A005917 -id:A005917 - cp's OEIS frontend

A063493 a(n) = (2n-1)(13n^2-13n+6)/6.

1, 16, 70, 189, 399, 726, 1196, 1835, 2669, 3724, 5026, 6601, 8475, 10674, 13224, 16151, 19481, 23240, 27454, 32149, 37351, 43086, 49380, 56259, 63749, 71876, 80666, 90145, 100339, 111274, 122976, 135471, 148785, 162944, 177974, 193901, 210751, 228550, 247324, 267099

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n-1)*(13*n^2-13*n+6)/6: n in [1..40]]; // Vincenzo Librandi, Dec 16 2015

Table[(2 n - 1) (13 n^2 - 13 n + 6)/6, {n, 1, 40}] (* Bruno Berselli, Dec 16 2015 *)
LinearRecurrence[{4,-6,4,-1}, {1,16,70,189}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(13*n^2 - 13*n + 6)/6 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-6+12*x+39*x^2+26*x^3)*exp(x)/6 + 1)) \\ G. C. Greubel, Dec 01 2017

A063493_list, m = [], [26, -13, 2, 1]
for _ in range(10**2):
    A063493_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

G.f.: x*(1+x)*(1+11*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Dec 16 2015
E.g.f.: (-6 + 12*x + 39*x^2 + 26*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

A063495 a(n) = (2n-1)(5n^2-5n+2)/2.

Original entry on oeis.org

1, 18, 80, 217, 459, 836, 1378, 2115, 3077, 4294, 5796, 7613, 9775, 12312, 15254, 18631, 22473, 26810, 31672, 37089, 43091, 49708, 56970, 64907, 73549, 82926, 93068, 104005, 115767, 128384, 141886, 156303, 171665, 188002, 205344, 223721, 243163, 263700, 285362

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n-1)*(5*n^2-5*n+2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2n-1)(5n^2-5n+2)/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,18,80,217},40] (* Harvey P. Dale, Dec 18 2011 *)

a(n) = (2*n - 1)*(5*n^2 - 5*n + 2)/2 \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2+4*x+15*x^2+10*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

From Harvey P. Dale, Dec 18 2011: (Start)
a(1)=1, a(2)=18, a(3)=80, a(4)=217, a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) - a(n-4).
G.f.: (x^3+14*x^2+14*x+1)/(1-x)^4. (End)
E.g.f.: (-2 + 4*x + 15*x^2 + 10*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017

A049480 a(n) = (2n-1)(n^2 -n +6)/6.

Original entry on oeis.org

1, 4, 10, 21, 39, 66, 104, 155, 221, 304, 406, 529, 675, 846, 1044, 1271, 1529, 1820, 2146, 2509, 2911, 3354, 3840, 4371, 4949, 5576, 6254, 6985, 7771, 8614, 9516, 10479, 11505, 12596, 13754, 14981, 16279, 17650, 19096, 20619, 22221

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harvey P. Dale, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n-1)*(n^2-n+6)/6: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2n-1)(n^2-n+6)/6,{n,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,4,10,21},50] (* Harvey P. Dale, Jan 01 2012 *)

a(n)=(2*n-1)*(n^2-n+6)/6 \\ Charles R Greathouse IV, Sep 24 2015

x='x+O('x^30); Vec(serlaplace((-6 + 12*x + 3*x^2 + 2*x^3)*exp(x)/6 + 1)) \\ G. C. Greubel, Dec 01 2017

From Harvey P. Dale, Jan 01 2012: (Start)
G.f.: x*(x^3 + 1)/(x-1)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4); a(1)=1, a(2)=4, a(3)=10, a(4)=21. (End)
E.g.f.: (-6 + 12*x + 3*x^2 + 2*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

A005914 Number of points on surface of hexagonal prism: 12*n^2 + 2 for n > 0 (coordination sequence for W(2)).

Original entry on oeis.org

1, 14, 50, 110, 194, 302, 434, 590, 770, 974, 1202, 1454, 1730, 2030, 2354, 2702, 3074, 3470, 3890, 4334, 4802, 5294, 5810, 6350, 6914, 7502, 8114, 8750, 9410, 10094, 10802, 11534, 12290, 13070, 13874, 14702, 15554, 16430, 17330, 18254, 19202, 20174, 21170

Offset: 0

N. J. A. Sloane and Ralf W. Grosse-Kunstleve

For n >= 1, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n,n+1} such that Im(f) contains 2 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Feb 24 2007
Equals binomial transform of [1, 13, 23, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Apr 22 2008
First bisection of A005918. After 1, all terms are in A000408 (see Formula section). - Bruno Berselli, Feb 07 2012
Also sequence found by reading the segment (1, 14) together with the line from 14, in the direction 14, 50, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Nov 02 2012
Unique sequence such that for all n > 0, n*a(1) + (n-1)*a(2) + (n-3)*a(3) + ... + 2*a(2) + a(1) = n^4. - Warren Breslow, Dec 12 2014

Gmelin Handbook of Inorganic and Organometallic Chemistry, 8th Ed., 1994, TYPIX search code (229) cI2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Ovidiu Bagdasar, On Some Functions Involving the lcm and gcd of Integer Tuples, Scientific Publications of the State University of Novi Pazar, Appl. Maths. Inform. and Mech., Vol. 6, No. 2 (2014), pp. 91-100.
Ralf W. Grosse-Kunstleve, Coordination Sequences and Encyclopedia of Integer Sequences, 1996.
Ralf W. Grosse-Kunstleve, G. O. Brunner and N. J. A. Sloane, Algebraic Description of Coordination Sequences and Exact Topological Densities for Zeolites, Acta Cryst., A52 (1996), pp. 879-889.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013), Article 13.5.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Boon K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem., Vol. 24 (1985), pp. 4545-4558.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First differences of A005917.

Cf. A000408, A001082, A003154, A005918, A206399.

A005914:=-(z+1)*(z**2+10*z+1)/(z-1)**3; # Simon Plouffe in his 1992 dissertation.

Table[If[n == 0, 1, 12*n^2 + 2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
Join[{1},LinearRecurrence[{3,-3,1},{14,50,110},50]] (* Harvey P. Dale, Oct 09 2012 *)

a(n)=12*n^2+2 \\ Charles R Greathouse IV, Jan 31 2012

G.f.: (1+x)*(1+10*x+x^2)/(1-x)^3. - Simon Plouffe (see MAPLE line)
a(n) = (2n-1)^2 + (2n)^2 + (2n+1)^2 for n > 0. - Bruno Berselli, Jan 30 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=14, a(2)=50, a(3)=110. - Harvey P. Dale, Oct 09 2012
E.g.f.: exp(x)*(12*x^2 + 12*x + 2) - 1. - Alois P. Heinz, Sep 10 2013
From Bruce J. Nicholson, Jan 19 2019: (Start)
Sum_{i=1..n} a(i) = A005917(n+1).
a(n) = A003154(n) + A003154(n+1). (End)
From Amiram Eldar, Jan 27 2022: (Start)
Sum_{n>=0} 1/a(n) = ((Pi/sqrt(6))*coth(Pi/sqrt(6)) + 3)/4.
Sum_{n>=0} (-1)^n/a(n) = ((Pi/sqrt(6))*cosech(Pi/sqrt(6)) + 3)/4. (End)

A132366 Partial sum of centered tetrahedral numbers A005894.

Original entry on oeis.org

1, 6, 21, 56, 125, 246, 441, 736, 1161, 1750, 2541, 3576, 4901, 6566, 8625, 11136, 14161, 17766, 22021, 27000, 32781, 39446, 47081, 55776, 65625, 76726, 89181, 103096, 118581, 135750, 154721, 175616, 198561, 223686, 251125, 281016, 313501, 348726, 386841

Offset: 0

Jonathan Vos Post, Nov 09 2007

From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) is the number of achiral colorings of the 5 tetrahedral facets (or vertices) of a regular 4-dimensional simplex using n or fewer colors. An achiral arrangement is identical to its reflection. The 4-dimensional simplex is also called a 5-cell or pentachoron. Its Schläfli symbol is {3,3,3}.
There are 60 elements in the automorphism group of the 4-dimensional simplex that are not in its rotation group. Each is an odd permutation of the vertices and can be associated with a partition of 5 based on the conjugacy class of the permutation. The first formula for a(n-1) is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
  Partition  Count  Odd Cycle Indices
  41         30     x_1x_4^1
  32         20     x_2^1x_3^1
  2111       10     x_1^3x_2^1 (End)

Nathaniel Johnston, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000292, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Cf. A337895 (oriented), A000389(n+4) (unoriented), A000389 (chiral), A331353 (5-cell edges, faces), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell), A338967 (120-cell, 600-cell).
a(n-1) = A325001(4,n).

Do[Print[n, " ", (n^4 + 4 n^3 + 11 n^2 + 14 n + 6)/6 ], {n, 0, 10000}]
Accumulate[Table[(2n+1)(n^2+n+3)/3,{n,0,40}]] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,6,21,56,125},40] (* Harvey P. Dale, Feb 26 2020 *)

a(n) = (n^4 + 4*n^3 + 11*n^2 + 14*n + 6)/6 = (n^2+2*n+6)*(n+1)^2/6.
G.f.: -(x+1)*(x^2+1) / (x-1)^5. - Colin Barker, May 04 2013
From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) = n^2 * (5 + n^2) / 6.
a(n-1) = binomial(n+4,5) - binomial(n,5) = A000389(n+4) - A000389(n).
a(n-1) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n-1) = 2*A000389(n+4) - A337895(n) = A337895(n) - 2*A000389(n) .
G.f. for a(n-1): x * (x+1) * (x^2+1) / (1-x)^5. (End)
From Amiram Eldar, Feb 14 2023: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/5 + 3/25 - 3*Pi*coth(sqrt(5)*Pi)/(5*sqrt(5)).
Sum_{n>=0} (-1)^n/a(n) = Pi^2/10 - 3/25 + 3*Pi*cosech(sqrt(5)*Pi)/(5*sqrt(5)). (End)
a(n) = A006007(n) + A006007(n+1) = A002415(n) + A002415(n+2). - R. J. Mathar, Jun 05 2025

Corrected offset, Mathematica program by Tomas J. Bulka (tbulka(AT)rodincoil.com), Sep 02 2009

A195522 T(n,k) = Number of lower triangles of an n X n -k..k array with all row and column sums zero.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 5, 15, 1, 1, 7, 65, 199, 1, 1, 9, 175, 3753, 6247, 1, 1, 11, 369, 27267, 860017, 505623, 1, 1, 13, 671, 121367, 23663523, 839301197, 105997283, 1, 1, 15, 1105, 401565, 286168923, 122092290831, 3535646416019, 58923059879, 1, 1, 17

Offset: 1

R. H. Hardin, Sep 20 2011

Table starts
....1......1........1.........1..........1...........1...........1.......1....1
....1......1........1.........1..........1...........1...........1.......1....1
....3......5........7.........9.........11..........13..........15......17...19
...15.....65......175.......369........671........1105........1695....2465.3439
..199...3753....27267....121367.....401565.....1089411.....2563933.5423365
.6247.860017.23663523.286168923.2106810049.11131321791.46387885537

			Some solutions for n=5 k=6
..0..........0..........0..........0..........0..........0..........0
..0.0.......-2.2........6-6.......-1.1........5-5.......-4.4.......-4.4
.-1.3-2.....-6.0.6.....-6.6.0.....-1.5-4.....-6.4.2......3-6.3.....-4.1.3
..6-3-2-1....4-4-4.4....5.3-5-3....0-5.3.2....0.4-3-1...-5.5.1-1....5-2-4.1
.-5.0.4.1.0..4.2-2-4.0.-5-3.5.3.0..2-1.1-2.0..1-3.1.1.0..6-3-4.1.0..3-3.1-1.0

R. H. Hardin, Table of n, a(n) for n = 1..75

Row 4 is A005917(n+1).

Empirical for rows:
T(2,k) = 1
T(3,k) = 2*k + 1
T(4,k) = 4*k^3 + 6*k^2 + 4*k + 1
T(5,k) = (643/45)*k^6 + (643/15)*k^5 + (2165/36)*k^4 + (293/6)*k^3 + (4423/180)*k^2 + (73/10)*k + 1
T(6,k) = (7389349/90720)*k^10 + (7389349/18144)*k^9 + (836251/864)*k^8 + (4318165/3024)*k^7 + (6254923/4320)*k^6 + (4563293/4320)*k^5 + (10247161/18144)*k^4 + (249983/1134)*k^3 + (21959/360)*k^2 + (3469/315)*k + 1

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

Clark Kimberling, Jan 01 2013

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

			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...

Clark Kimberling, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A123865, A219085, A234459, A000217, A001844.

Cf. A317297, A006003, A005917, A000583, A011863, A019298.

A219086:=n->floor((n + (1/2))^4); seq(A219086(n), n=0..50); # Wesley Ivan Hurt, Apr 05 2014

Table[Floor[(n + 1/2)^4], {n, 0, 100}]
LinearRecurrence[{5,-10,10,-5,1},{0,5,39,150,410},40] (* Harvey P. Dale, Jan 15 2023 *)

a(n)=floor((n + 1/2)^4) \\ Charles R Greathouse IV, Apr 15 2014

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

A007606 Take 1, skip 2, take 3, etc.

Original entry on oeis.org

1, 4, 5, 6, 11, 12, 13, 14, 15, 22, 23, 24, 25, 26, 27, 28, 37, 38, 39, 40, 41, 42, 43, 44, 45, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 137, 138

Offset: 1

N. J. A. Sloane, Robert G. Wilson v, Mira Bernstein

List the natural numbers: 1, 2, 3, 4, 5, 6, 7, ... . Keep the first number (1), delete the next two numbers (2, 3), keep the next three numbers (4, 5, 6), delete the next four numbers (7, 8, 9, 10) and so on.
a(A000290(n)) = A000384(n). - Reinhard Zumkeller, Feb 12 2011
A057211(a(n)) = 1. - Reinhard Zumkeller, Dec 30 2011
Numbers k with the property that the smallest Dyck path of the symmetric representation of sigma(k) has a central valley. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
Union of nonzero terms of A000384 and A317304. - Omar E. Pol, Aug 29 2018
The values of k such that, in a listing of all congruence classes of positive integers, the k-th congruence class contains k. Here the class r mod m (with r in {1,...,m}) precedes the class r' mod m' (with r' in {1,...,m'}) iff mA360418. - James Propp, Feb 10 2023

			From _Omar E. Pol_, Aug 29 2018: (Start)
Written as an irregular triangle in which the row lengths are the odd numbers the sequence begins:
    1;
    4,   5,   6;
   11,  12,  13,  14,  15;
   22,  23,  24,  25,  26,  27,  28;
   37,  38,  39,  40,  41,  42,  43,  44,  45;
   56,  57,  58,  59,  60,  61,  62 , 63,  64,  65,  66;
   79,  80,  81,  82 , 83,  84,  85,  86,  87,  88,  89,  90,  91;
  106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120;
...
Row sums give A005917.
Column 1 gives A084849.
Column 2 gives A096376, n >= 1.
Right border gives A000384, n >= 1.
(End)

C. Dumitrescu & V. Seleacu, editors, Some Notions and Questions in Number Theory, Vol. I, Erhus Publ., Glendale, 1994.
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).
F. Smarandache, Properties of Numbers, 1972.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
C. Dumitrescu & V. Seleacu, editors, Some Notions and Questions in Number Theory, Vol. I.
Index entries for sequences generated by sieves

Complement of A007607.
Cf. A007950, A007951, A007952, A048859, A004201.
Cf. A000384, A005917, A084849, A096376.

a007606 n = a007606_list !! (n-1)
a007606_list = takeSkip 1 [1..] where
   takeSkip k xs = take k xs ++ takeSkip (k + 2) (drop (2*k + 1) xs)
-- Reinhard Zumkeller, Feb 12 2011

Flatten[ Table[i, {j, 1, 17, 2}, {i, j(j - 1)/2 + 1, j(j + 1)/2}]] (* Robert G. Wilson v, Mar 11 2004 *)
Join[{1},Flatten[With[{nn=20},Range[#[[1]],Total[#]]&/@Take[Thread[ {Accumulate[ Range[nn]]+1,Range[nn]}],{2,-1,2}]]]] (* Harvey P. Dale, Jun 23 2013 *)
With[{nn=20},Take[TakeList[Range[(nn(nn+1))/2],Range[nn]],{1,nn,2}]]//Flatten (* Harvey P. Dale, Feb 10 2023 *)

for(n=1,66,m=sqrtint(n-1);print1(n+m*(m+1),","))

a(n) = n + m*(m+1) where m = floor(sqrt(n-1)). - Klaus Brockhaus, Mar 26 2004

a(n+1) = a(n) + if n=k^2 then 2*k+1 else 1; a(1) = 1. - Reinhard Zumkeller, May 13 2009

A254681 Fifth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 21, 176, 936, 3750, 12342, 35112, 89232, 207207, 446875, 906048, 1743248, 3206268, 5670588, 9690000, 16062144, 25912029, 40797009, 62837104, 94875000, 140670530, 205134930, 294610680, 417203280, 583171875, 805386231

Offset: 1

Luciano Ancora, Feb 12 2015

			Fourth differences:  1, 12,  23,  24, (repeat 24)  ...   (A101104)
Third differences:   1, 13,  36,  60,   84,   108, ...   (A101103)
Second differences:  1, 14,  50, 110,  194,   302, ...   (A005914)
First differences:   1, 15,  65, 175,  369,   671, ...   (A005917)
-------------------------------------------------------------------------
The fourth powers:   1, 16,  81, 256,  625,  1296, ...   (A000583)
-------------------------------------------------------------------------
First partial sums:  1, 17,  98, 354,  979,  2275, ...   (A000538)
Second partial sums: 1, 18, 116, 470, 1449,  3724, ...   (A101089)
Third partial sums:  1, 19, 135, 605, 2054,  5778, ...   (A101090)
Fourth partial sums: 1, 20, 155, 760, 2814,  8592, ...   (A101091)
Fifth partial sums:  1, 21, 176, 936, 3750, 12342, ...   (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers .
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000538, A000583, A005914, A005917, A101089, A101090, A101091, A101103, A101104, A254682, A254683, A254684.

[Binomial(n+5,6)*n*(n+5)*(2*n+5)/42: n in [1..30]]; // G. C. Greubel, Dec 01 2018

seq(coeff(series((x+11*x^2+11*x^3+x^4)/(1-x)^10,x,n+1), x, n), n = 1 .. 30); # Muniru A Asiru, Dec 02 2018

Table[n^2(1+n)(2+n)(3+n)(4+n)(5+n)^2(5+2n)/30240, {n,26}] (* or *)
CoefficientList[Series[(1 + 11 x + 11 x^2 + x^3)/(1-x)^10, {x,0,25}], x]
CoefficientList[Series[(1/30240)E^x (30240 + 604800 x + 2041200 x^2 + 2368800 x^3 + 1233540 x^4 + 326592 x^5 + 46410 x^6 + 3540 x^7 + 135 x^8 + 2 x^9), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* Stefano Spezia, Dec 02 2018 *)
Nest[Accumulate[#]&,Range[30]^4,5] (* Harvey P. Dale, Jan 03 2022 *)

my(x='x+O('x^30)); Vec((x+11*x^2+11*x^3+x^4)/(1-x)^10) \\ G. C. Greubel, Dec 01 2018

[binomial(n+5,6)*n*(n+5)*(2*n+5)/42 for n in (1..30)] # G. C. Greubel, Dec 01 2018

G.f.:(x + 11*x^2 + 11*x^3 + x^4)/(1 - x)^10.
a(n) = n^2*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(5 + 2*n)/30240.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^4.
E.g.f.: (1/30240)*exp(x)*(30240 + 604800*x + 2041200*x^2 + 2368800*x^3 + 1233540*x^4 + 326592*x^5 + 46410*x^6 + 3540*x^7 + 135*x^8 + 2*x^9). - Stefano Spezia, Dec 02 2018
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 172032*log(2)/125 - 2382233/2500.
Sum_{n>=1} (-1)^(n+1)/a(n) = 42*Pi^2/25 - 43008*Pi/125 + 2663213/2500. (End)

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

cp's OEIS Frontend

A063493 a(n) = (2*n-1)*(13*n^2-13*n+6)/6.

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A063495 a(n) = (2*n-1)*(5*n^2-5*n+2)/2.

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A049480 a(n) = (2*n-1)*(n^2 -n +6)/6.

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A005914 Number of points on surface of hexagonal prism: 12*n^2 + 2 for n > 0 (coordination sequence for W(2)).

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A132366 Partial sum of centered tetrahedral numbers A005894.

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A195522 T(n,k) = Number of lower triangles of an n X n -k..k array with all row and column sums zero.

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

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A063493 a(n) = (2n-1)(13n^2-13n+6)/6.

A063495 a(n) = (2n-1)(5n^2-5n+2)/2.

A049480 a(n) = (2n-1)(n^2 -n +6)/6.