A057769 -id:A057769 - cp's OEIS frontend

A007204 Crystal ball sequence for D_4 lattice.

1, 25, 169, 625, 1681, 3721, 7225, 12769, 21025, 32761, 48841, 70225, 97969, 133225, 177241, 231361, 297025, 375769, 469225, 579121, 707281, 855625, 1026169, 1221025, 1442401, 1692601, 1974025, 2289169, 2640625, 3031081, 3463321, 3940225, 4464769, 5040025

Offset: 0

N. J. A. Sloane and J. H. Conway, Apr 28 1994

Equals binomial transform of [1, 24, 120, 192, 96, 0, 0, 0, ...]. - Gary W. Adamson, Aug 13 2009
Hypotenuse of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + A069074(n-1)^2 = a(n)^2. - Martin Renner, Nov 12 2011
Numbers n such that n*x^4 + x^2 + 1 is reducible. - Arkadiusz Wesolowski, Nov 04 2013

Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 53.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
Index entries for crystal ball sequences
Index entries for sequences related to D_4 lattice
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A016754, A057769, A060300, A069074.

Cf. A000290, A000583, A001844, A005563, A099761.

[(2*n^2+2*n+1)^2: n in [0..40]]; // Vincenzo Librandi, Nov 18 2016

A007204:=n->(2*n^2+2*n+1)^2; seq(A007204(n), n=0..30);

Table[(2n^2+2n+1)^2,{n,0,30}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,25,169,625,1681},40] (* Harvey P. Dale, Mar 03 2013 *)

a(n)=(2*n^2+2*n+1)^2 \\ Charles R Greathouse IV, Feb 08 2017

G.f.: (1 + 54*x^2 + 20*x + 20*x^3 + x^4)/(1-x)^5.
a(0)=1, a(1)=25, a(2)=169, a(3)=625, a(4)=1681, a(n)=5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Mar 03 2013
Sum_{n>=0} 1/a(n) = Pi*(sinh(Pi) - Pi)/(2*(cosh(Pi) + 1)) = 1.0487582722070177... - Ilya Gutkovskiy, Nov 18 2016
a(n) = A016754(n) + A060300(n). - Bruce J. Nicholson, Apr 14 2017
a(n) = A001844(n)^2 = (2*n^2+2*n+1)^2. - Bruce J. Nicholson, May 15 2017
a(n) = A000583(n+1) + A099761(n) + 2*A005563(n-1)*A000290(n). - Charlie Marion, Jan 14 2021
E.g.f.: exp(x)*(1 + 24*x + 60*x^2 + 32*x^3 + 4*x^4). - Stefano Spezia, Jun 06 2021

More terms from Harvey P. Dale, Mar 03 2013

A069074 a(n) = (2n+2)(2n+3)(2n+4) = 24A000330(n+1).

Original entry on oeis.org

24, 120, 336, 720, 1320, 2184, 3360, 4896, 6840, 9240, 12144, 15600, 19656, 24360, 29760, 35904, 42840, 50616, 59280, 68880, 79464, 91080, 103776, 117600, 132600, 148824, 166320, 185136, 205320, 226920, 249984, 274560, 300696, 328440, 357840

Offset: 0

Benoit Cloitre, Apr 05 2002

sqrt((Sum_{k=0..n} 2*a(k)) + 1) = A056220(n+2). - Doug Bell, Mar 09 2009
Second leg of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + a(n-1)^2 = A007204(n)^2. - Martin Renner, Nov 12 2011
Numbers which are both the sum of 2*n + 4 consecutive odd integers and the sum of the 2*n + 2 immediately higher consecutive odd integers.  In general, let f(k,n) = 3*k^3*A000330(n).  Then f(k,n) is both the sum of k*n + k consecutive terms from the arithmetic progression with first term A000217(k) and constant difference k and the immediately higher k*n terms from the same progression.  When k = 1, f(k,n) = A059270(n). - Charlie Marion, Aug 23 2021

Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 53.
T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 190.
Jolley, Summation of Series, Dover (1961).
Konrad Knopp, Theory and application of infinite series, Dover, p. 269

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001844. A001844(n+1)^2 - a(n) and A001844(n+1)^2 + a(n) are both square numbers. - Doug Bell, Mar 08 2009

Cf. A000466. a(n) = Sum_{k=0..2n+3} (A000466(n+1) + 2k) which is the sum of 2n+4 consecutive odd integers starting at A000466(n+1). - Doug Bell, Mar 08 2009

[(2*n+2)*(2*n+3)*(2*n+4): n in [0..40]]; // Vincenzo Librandi, Oct 04 2011

LinearRecurrence[{4,-6,4,-1},{24,120,336,720},40] (* Harvey P. Dale, Apr 10 2017 *)

a(n)=6*binomial(2*n+4,3) \\ Charles R Greathouse IV, Mar 21 2015

Sum_{n>=0} (-1)^n/a(n) = (Pi-3)/4 = 0.03539816339... [Jolley, eq. 244]
Sum_{n>=0} 1/a(n) = 3/4 - log(2) = 0.05685281... [Jolley, eq. 249]
G.f.: ( 24+24*x ) / (x-1)^4. - R. J. Mathar, Oct 03 2011

A075972 Positions of check bits in code in A075970.

Original entry on oeis.org

131071, 33423871, 505290271, 1717986919, 2863311531, 3579139413, 5055419121, 10094819116, 22645810751, 41800423686, 78201007705, 120750351308, 163539395911, 320531740310

Offset: 0

Bob Jenkins (bob_jenkins(AT)burtleburtle.net)

nonn

J. H. Conway and N. J. A. Sloane, Lexicographic codes: error-correcting codes from game theory, IEEE Transactions on Information Theory, 32:337-348, 1986.

Bob Jenkins, Tables of Binary Lexicodes
Ari Trachtenberg, Error-Correcting Codes on Graphs: Lexicodes, Trellises and Factor Graphs

Cf. A075928, A075929, A075930, A075970, A075971, A057769, A075966, A075978.

A086302 a(n) = 4n^4 + 24n^3 + 48n^2 + 36n + 8.

Original entry on oeis.org

8, 120, 528, 1520, 3480, 6888, 12320, 20448, 32040, 47960, 69168, 96720, 131768, 175560, 229440, 294848, 373320, 466488, 576080, 703920, 851928, 1022120, 1216608, 1437600, 1687400, 1968408, 2283120, 2634128, 3024120, 3455880, 3932288, 4456320, 5031048

Offset: 0

Neven Juric (neven.juric(AT)apis-it.hr), Aug 29 2003

Suppose one wishes to find sets of four positive integers (a,b,c,d) such that ab+1, ac+1, ad+1, bc+1, bd+1, cd+1 are perfect squares. Then one may take a = 1, b = x^2 + 2x, c = x^2 + 4x + 3, d = 4x^4 + 24x^3 + 48x^2 + 36x + 8.

			(a,b,c,d) = (1,3,8,120), (1,8,15,528), (1,15,24,1520), (1,24,35,3480), ...

Paolo Xausa, Table of n, a(n) for n = 0..10000
Philip Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
Eric Weisstein's World of Mathematics, Diophantus Property.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A057769, A062392.

LinearRecurrence[{5, -10, 10, -5, 1}, {8, 120, 528, 1520, 3480}, 50] (* or *)
A086302[n_] := 4 (n + 1) (n + 2) (n^2 + 3 n + 1);
Array[A086302, 50, 0] (* Paolo Xausa, Jan 16 2024 *)

a(n) = A057769(n+1) +  1. - N. J. A. Sloane, Jun 12 2004
G.f.: 8*(1 + 10*x + x^2)/(1 - x)^5. - Colin Barker, Mar 26 2012
a(n) = 4 * (n+1) * (n+2) * (n^2 + 3*n + 1). - Bruno Berselli, Mar 26 2012
a(n) = 8*A062392(n+1). - Bruce J. Nicholson, Jun 05 2017
Sum_{n>=0} 1/a(n) = tan(sqrt(5)*Pi/2)*Pi/(4*sqrt(5)). - Amiram Eldar, Jan 22 2024
E.g.f.: 4*exp(x)*(2 + 28*x + 37*x^2 + 12*x^3 + x^4). - Stefano Spezia, Apr 27 2025

cp's OEIS Frontend

A007204 Crystal ball sequence for D_4 lattice.

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A069074 a(n) = (2*n+2)*(2*n+3)*(2*n+4) = 24*A000330(n+1).

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A075972 Positions of check bits in code in A075970.

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A086302 a(n) = 4*n^4 + 24*n^3 + 48*n^2 + 36*n + 8.

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Mathematica

Formula

A069074 a(n) = (2n+2)(2n+3)(2n+4) = 24A000330(n+1).

A086302 a(n) = 4n^4 + 24n^3 + 48n^2 + 36n + 8.