A260810 -id:A260810 - cp's OEIS frontend

A000583 Fourth powers: a(n) = n^4.

0, 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736, 28561, 38416, 50625, 65536, 83521, 104976, 130321, 160000, 194481, 234256, 279841, 331776, 390625, 456976, 531441, 614656, 707281, 810000, 923521, 1048576, 1185921

Offset: 0

N. J. A. Sloane

Figurate numbers based on 4-dimensional regular convex polytope called the 4-measure polytope, 4-hypercube or tesseract with Schlaefli symbol {4,3,3}. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004
Totally multiplicative sequence with a(p) = p^4 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A058649. The inverse binomial transforms yields the (finite) 0, 1, 14, 36, 24, the 4th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Generate Pythagorean triangles with parameters a and b to get sides of lengths x = b^2-a^2, y = 2*a*b, and z = a^2 + b^2. In particular use a=n-1 and b=n for a triangle with sides (x1,y1,z1) and a=n and b=n+1 for another triangle with sides (x2,y2,z2). Then x1*x2 + y1*y2 + z1*z2 = 8*a(n). - J. M. Bergot, Jul 22 2013
For n > 0, a(n) is the largest integer k such that k^4 + n is a multiple of k + n. Also, for n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n^2. - Derek Orr, Sep 04 2014
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
a(n+2)/2 is the area of a trapezoid with vertices at (T(n), T(n+1)), (T(n+1), T(n)), (T(n+1), T(n+2)), and (T(n+2), T(n+1)) with T(n)=A000292(n) for n >= 0. - J. M. Bergot, Feb 16 2018

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
Dov Juzuk, Curiosa 56: An interesting observation, Scripta Mathematica 6 (1939), 218.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Page 47.

T. D. Noe, Table of n, a(n) for n = 0..1000
Henry Bottomley, Illustration of initial terms
Henry Bottomley, Some Smarandache-type multiplicative sequences
Ralph Greenberg, Math for Poets.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Sameen Ahmed Khan, Sums of the powers of reciprocals of polygonal numbers, Int'l J. of Appl. Math. (2020) Vol. 33, No. 2, 265-282.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., Vol. 131, No. 1 (2002), pp. 65-75.
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
Kenneth A. Ross, First Digits of Squares and Cubes, Math. Mag. 85 (2012) 36-42.
Eric Weisstein's World of Mathematics, Biquadratic Number.
Index entries for "core" sequences
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Index entries for sequences related to Benford's law

Cf. A000538, A005917 (first differences), A000332, A014820, A092181, A092182, A092183.
Cf. A004831, A002646.
Cf. A002593, A260810. - Bruno Berselli, Jul 31 2015
Cf. A002415, A000290, A006008, A132366, A039623, A139584, A071270, A047928, A187756.
Cf. A062392, A231303, A267315.

a000583 = (^ 4)
a000583_list = scanl (+) 0 a005917_list
-- Reinhard Zumkeller, Nov 13 2014, Nov 11 2012

[n^4 : n in [0..50]]; // Wesley Ivan Hurt, Sep 05 2014

A000583 := n->n^4: seq(A000583(n), n=0..50);
A000583:=-(z+1)*(z**2+10*z+1)/(z-1)**5; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
with (combinat):seq(fibonacci(3, n^2)-1, n=0..33); # Zerinvary Lajos, May 25 2008

Range[0,100]^4 (* Vladimir Joseph Stephan Orlovsky, Mar 14 2011 *)

makelist(n^4,n,0,30); /* Martin Ettl, Nov 12 2012 */

A000583(n) = n^4 \\ Michael B. Porter, Nov 09 2009

def a(n): return n**4
print([a(n) for n in range(34)]) # Michael S. Branicky, Nov 10 2022

a(n) = A123865(n)+1 = A002523(n)-1.
Multiplicative with a(p^e) = p^(4e). - David W. Wilson, Aug 01 2001
G.f.: x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5. More generally, g.f. for n^m is Euler(m, x)/(1-x)^(m+1), where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292).
Dirichlet generating function: zeta(s-4). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: (x + 7*x^2 + 6*x^3 + x^4)*e^x. More generally, the general form for the e.g.f. for n^m is phi_m(x)*e^x, where phi_m is the exponential polynomial of order n. - Franklin T. Adams-Watters, Sep 11 2005
Sum_{k>0} 1/a(k) = Pi^4/90 = A013662. - Jaume Oliver Lafont, Sep 20 2009
a(n) = C(n+3,4) + 11*C(n+2,4) + 11*C(n+1,4) + C(n,4). [Worpitzky's identity for powers of 4. See, e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n*A177342(n) - Sum_{i=1..n-1} A177342(i) - (n - 1), with n > 1. - Bruno Berselli, May 07 2010
a(n) + a(n+1) + 1 = 2*A002061(n+1)^2. - Charlie Marion, Jun 13 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 24. - Ant King, Sep 23 2013
From Amiram Eldar, Jan 20 2021: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/720 (A267315).
Product_{n>=2} (1 - 1/a(n)) = sinh(Pi)/(4*Pi). (End)

A002593 a(n) = n^2(2n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.

Original entry on oeis.org

0, 1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600, 388521, 468028, 559153, 662976, 780625, 913276, 1062153, 1228528, 1413721, 1619100, 1846081

Offset: 0

N. J. A. Sloane

The m-th term, for m = A065549(n), is perfect (A000396). - Lekraj Beedassy, Jun 04 2002
Partial sums of A016755. - Lekraj Beedassy, Jan 06 2004
Also, the k-th triangular number, where k = 2n^2 - 1 = A056220(n), i.e., a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004
Also, the j-th hexagonal number, where j = n^2 = A000290(n), i.e., a(n) = A000384(A000290(n)) and a(n) = A056220(n) * A000290(n) or j * k. This sequence is a subsequence of the hexagonal number sequence and retains the aspect intrinsic to the hexagonal number sequence that each number in this sequence can be found by multiplying its triangular number by its hexagonal number. - Bruce J. Nicholson, Aug 22 2017
Odd numbers and their squares both having the form 2x-+1, we may write (2r+1)^3 = (2r+1)*(2s-1), where s = centered squares = (r+1)^2 + r^2. Since 2r+1 = (r+1)^2 - r^2, it follows immediately from summing telescopingly over n-1, the product 2*{(r+1)^4 - r^4} - {(r+1)^2 - r^2}, that Sum_{r=0..n-1} (2r+1)^3 = 2*n^4 - n^2 = n^2*(2n^2 - 1). - Lekraj Beedassy, Jun 16 2004
a(n) is also the starting term in the sum of a number M(n) of consecutive cubed integers equaling a squared integer (A253724) for M(n) equal to twice a squared integer (A001105). Numbers a(n) such that a^3 + (a+1)^3 + ... + (a+M-1)^3 = c^2 has nontrivial solutions over the integers for M equal to twice a squared integer (A001105). If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting from a^3 and equaling a squared integer c^2. For n >= 1, M(n) = 2n^2 (A001105), a(n) = M(M-1)/2 = n^2(2n^2 - 1), and c(n) = sqrt(M/2) (M(M^2-1)/2) = n^3(4n^4 - 1). The trivial solutions with M < 1 and a < 2 are not considered. - Vladimir Pletser, Jan 10 2015
Binomial transform of the sequence with offset 1 is (1, 27, 98, 120, 48, 0, 0, 0, ...). - Gary W. Adamson, Jul 23 2015

Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 169, #31.
F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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
F. E. Croxton and D. J. Cowden, Applied General Statistics, 2nd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1955. [Annotated scans of just pages 742-743]
Neslihan Kilar, Abdelmejid Bayad, and Yilmaz Simsek, Finite sums involving trigonometric functions and special polynomials: analysis of generating functions and p-adic integrals, Appl. Anal. Disc. Math., hal-04535748, 2024. See p. 22.
Vladimir Pletser, File Triplets (M,a,c) for M=2n^2
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
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.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
G. Xiao, Sigma Server, Operate on "(2*n-1)^3".
M. J. Zerger, Proof without words: The sum of consecutive odd cubes is a triangular number, Math. Mag., 68 (1995), 371.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000290, A000384, A000447, A000583, A002309, A253724, A253725, A260810.

[n^2*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Sep 07 2011

A002593:=-z*(z+1)*(z**2+22*z+1)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation
a:= n-> n^2*(2*n^2-1): seq(a(n), n=0..50);  # Vladimir Pletser, Jan 10 2015

CoefficientList[Series[(-x^4-23x^3-23x^2-x)/(x-1)^5,{x,0, 80}],x]  (* or *)
Table[ n^2 (2n^2-1),{n,0,80}]  (* Harvey P. Dale, Mar 28 2011 *)
Join[{0},Accumulate[Range[1,91,2]^3]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,28,153,496},40] (* Harvey P. Dale, Mar 22 2017 *)

a(n) = n^2*(2*n^2 - 1) \\ Charles R Greathouse IV, Feb 07 2017

a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004
G.f.: (-x^4 - 23*x^3 - 23*x^2 - x)/(x - 1)^5. - Harvey P. Dale, Mar 28 2011
a(n) = n^2*(2n^2 - 1). - Vladimir Pletser, Jan 10 2015
E.g.f.: exp(x)*x*(1 + 13*x + 24*x^2/2! + 12*x^3/3!). - Wolfdieter Lang, Mar 11 2017
a(n) = A000384(A000290(n)) = A056220(n) * A000290(n). - Bruce J. Nicholson, Aug 22 2017
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 1 - Pi^2/6 - cot(Pi/sqrt(2))*Pi/sqrt(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = cosec(Pi/sqrt(2))*Pi/sqrt(2) - Pi^2/12 - 1. (End)

A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.

Original entry on oeis.org

0, 1, 12, 51, 145, 330, 651, 1162, 1926, 3015, 4510, 6501, 9087, 12376, 16485, 21540, 27676, 35037, 43776, 54055, 66045, 79926, 95887, 114126, 134850, 158275, 184626, 214137, 247051, 283620, 324105, 368776, 417912, 471801, 530740, 595035, 665001, 740962

Offset: 0

Bruno Berselli, Jan 31 2014

After 0, first trisection of A011779 and right border of A177708.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Partial sums of A004188.
Cf. A000217, A000332, A011779, A177708.
Cf. similar sequences on the polygonal numbers: A002817(n) = A000217(A000217(n)); A000537(n) = A000290(A000217(n)); A037270(n) = A000217(A000290(n)); A062392(n) = A000384(A000217(n)).
Cf. sequences of the form A000217(m)+k*A000332(m+2): A062392 (k=12); A264854 (k=11); A264853 (k=10); this sequence (k=9); A006324 (k=8); A006323 (k=7); A000537 (k=6); A006322 (k=5); A006325 (k=4), A002817 (k=3), A006007 (k=2), A006522 (k=1).
Cf. A232713, A260810.

[n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];

Table[n (n + 1) (3 n^2 + 3 n - 2)/8, {n, 0, 40}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,12,51,145},40] (* Harvey P. Dale, Aug 22 2016 *)

for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));

G.f.: x*(1 + 7*x + x^2)/(1 - x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A000326(A000217(n)).
a(n) = A000217(n) + 9*A000332(n+2).
Sum_{n>=1} 1/a(n) = 2 + 4*sqrt(3/11)*Pi*tan(sqrt(11/3)*Pi/2) = 1.11700627139319... . - Vaclav Kotesovec, Apr 27 2016

A232713 Doubly pentagonal numbers: a(n) = n(3n-2)(3n-1)(3n+1)/8.

Original entry on oeis.org

0, 1, 35, 210, 715, 1820, 3876, 7315, 12650, 20475, 31465, 46376, 66045, 91390, 123410, 163185, 211876, 270725, 341055, 424270, 521855, 635376, 766480, 916895, 1088430, 1282975, 1502501, 1749060, 2024785, 2331890, 2672670, 3049501, 3464840, 3921225, 4421275

Offset: 0

Bruno Berselli, Nov 28 2013

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000326, A000332.
Cf. similar sequences: A000583 for A000290(A000290(n)); A002817 for A000217(A000217(n)); A063249 for A000384(A000384(n)).
Cf. A236770, A260810.

[n*(3*n-2)*(3*n-1)*(3*n+1)/8: n in [0..40]];

Table[n (3 n - 2) (3 n - 1) (3 n + 1)/8, {n, 0, 40}]

a(n)=n*(3*n-2)*(3*n-1)*(3*n+1)/8 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(1 + 30*x + 45*x^2 + 5*x^3) / (1 - x)^5.
a(n) = A000326(A000326(n)) = A000332(3n+1).
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 4 + 2*Pi/sqrt(3) - 6*log(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 32*log(2)/3 - 4*Pi/(3*sqrt(3)) - 4. (End)

A193218 Number of vertices in truncated tetrahedron with faces that are centered polygons.

Original entry on oeis.org

1, 21, 95, 259, 549, 1001, 1651, 2535, 3689, 5149, 6951, 9131, 11725, 14769, 18299, 22351, 26961, 32165, 37999, 44499, 51701, 59641, 68355, 77879, 88249, 99501, 111671, 124795, 138909, 154049, 170251, 187551, 205985, 225589, 246399, 268451, 291781, 316425

Offset: 1

Craig Ferguson, Jul 18 2011

The sequence starts with a central vertex and expands outward with (n-1) centered polygonal pyramids producing a truncated tetrahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon in each face. For centered triangles see A005448 and centered hexagons A003215.
This sequence is the 18th in the series (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 and t = 36. While adjusting for offsets, the beginning sequence A049480 is generated by adding the square pyramidal numbers A000330 to the odd numbers A005408 and each subsequent sequence is found by adding another set of square pyramidals A000330. (T/2) * A000330(n) + A005408(n). At 30 * A000330 + A005408 = centered dodecahedral numbers, 36 * A000330 + A005408 = A193228 truncated octahedron and 90 * A000330 + A005408 = A193248 = truncated icosahedron and dodecahedron. All five of the "Centered Platonic Solids" numbers sequences are in this series of sequences. Also 4 out of five of the "truncated" platonic solid number sequences are in this series. - Bruce J. Nicholson, Jul 06 2018
It would be good to have a detailed description of how the sequence is constructed. Maybe in the Examples section? - N. J. A. Sloane, Sep 07 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
OEIS, (Centered_polygons) pyramidal numbers
Wikipedia, Tetrahedral number
Wikipedia, Triangular number
Wikipedia, Centered polygonal number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A260810 (partial sums).

[6*n^3-9*n^2+5*n-1: n in [1..40]]; // Vincenzo Librandi, Aug 30 2011

Table[6 n^3 - 9 n^2 + 5 n - 1, {n, 35}] (* Alonso del Arte, Jul 18 2011 *)
CoefficientList[Series[(1+x)*(x^2+16*x+1)/(1-x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 04 2018 *)

a(n) = 6*n^3 - 9*n^2 + 5*n - 1.
G.f.: x*(1+x)*(x^2+16*x+1) / (1-x)^4. - R. J. Mathar, Aug 26 2011
a(n) = 18 * A000330(n-1) + A005408(n-1) = A063496(n) + A006331(n-1). - Bruce J. Nicholson, Jul 06 2018

cp's OEIS Frontend

A000583 Fourth powers: a(n) = n^4.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

Maxima

PARI

Python

Formula

A002593 a(n) = n^2*(2*n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A236770 a(n) = n*(n + 1)*(3*n^2 + 3*n - 2)/8.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A232713 Doubly pentagonal numbers: a(n) = n*(3*n-2)*(3*n-1)*(3*n+1)/8.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A193218 Number of vertices in truncated tetrahedron with faces that are centered polygons.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A002593 a(n) = n^2(2n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.

A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.

A232713 Doubly pentagonal numbers: a(n) = n(3n-2)(3n-1)(3n+1)/8.