1] = 1; a[2] = 2; a[3] = 4; a[n_] := 6 a[n - 2] - a[n - 4]; t = Array[a, 30, 0]; Join[{0}, Drop[ Floor[ Sqrt[2 #^2 - 1]] & /@ t, 2]] (* Robert G. Wilson v, Jun 11 2010 *)

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 *)

PROG

(PARI) for(n=1, 10000, t=n*(n+1)/2+1; if(issquare(t), print1(n, ", "))) \\ Joerg Arndt, Oct 10 2009

(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

CROSSREFS

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

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

Sequence in context: A078528 A077686 A034499 * A226103 A000962 A118387

Adjacent sequences:  A006448 A006449 A006450 * A006452 A006453 A006454

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane and Jeffrey Shallit

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

STATUS

approved

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Last modified August 20 01:56 EDT 2017. Contains 290821 sequences.