0,2

Erdos conjectured that n^2 - 1 = k! has a solution iff n is 5, 11 or 71 (when k is 4, 5 or 7).

Second order linear recurrences y(m)=2y(m-1)+A005563(n)y(m-2),y(0)=y(1)=1, have closed form solutions involving only powers of integers. - Len Smiley (smiley(AT)math.uaa.alaska.edu), Dec 08 2001

Number of edges in the join of two cycle graphs, both of order n, C_n * C_n - Roberto E. Martinez II (remartin(AT)fas.harvard.edu), Jan 07 2002

Let k be a positive integer, M_n be the n X n matrix m_(i,j)=k^abs(i-j) then det(M_n)=(-1)^(n-1)*a(k-1)^(n-1) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 28 2002

Also numbers n such that 4n + 4 is a square. - Cino Hilliard (hillcino368(AT)gmail.com), Dec 18 2003

The function sqrt(x^2 + 1), starting with 1, produces an integer after n(n+2) iterations. - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Aug 19 2004

a(n) mod 3 = 0 iff n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 16 2006

a(n)=A067725/3 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 06 2007

A129296(n) = number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Apr 09 2007

Sequence allows us to find X values of the equation: X^3 + X^2 = Y^2. To find Y values: b(n)=n(n+1)(n+2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 06 2007

Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007

Comment from R. Guy (rkg(AT)cpsc.ucalgary.ca), Feb 01 2008 (Start):

This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n+1 squares (or positions, or lilypads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by

T T - F F

T - T F F

T F T - F

T F T F -

T F - F T

- F T F T

F - T F T

F F T - T

F F - T T

I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.

Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)

For n>0: A143053(a(n)) = A000290(n+1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 20 2008

a(n+1)=terms of rank 0,1,3,6,10=A000217 of A120072 (3,8,5,15,). [From Paul Curtz (bpcrtz(AT)free.fr), Oct 28 2008]

For all n=A (0,3,8,15,24,..), X=[A000027] (1,2,3,4,5,...,), Y=[A000012] (1,1,1,1,...,) we have the Pell's equation X^2-A*Y^2=1. Example: 1-0*1=1; 2^2-3*1=1; 3^2-8*1=1; 4^2-15*1=1; and so on. [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Mar 08 2009]

A053186(a(n)) = 2*n. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 20 2009]

R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.

Derek Holton, Math in School, 37 #1 (Jan 2008) 20-22,

Edouard Lucas, Recreations Mathematiques, Gauthier-Villars, Vol. 2 (1883) 141-143.

Index entries for sequences related to linear recurrences with constant coefficients

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets

R. K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seqs., Vol. 3 (2000), #00.1.6

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Eric Weisstein's World of Mathematics, Near-Square Prime

G.f.: ( 3 - x ) / ( 1 - x )^3.

C(n+1, 1)*C(n+3, 1). - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 06 2005

A002378(a(n))=A002378(n)*A002378(n+1); e.g. A002378(15)=240=12*20 - Charlie Marion (charliemath(AT)verizon.net), Dec 29 2003

a(n)=sum(sum(j-k, j=2..n),k=0..n), n>=1. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 11 2007

a(n)==A134582(n+1)/4 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 01 2008

a(n)=Real((n+1+i)^2) [From Gerald Hillier (adr.rabbicat(AT)gmail.com), Oct 12 2008]

G.f.: x*(1+x)/exp(x) . [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 03 2009]

[seq(2*binomial(n, 2)-n, n=2..43)]; - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 25 2006

a:=n->sum (n, j=3..n): seq(a(n), n=2..43); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 21 2007

a:=n->sum(sum(j-k, j=2..n), k=0..n): seq(a(n), n=1..42); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 11 2007

a:=n->sum(sum(3, j=3..n)/3, k=1..n): seq(a(n), n=2..43); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 30 2007

A005563:=(-3+z)/(z-1)**3; [S. Plouffe in his 1992 dissertation.]

with(combinat, fibonacci):seq(fibonacci(3, i)-2, i=1..42); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 20 2008

a:=n->sum(1+sum(1, k=4..n), k=1..n):seq(a(n), n=2...43); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 11 2008

restart: G(x):=x*(1+x)/exp(x): f[0]:=G(x): for n from 1 to 43 do f[n]:=diff(f[n-1], x) od: x:=0: seq(abs(f[n]), n=2..43); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 03 2009]

Table[(m^2 - 1), {m, 42}] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 21 2007

(Other) sage: [lucas_number1(3, n, 1) for n in xrange(1, 43)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 16 2009]

A column of triangle A102537.

a(n+1), n>=2, first column (used for the Lyman series of the hydrogen atom) of triangle A120070.

Cf. A013468, A007531, A062196, A002378, A000290(n) = a(n-1) + 1.

Cf. A046092, A067725, A123865, A123866, A123867, A123868.

Cf. A028560.

Adjacent sequences: A005560 A005561 A005562 this_sequence A005564 A005565 A005566

Sequence in context: A086959 A083656 A013648 this_sequence A132411 A147998 A067998

nonn,easy

N. J. A. Sloane (njas(AT)research.att.com).