cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A006003 a(n) = n*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775
Offset: 0

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Author

Keywords

Comments

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

Examples

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
  • F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

Programs

  • GAP
    a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018
    
  • Haskell
    a006003 n = n * (n ^ 2 + 1) `div` 2
    a006003_list = scanl (+) 0 a005448_list
    -- Reinhard Zumkeller, Jun 20 2013
    
  • MATLAB
    % Also works with FreeMat.
    for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
    % Stefano Spezia, Aug 12 2018
    
  • Magma
    [n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    [Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015
    
  • Mathematica
    Table[ n(n^2 + 1)/2, {n, 0, 45}]
    LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
    CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
    With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)
  • Maxima
    a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */
    
  • PARI
    {a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016
    
  • Python
    def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

Formula

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Extensions

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

A317614 a(n) = (1/2)*(n^3 + n*(n mod 2)).

Original entry on oeis.org

1, 4, 15, 32, 65, 108, 175, 256, 369, 500, 671, 864, 1105, 1372, 1695, 2048, 2465, 2916, 3439, 4000, 4641, 5324, 6095, 6912, 7825, 8788, 9855, 10976, 12209, 13500, 14911, 16384, 17985, 19652, 21455, 23328, 25345, 27436, 29679, 32000, 34481, 37044, 39775, 42592
Offset: 1

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Author

Stefano Spezia, Aug 01 2018

Keywords

Comments

Terms are obtained as partial sums in an algorithm for the generation of the sequence of the fourth powers (A000583). Starting with the sequence of the positive integers (A000027), it is necessary to delete every 4th term and to consider the partial sums of the obtained sequence, then to delete every 3rd term, and lastly to consider again the partial sums (see References).
a(n) is the trace of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern as shown in the examples below. Specifically, M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even, and it has det(M(n)) = 0 for n > 2 (proved).
From Saeed Barari, Oct 31 2021: (Start)
Also the sum of the entries in an n X n matrix whose elements start from 1 and increase as they approach the center. For instance, in case of n=5, the entries of the following matrix sum to 65:
1 2 3 2 1
2 3 4 3 2
3 4 5 4 3
2 3 4 3 2
1 2 3 2 1. (End)
The n X n square matrix of the preceding comment is defined as: A[i,j,n] = n - abs((n + 1)/2 - j) - abs((n + 1)/2 - i). - Stefano Spezia, Nov 05 2021

Examples

			For n = 1 the matrix M(1) is
  1
with trace Tr(M(1)) = a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with Tr(M(2)) = a(2) = 4.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with Tr(M(3)) = a(3) = 15.
		

References

  • Edward A. Ashcroft, Anthony A. Faustini, Rangaswami Jagannathan, and William W. Wadge, Multidimensional Programming, Oxford University Press 1995, p. 12.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
  • G. Polya, Mathematics and Plausible Reasoning: Induction and analogy in mathematics, Princeton University Press 1990, p. 118.
  • Shailesh Shirali, A Primer on Number Sequences, Universities Press (India) 2004, p. 106.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.7.3 on pages 122-123.

Crossrefs

Cf. A000583, A000027, A186424 (first differences).
Cf. related to the M matrices: A074147 (antidiagonals), A130130 (rank), A241016 (row sums), A317617 (column sums), A322277 (permanent), A323723 (subdiagonal sums), A323724 (superdiagonal sums).

Programs

  • GAP
    a_n:=List([1..nmax], n->(1/2)*(n^3 + n*RemInt(n, 2)));
    
  • GAP
    List([1..50],n->(1/2)*(n^3+n*(n mod 2))); # Muniru A Asiru, Aug 24 2018
  • Magma
    [IsEven(n) select n^3/2 else (n^3+n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 07 2018
    
  • Maple
    a:=n->(1/2)*(n^3+n*modp(n,2)): seq(a(n),n=1..50); # Muniru A Asiru, Aug 24 2018
  • Mathematica
    CoefficientList[Series[1/4 E^-x (1 + 3 E^(2 x) + 6 E^(2 x) x + 2 E^(2 x) x^2), {x, 0, 45}], x]*Table[(k + 1)!, {k, 0, 45}]
    CoefficientList[Series[-(1 + x^2)/((-1 + x)*(1 + x)^3), {x, 0, 45}], x]*Table[(k + 1)*(-1)^k, {k, 0, 45}]
    CoefficientList[Series[-(1 + x^2)/((-1 + x)^3*(1 + x)), {x, 0, 45}], x]*Table[(k + 1), {k, 0, 45}]
    From Robert G. Wilson v, Aug 01 2018: (Start)
    a[i_, j_, n_] := If[OddQ@ i, j + n (i - 1), n*i - j + 1]; f[n_] := Tr[Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 45]
    CoefficientList[Series[(x^4 + 2x^3 + 6x^2 + 2x + 1)/((x - 1)^4 (x + 1)^2), {x, 0, 45}], x]
    LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 4, 15, 32, 65, 108}, 45]
    (End)
  • Maxima
    a(n):=(1/2)*(n^3 + n*mod(n,2))$ makelist(a(n), n, 1, nmax);
    
  • PARI
    Vec(x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2) + O(x^40)) \\ Colin Barker, Aug 02 2018
    
  • PARI
    M(i, j, n) = if (i % 2, j + n*(i-1), n*i - j + 1);
    a(n) = sum(k=1, n, M(k, k, n)); \\ Michel Marcus, Aug 07 2018
    
  • R
    for (n in 1:nmax){
       a <- (n^3+n*n%%2)/2
       output <- c(n, a)
       cat(output, "\n")
    }
    (MATLAB and FreeMat)
    for(n=1:nmax); a=(n^3+n*mod(n,2))/2; fprintf('%d\t%0.f\n',n,a); end
    

Formula

a(n) = (1/2)*(A000578(n) + n*A000035(n)).
a(n) = A006003(n) - (n/2)*(1 - (n mod 2)).
a(n) = Sum_{k=1..n} T(n,k), where T(n,k) = ((n + 1)*k - n)*(n mod 2) + ((n - 1)*k + 1)*(1 - (n mod 2)).
E.g.f.: E(x) = (1/4)*exp(-x)*x*(1 + 3*exp(2*x) + 6*exp(2*x)*x + 2*exp(2*x)*x^2).
L.g.f.: L(x) = -x*(1 + x^2)/((-1 + x)*(1 + x)^3).
H.l.g.f.: LH(x) = -x*(1 + x^2)/((-1 + x)^3*(1 + x)).
Dirichlet g.f.: (1/2)*(Zeta(-3 + s) + 2^(-s)*(-2 + 2^s)*Zeta(-1 + s)).
From Colin Barker, Aug 02 2018: (Start)
G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = n^3/2 for n even.
a(n) = (n^3+n)/2 for n odd. (End)
a(2*n) = A317297(n+1) + A001489(n). - Stefano Spezia, Dec 28 2018
Sum_{n>0} 1/a(n) = (1/2)*(-2*polygamma(0, 1/2) + polygamma(0, (1-i)/2)+ polygamma(0, (1+i)/2)) + zeta(3)/4 approximately equal to 1.3959168891658447368440622669882813003351669... - Stefano Spezia, Feb 11 2019
a(n) = (A000578(n) + A193356(n))/2. - Stefano Spezia, Jun 27 2022
a(n) = A210378(n-1)/n. - Stefano Spezia, Jul 15 2024

A219086 a(n) = floor((n + 1/2)^4).

Original entry on oeis.org

0, 5, 39, 150, 410, 915, 1785, 3164, 5220, 8145, 12155, 17490, 24414, 33215, 44205, 57720, 74120, 93789, 117135, 144590, 176610, 213675, 256289, 304980, 360300, 422825, 493155, 571914, 659750, 757335, 865365, 984560, 1115664
Offset: 0

Views

Author

Clark Kimberling, Jan 01 2013

Keywords

Comments

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/4 and { } = fractional part. Equivalently, the jump sequence of f(x) = x^(1/4), in the sense that these are the nonnegative integers k for which round(k^p) < round((k+1)^p). For details and a guide to related sequences, see A219085.
-4*a(n) gives the real part of (n+n*i)*((n+1)+n*i)*(n+(n+1)*i)*((n+1)+(n+1)*i). The imaginary part is always zero. - Jon Perry, Feb 05 2014
Numbers k such that 16*k+1 is a fourth power. - Bruno Berselli, May 29 2018
The row sums of "Floyd's Triangle", which is a triangular array of natural numbers beginning with the number 1, produce the sequence A006003. A006003 can be bisected to get the Rhombic Dodecahedron Sequence A005917, whose n-th partial sum is n^4, and A317297, whose n-th partial sum is a(n). Interleave n^4 or A000583 back with {a(n)} to get A011863, whose first differences are A019298. Finally, A011863(n)-A011863(n-2) = A006003(n-1). - Bruce J. Nicholson, Dec 22 2019

Examples

			0^(1/4) = 0.000...; 1^(1/4) = 1.000...
5^(1/4) = 1.495...; 6^(1/4) = 1.565...
39^(1/4) = 2.499...; 40^(1/4) = 2.514...
		

Crossrefs

Programs

Formula

G.f.: (5*x^3 + 14*x^2 + 5*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = (2*n^4 + 4*n^3 + 3*n^2 + n)/2. - J. M. Bergot, Apr 05 2014
a(n) = Sum_{i=0..n} i*(4*i^2 + 1) = n*(n + 1)*(2*n^2 + 2*n + 1)/2. - Bruno Berselli, Feb 09 2017
a(n) = lcm((2*n + 1)^2 - 1, (2*n + 1)^2 + 1)/8 for n>=1. - Lechoslaw Ratajczak, Mar 26 2017
a(n) = A000217(n) * A001844(n). - Bruce J. Nicholson, May 14 2017
E.g.f.: (1/2)*exp(x)*x*(10 + 29*x + 16*x^2 + 2*x^3). - Stefano Spezia, Dec 27 2019
a(n) = ((2*n+1)^4 - 1)/16. - Jianing Song, Jan 03 2023
Sum_{n>=1} 1/a(n) = 6 - 2*Pi*tanh(Pi/2). - Amiram Eldar, Jan 08 2023

A007607 Skip 1, take 2, skip 3, etc.

Original entry on oeis.org

2, 3, 7, 8, 9, 10, 16, 17, 18, 19, 20, 21, 29, 30, 31, 32, 33, 34, 35, 36, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130
Offset: 1

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Author

Keywords

Comments

Numbers k with the property that the smallest Dyck path of the symmetric representation of sigma(k) has a central peak. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
Union of A317303 and A014105. - Omar E. Pol, Aug 29 2018

Examples

			From _Omar E. Pol_, Aug 29 2018: (Start)
Written as an irregular triangle in which the row lengths are the nonzero even numbers the sequence begins:
    2,   3;
    7,   8,   9,  10;
   16,  17,  18,  19,  20,  21;
   29,  30,  31,  32,  33,  34,  35,  36;
   46,  47,  48,  49,  50,  51,  52,  53,  54,  55;
   67,  68,  69,  70,  71,  72,  73,  74,  75,  76,  77,  78;
   92,  93,  94,  95,  96,  97,  98,  99, 100, 101, 102, 103, 104, 105;
  121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136;
...
Row sums give the nonzero terms of A317297.
Column 1 gives A130883, n >= 1.
Right border gives A014105, n >= 1.
(End)
		

References

  • R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 177.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Complement of A007606.
Similar to A360418.

Programs

  • Haskell
    a007607 n = a007607_list !! (n-1)
    a007607_list = skipTake 1 [1..] where
       skipTake k xs = take (k + 1) (drop k xs)
                       ++ skipTake (k + 2) (drop (2*k + 1) xs)
    -- Reinhard Zumkeller, Feb 12 2011
    
  • Haskell
    a007607_list' = f $ tail $ scanl (+) 0 [1..] where
       f (t:t':t'':ts) = [t+1..t'] ++ f (t'':ts)
    -- Reinhard Zumkeller, Feb 12 2011
  • Mathematica
    Flatten[ Table[i, {j, 2, 16, 2}, {i, j(j - 1)/2 + 1, j(j + 1)/2}]] (* Robert G. Wilson v, Mar 11 2004 *)
    With[{t=20},Flatten[Take[TakeList[Range[(t(t+1))/2],Range[t]],{2,-1,2}]]] (* Harvey P. Dale, Sep 26 2021 *)
  • PARI
    for(m=0,10,for(n=2*m^2+3*m+2,2*m^2+5*m+3,print1(n", "))) \\ Charles R Greathouse IV, Feb 12 2011
    

Formula

G.f.: 1/(1-x) * (1/(1-x) + x*Sum_{k>=1} (2k+1)*x^(k*(k+1))). - Ralf Stephan, Mar 03 2004
a(A000290(n)) = A001105(n). - Reinhard Zumkeller, Feb 12 2011
A057211(a(n)) = 0. - Reinhard Zumkeller, Dec 30 2011
a(n) = floor(sqrt(n) + 1/2)^2 + n = A053187(n) + n. - Ridouane Oudra, May 04 2019

A144965 a(n) = 4*n*(4*n^2 + 1).

Original entry on oeis.org

0, 20, 136, 444, 1040, 2020, 3480, 5516, 8224, 11700, 16040, 21340, 27696, 35204, 43960, 54060, 65600, 78676, 93384, 109820, 128080, 148260, 170456, 194764, 221280, 250100, 281320, 315036, 351344, 390340, 432120, 476780, 524416, 575124, 629000, 686140, 746640
Offset: 0

Views

Author

Luc Comeau-Montasse, Sep 27 2008

Keywords

Comments

(a(n))^2 + (n*a(n)+1)^2 is always a perfect square.

Crossrefs

Programs

  • Magma
    I:=[0, 20, 136, 444]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 30 2012
  • Mathematica
    CoefficientList[Series[4*x*(5+14*x+5*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 30 2012 *)
    LinearRecurrence[{4,-6,4,-1},{0,20,136,444},50] (* Harvey P. Dale, Aug 07 2022 *)

Formula

G.f.: 4*x*(5+14*x+5*x^2)/(1-x)^4. - Colin Barker, May 24 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 30 2012
From Elmo R. Oliveira, Aug 07 2025: (Start)
E.g.f.: 4*x*(5 + 2*x)*(1 + 2*x)*exp(x).
a(n) = 4*A317297(n+1) = A008586(n)*A053755(n). (End)
Showing 1-5 of 5 results.