A272399 -id:A272399 - cp's OEIS frontend

A016814 a(n) = (4*n + 1)^2.

1, 25, 81, 169, 289, 441, 625, 841, 1089, 1369, 1681, 2025, 2401, 2809, 3249, 3721, 4225, 4761, 5329, 5929, 6561, 7225, 7921, 8649, 9409, 10201, 11025, 11881, 12769, 13689, 14641, 15625, 16641, 17689, 18769, 19881, 21025, 22201, 23409, 24649, 25921, 27225, 28561, 29929

Offset: 0

N. J. A. Sloane

A bisection of A016754. Sequence arises from reading the line from 1, in the direction 1, 25, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

G. C. Greubel, Table of n, a(n) for n = 0..10000 (terms 0..200 from Ivan Panchenko).
Leo Tavares, Illustration: Square trapeziums
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m-3), this sequence (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).
Cf. A001539, A006752, A016742, A016802, A016813.
Cf. A016826, A016838, A017126, A068467, A087197.
Cf. A272399, A014105.

[(4*n+1)^2: n in [0..40]]; // G. C. Greubel, Dec 28 2022

A016814:=n->(4*n+1)^2; seq(A016814(k), k=0..100); # Wesley Ivan Hurt, Nov 02 2013

(4*Range[0,40] +1)^2 (* or *) LinearRecurrence[{3,-3,1}, {1,25,81}, 40] (* Harvey P. Dale, Nov 20 2012 *)
Accumulate[32Range[0, 47] - 8] + 9 (* Alonso del Arte, Aug 19 2017 *)

a(n)=(4*n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015

[(4*n+1)^2 for n in range(41)] # G. C. Greubel, Dec 28 2022

a(n) = a(n-1) + 32*n - 8, n > 0. - Vincenzo Librandi, Dec 15 2010
From George F. Johnson, Sep 28 2012: (Start)
G.f.: (1 + 22*x + 9*x^2)/(1 - x)^3.
a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
a(n-1)*a(n+1) = (a(n) - 16)^2 ; a(n+1) - a(n-1) = 16*sqrt(a(n)).
a(n) = A016754(2*n) = (A016813(n))^2. (End)
Sum_{n>=0} 1/a(n) = G/2 + Pi^2/16, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=1} (1 - 1/a(n)) = 2*Gamma(5/4)^2/sqrt(Pi) = 2 * A068467^2 * A087197. - Amiram Eldar, Feb 01 2021
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A017078(n).
a(2*n+1) = A017126(n).
E.g.f.: (1 + 24*x + 16*x^2)*exp(x). (End)
a(n) = A272399(n+1) - A014105(n). - Leo Tavares, Dec 24 2023

A063436 Write 1, 2, 3, 4, ... counterclockwise in a hexagonal spiral around 0 starting left down, then a(n) is the sequence found by reading from 0 in the vertical upward direction.

Original entry on oeis.org

0, 15, 54, 117, 204, 315, 450, 609, 792, 999, 1230, 1485, 1764, 2067, 2394, 2745, 3120, 3519, 3942, 4389, 4860, 5355, 5874, 6417, 6984, 7575, 8190, 8829, 9492, 10179, 10890, 11625, 12384, 13167, 13974, 14805, 15660, 16539, 17442, 18369, 19320, 20295, 21294, 22317

Offset: 0

Floor van Lamoen, Jul 21 2001

Related to parity of Beatty sequences for exp(-(1/2)/n). Let f(k,n)=-sum(i=1,n,sum(j=1,i,(-1)^floor(j*exp(-(1/2)/n)))), then a(n)=Max{f(k,n) : 1<=k<=4*a(n)-2} and for 0<=i<=4*a(n)-3, f(i,n)=f(4*a(n)-2-i,n). - Benoit Cloitre, May 26 2004

Or, sum of multiples of 2 and 3 from 0 to 6*n. - Zak Seidov, Aug 06 2016

			The spiral begins:
.
        16--15--14
        /         \
      17   5---4  13
      /   /     \   \
    18   6   0   3  12
    /   /   /   /   /
  19   7   1---2  11  26
    \   \         /   /
    20   8---9--10  25
      \             /
      21--22--23--24

Harry J. Smith, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Stellar Triangles.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A003154, A007742, A062783, A272399.

a[n_] := 3*n*(4*n + 1); Array[a, 40, 0] (* Amiram Eldar, Mar 27 2022 *)

{ for (n=0, 1000, write("b063436.txt", n, " ", n*(12*n + 3)) ) } \\ Harry J. Smith, Aug 21 2009

a(n) = 3*n*(4*n+1) = 3*A007742(n).
a(n) = 24*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
From Colin Barker, Jul 07 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 3*x*(5 + 3*x)/(1-x)^3. (End)
a(n) = A272399(n+1) - A003154(n+1). - Leo Tavares, Mar 24 2022
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=1} 1/a(n) = 4/3 - Pi/6 - log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(2)) + log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3 - 4/3. (End)
E.g.f.: 3*x*(5 + 4*x)*exp(x). - Elmo R. Oliveira, Oct 31 2024

A272398 The union of hexagonal numbers (A000384) and centered 9-gonal numbers (A060544).

Original entry on oeis.org

1, 6, 10, 15, 28, 45, 55, 66, 91, 120, 136, 153, 190, 231, 253, 276, 325, 378, 406, 435, 496, 561, 595, 630, 703, 780, 820, 861, 946, 1035, 1081, 1128, 1225, 1326, 1378, 1431, 1540, 1653, 1711, 1770, 1891, 2016, 2080, 2145, 2278, 2415, 2485, 2556, 2701, 2850

Offset: 1

Colin Barker, Apr 28 2016

nonn

The construction of the g.f. works basically as follows every third entry of A000384 equals every second entry of A060544, A000384(3n+1) = A060544(2n+1) = (3*n+1)*(6*n+1), which is an immediate consequence of their polynomial representations. So the sequence is the union of A000384 and the bisection 10, 55, 136, 253,... of A060544. Following Section 4.3 of Riordan's book "Combinatorial identities", subsampling and "aering" are done by replacing the independent variable of the g.f. by roots of the independent variable. So this sequence has rational g.f. because it is derived by regular interlacing of the two original sequences which also have rational g.f.'s. - R. J. Mathar, Jul 15 2016

Colin Barker, Table of n, a(n) for n = 1..1000

Cf. A000384, A060544, A272399 (intersection).

a(4*n-3) = A272399(n).
Conjectures:
  a(n) = (-1+(-1)^n-6*((-i)^n+i^n)*n+18*n^2)/16 where i is the imaginary unit.
  a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-2*a(n-5)+2*a(n-6)-2*a(n-7)+a(n-8) for n>8.
  G.f.: x*(1+4*x+5*x^3+6*x^4+x^5+x^6) / ((1-x)^3*(1+x)*(1+x^2)^2).

A368692 a(n) = (12n + 6)!(6n + 9)!/(108(4n + 2)!(2n + 3)!((6*n + 5)!)^2).

Original entry on oeis.org

14, 563108, 54231252075, 6700034035890000, 928978310614152999200, 137569863175651804211692560, 21253098849879053645154605945160, 3381375421559384124434964404229384000, 549714622911935710495977183989400234273000

Offset: 0

Karol A. Penson, Jan 03 2024

nonn

According to A. Adolphson and S. Sperber, "On the integrality of hypergeometric series whose coefficients are factorial ratios", ArXiv: 2001.03296, s.page 14, first equation after Eq.(7.4): for any two integers K, L, the ratios (3*K)!*(3*L)!/(K!*L!*((K+L)!)^2) are proven to be integers. 108*a(n) results from K = 4*n+2 and L = 2*n+3, n>=0. It is conjectured here that a(n) are integers.

A. Adolphson and S. Sperber, On the integrality of hypergeometric series whose coefficients are factorial ratios, Acta Arithmetica 200 (2021), no.1, 39-59.

Cf. A368650, A304126, A368545, A082368, A113424, A368875.

seq((12*n + 6)!*(6*n + 9)!/(108*(4*n + 2)!*(2*n + 3)!*((6*n + 5)!)^2),n=0..9);

G.f.:  14*hypergeometric8F7([7/12, 2/3, 5/6, 11/12, 13/12, 17/12, 13/6, 7/3], [1, 7/6, 4/3, 3/2, 3/2, 5/3, 11/6], 186624*z).
E.g.f.: 14*hypergeometric8F8([7/12, 2/3, 5/6, 11/12, 13/12, 17/12, 13/6, 7/3], [1, 1, 7/6, 4/3, 3/2, 3/2, 5/3, 11/6], 186624*z).
a(n) = Integral_{x=0..186624} x^n*W(x) dx, n>=0, where W(x) = (1/(20736*Pi))*MeijerG([[], [0, 0, 1/6, 1/3, 1/2, 1/2, 2/3, 5/6]], [[-5/12, -1/3, -1/6, -1/12, 1/12, 5/12, 7/6, 4/3], []], x/186624). MeijerG is the Meijer G - function.  W(x) can be represented as an expression containing the sum of 4 generalized hypergeometric functions of type 8F7. W(x) is a positive function in the interval [0, 186624], is singular at x=0 and monotonically decreases to zero at x = 186624. This integral representation as the n-th power moment of the positive function  W(x) in the interval [0, 186624] is unique, as W(x) is the solution of the Hausdorff moment problem.
Let b(n) = Gamma(7+ 12*n)/(6*Gamma(2 + 2*n)*Gamma(3 + 4*n)*Gamma(6 + 6*n)), then a(n) = b(n) * A272399(n+2). - Peter Luschny, Jan 06 2024

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A016814 a(n) = (4*n + 1)^2.

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A063436 Write 1, 2, 3, 4, ... counterclockwise in a hexagonal spiral around 0 starting left down, then a(n) is the sequence found by reading from 0 in the vertical upward direction.

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A272398 The union of hexagonal numbers (A000384) and centered 9-gonal numbers (A060544).

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A368692 a(n) = (12*n + 6)!*(6*n + 9)!/(108*(4*n + 2)!*(2*n + 3)!*((6*n + 5)!)^2).

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A368692 a(n) = (12n + 6)!(6n + 9)!/(108(4n + 2)!(2n + 3)!((6*n + 5)!)^2).