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.

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A259108 a(n) = 2 * A000538(n).

Original entry on oeis.org

0, 2, 34, 196, 708, 1958, 4550, 9352, 17544, 30666, 50666, 79948, 121420, 178542, 255374, 356624, 487696, 654738, 864690, 1125332, 1445332, 1834294, 2302806, 2862488, 3526040, 4307290, 5221242, 6284124, 7513436, 8927998, 10547998, 12395040, 14492192, 16864034, 19536706
Offset: 0

Views

Author

N. J. A. Sloane, Jun 24 2015

Keywords

Crossrefs

Cf. A000538.

Programs

  • Mathematica
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,2,34,196,708,1958},40] (* Harvey P. Dale, Aug 16 2018 *)
  • PARI
    concat(0, Vec(2*x*(x+1)*(x^2+10*x+1)/(x-1)^6 + O(x^100))) \\ Colin Barker, Jun 28 2015
    
  • Python
    def A259108(n): return n*(n**2*(n*(3*(2*n+5))+10)-1)//15 # Chai Wah Wu, Oct 03 2024

Formula

a(n) = -n/15+(2*n^3)/3+n^4+(2*n^5)/5. - Colin Barker, Jun 28 2015
G.f.: 2*x*(x+1)*(x^2+10*x+1) / (x-1)^6. - Colin Barker, Jun 28 2015
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Wesley Ivan Hurt, Oct 01 2021

A259317 a(n) = 2*(2*n+1)*A000538(n) - 4*A000330(n)^2.

Original entry on oeis.org

0, 2, 70, 588, 2772, 9438, 26026, 61880, 131784, 257754, 471086, 814660, 1345500, 2137590, 3284946, 4904944, 7141904, 10170930, 14202006, 19484348, 26311012, 35023758, 46018170, 59749032, 76735960, 97569290, 122916222, 153527220, 190242668, 233999782
Offset: 0

Views

Author

N. J. A. Sloane, Jun 24 2015

Keywords

Examples

			n=3: 588 = 2*7*92-4*14^2.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,2,70,588,2772,9438,26026},30] (* Harvey P. Dale, Jul 12 2025 *)
  • PARI
    concat(0, Vec(-2*x*(x^4+28*x^3+70*x^2+28*x+1)/(x-1)^7 + O(x^100))) \\ Colin Barker, Jun 28 2015
    
  • Python
    def A259317(n): return n*(n*(n**2*(n*(16*n + 48) + 40) - 11) - 3)//45 # Chai Wah Wu, Dec 07 2021

Formula

Also a(n) = (2*n+1)*A259108(n) - A006331(n)^2.
a(n) = (n*(1+2*n)^2*(-3+n+8*n^2+4*n^3))/45. - Colin Barker, Jun 28 2015
G.f.: -2*x*(x^4+28*x^3+70*x^2+28*x+1) / (x-1)^7. - Colin Barker, Jun 28 2015

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

Original entry on oeis.org

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

Views

Author

Keywords

Comments

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

References

  • 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.

Crossrefs

Programs

Formula

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)

A000537 Sum of first n cubes; or n-th triangular number squared.

Original entry on oeis.org

0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, 4356, 6084, 8281, 11025, 14400, 18496, 23409, 29241, 36100, 44100, 53361, 64009, 76176, 90000, 105625, 123201, 142884, 164836, 189225, 216225, 246016, 278784, 314721, 354025, 396900, 443556, 494209, 549081
Offset: 0

Views

Author

Keywords

Comments

Number of parallelograms in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Or, number of orthogonal rectangles in an n X n checkerboard, or rectangles in an n X n array of squares. - Jud McCranie, Feb 28 2003. Compare A085582.
Also number of 2-dimensional cage assemblies (cf. A059827, A059860).
The n-th triangular number T(n) = Sum_{r=1..n} r = n(n+1)/2 satisfies the relations: (i) T(n) + T(n-1) = n^2 and (ii) T(n) - T(n-1) = n by definition, so that n^2*n = n^3 = {T(n)}^2 - {T(n-1)}^2 and by summing on n we have Sum_{ r = 1..n } r^3 = {T(n)}^2 = (1+2+3+...+n)^2 = (n*(n+1)/2)^2. - Lekraj Beedassy, May 14 2004
Number of 4-tuples of integers from {0,1,...,n}, without repetition, whose last component is strictly bigger than the others. Number of 4-tuples of integers from {1,...,n}, with repetition, whose last component is greater than or equal to the others.
Number of ordered pairs of two-element subsets of {0,1,...,n} without repetition.
Number of ordered pairs of 2-element multisubsets of {1,...,n} with repetition.
1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
a(n) is the number of parameters needed in general to know the Riemannian metric g of an n-dimensional Riemannian manifold (M,g), by knowing all its second derivatives; even though to know the curvature tensor R requires (due to symmetries) (n^2)*(n^2-1)/12 parameters, a smaller number (and a 4-dimensional pyramidal number). - Jonathan Vos Post, May 05 2006
Also number of hexagons with vertices in an hexagonal grid with n points in each side. - Ignacio Larrosa Cañestro, Oct 15 2006
Number of permutations of n distinct letters (ABCD...) each of which appears twice with 4 and n-4 fixed points. - Zerinvary Lajos, Nov 09 2006
With offset 1 = binomial transform of [1, 8, 19, 18, 6, ...]. - Gary W. Adamson, Dec 03 2008
The sequence is related to A000330 by a(n) = n*A000330(n) - Sum_{i=0..n-1} A000330(i): this is the case d=1 in the identity n*(n*(d*n-d+2)/2) - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)*(2*d*n-2*d+3)/6. - Bruno Berselli, Apr 26 2010, Mar 01 2012
From Wolfdieter Lang, Jan 11 2013: (Start)
For sums of powers of positive integers S(k,n) := Sum_{j=1..n}j^k one has the recurrence S(k,n) = (n+1)*S(k-1,n) - Sum_{l=1..n} S(k-1,l), n >= 1, k >= 1.
This was used for k=4 by Ibn al-Haytham in an attempt to compute the volume of the interior of a paraboloid. See the Strick reference where the trick he used is shown, and the W. Lang link.
This trick generalizes immediately to arbitrary powers k. For k=3: a(n) = (n+1)*A000330(n) - Sum_{l=1..n} A000330(l), which coincides with the formula given in the previous comment by Berselli. (End)
Regarding to the previous contribution, see also Matem@ticamente in Links field and comments on this recurrences in similar sequences (partial sums of n-th powers). - Bruno Berselli, Jun 24 2013
A rectangular prism with sides A000217(n), A000217(n+1), and A000217(n+2) has surface area 6*a(n+1). - J. M. Bergot, Aug 07 2013, edited with corrected indices by Antti Karttunen, Aug 09 2013
A formula for the r-th successive summation of k^3, for k = 1 to n, is (6*n^2+r*(6*n+r-1)*(n+r)!)/((r+3)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
Note that this sequence and its formula were known to (and possibly discovered by) Nicomachus, predating Ibn al-Haytham by 800 years. - Charles R Greathouse IV, Apr 23 2014
a(n) is the number of ways to paint the sides of a nonsquare rectangle using at most n colors. Cf. A039623. - Geoffrey Critzer, Jun 18 2014
For n > 0: A256188(a(n)) = A000217(n) and A256188(m) != A000217(n) for m < a(n), i.e., positions of first occurrences of triangular numbers in A256188. - Reinhard Zumkeller, Mar 26 2015
There is no cube in this sequence except 0 and 1. - Altug Alkan, Jul 02 2016
Also the number of chordless cycles in the complete bipartite graph K_{n+1,n+1}. - Eric W. Weisstein, Jan 02 2018
a(n) is the sum of the elements in the multiplication table [0..n] X [0..n]. - Michel Marcus, May 06 2021

Examples

			G.f. = x + 9*x^2 + 36*x^3 + 100*x^4 + 225*x^5 + 441*x^6 + ... - _Michael Somos_, Aug 29 2022
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 62, eq. (6.3) for k=3.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 110ff.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, pp. 36, 58.
  • Clifford Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Oxford University Press, 2001, p. 325.
  • 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).
  • H. K. Strick, Geschichten aus der Mathematik II, Spektrum Spezial 3/11, p. 13.
  • D. Wells, You Are A Mathematician, "Counting rectangles in a rectangle", Problem 8H, pp. 240; 254, Penguin Books 1995.

Crossrefs

Convolution of A000217 and A008458.
Row sums of triangles A094414 and A094415.
Second column of triangle A008459.
Row 3 of array A103438.
Cf. A236770 (see crossrefs).

Programs

  • GAP
    List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018
    
  • Haskell
    a000537 = a000290 . a000217  -- Reinhard Zumkeller, Mar 26 2015
    
  • Magma
    [(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
    
  • Maple
    a:= n-> (n*(n+1)/2)^2:
    seq(a(n), n=0..40);
  • Mathematica
    Accumulate[Range[0, 50]^3] (* Harvey P. Dale, Mar 01 2011 *)
    f[n_] := n^2 (n + 1)^2/4; Array[f, 39, 0] (* Robert G. Wilson v, Nov 16 2012 *)
    Table[CycleIndex[{{1, 2, 3, 4}, {3, 2, 1, 4}, {1, 4, 3, 2}, {3, 4, 1, 2}}, s] /. Table[s[i] -> n, {i, 1, 2}], {n, 0, 30}] (* Geoffrey Critzer, Jun 18 2014 *)
    Accumulate @ Range[0, 50]^2 (* Waldemar Puszkarz, Jan 24 2015 *)
    Binomial[Range[20], 2]^2 (* Eric W. Weisstein, Jan 02 2018 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 9, 36, 100}, 20] (* Eric W. Weisstein, Jan 02 2018 *)
    CoefficientList[Series[-((x (1 + 4 x + x^2))/(-1 + x)^5), {x, 0, 20}], x] (* Eric W. Weisstein, Jan 02 2018 *)
  • PARI
    a(n)=(n*(n+1)/2)^2
    
  • Python
    def A000537(n): return (n*(n+1)>>1)**2 # Chai Wah Wu, Oct 20 2023

Formula

a(n) = (n*(n+1)/2)^2 = A000217(n)^2 = Sum_{k=1..n} A000578(k), that is, 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
G.f.: (x+4*x^2+x^3)/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = Sum ( Sum ( 1 + Sum (6*n) ) ), rephrasing the formula in A000578. - Xavier Acloque, Jan 21 2003
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j, row sums of A127777. - Alexander Adamchuk, Oct 24 2004
a(n) = A035287(n)/4. - Zerinvary Lajos, May 09 2007
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=1. - Alexander R. Povolotsky, May 17 2008
G.f.: x*F(3,3;1;x). - Paul Barry, Sep 18 2008
Sum_{k > 0} 1/a(k) = (4/3)*(Pi^2-9). - Jaume Oliver Lafont, Sep 20 2009
a(n) = Sum_{1 <= k <= m <= n} A176271(m,k). - Reinhard Zumkeller, Apr 13 2010
a(n) = Sum_{i=1..n} J_3(i)*floor(n/i), where J_ 3 is A059376. - Enrique Pérez Herrero, Feb 26 2012
a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} min(i,j,k). - Enrique Pérez Herrero, Feb 26 2013 [corrected by Ridouane Oudra, Mar 05 2025]
a(n) = 6*C(n+2,4) + C(n+1,2) = 6*A000332(n+2) + A000217(n), (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..3} j*Stirling1(n+1,n+1-j)*Stirling2(n+3-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*(3-4*log(2)). - Vaclav Kotesovec, Feb 13 2015
a(n)*((s-2)*(s-3)/2) = P(3, P(s, n+1)) - P(s, P(3, n+1)), where P(s, m) = ((s-2)*m^2-(s-4)*m)/2 is the m-th s-gonal number. For s=7, 10*a(n) = A000217(A000566(n+1)) - A000566(A000217(n+1)). - Bruno Berselli, Aug 04 2015
From Ilya Gutkovskiy, Jul 03 2016: (Start)
E.g.f.: x*(4 + 14*x + 8*x^2 + x^3)*exp(x)/4.
Dirichlet g.f.: (zeta(s-4) + 2*zeta(s-3) + zeta(s-2))/4. (End)
a(n) = (Bernoulli(4, n+1) - Bernoulli(4, 1))/4, n >= 0, with the Bernoulli polynomial B(4, x) from row n=4 of A053382/A053383. See, e.g., the Ash-Gross reference, p. 62, eq. (6.3) for k=3. - Wolfdieter Lang, Mar 12 2017
a(n) = A000217((n+1)^2) - A000217(n+1)^2. - Bruno Berselli, Aug 31 2017
a(n) = n*binomial(n+2, 3) + binomial(n+2, 4) + binomial(n+1, 4). - Tony Foster III, Nov 14 2017
Another identity: ..., a(3) = (1/2)*(1*(2+4+6)+3*(4+6)+5*6) = 36, a(4) = (1/2)*(1*(2+4+6+8)+3*(4+6+8)+5*(6+8)+7*(8)) = 100, a(5) = (1/2)*(1*(2+4+6+8+10)+3*(4+6+8+10)+5*(6+8+10)+7*(8+10)+9*(10)) = 225, ... - J. M. Bergot, Aug 27 2022
Comment from Michael Somos, Aug 28 2022: (Start)
The previous comment expresses a(n) as the sum of all of the n X n multiplication table array entries.
For example, for n = 4:
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
This array sum can be split up as follows:
+---+---------------+
| 0 | 1 2 3 4 | (0+1)*(1+2+3+4)
| +---+-----------+
| 0 | 2 | 4 6 8 | (1+2)*(2+3+4)
| | +---+-------+
| 0 | 3 | 6 | 9 12 | (2+3)*(3+4)
| | | +---+---+
| 0 | 4 | 8 |12 |16 | (3+4)*(4)
+---+---+---+---+---+
This kind of row+column sums was used by Ramanujan and others for summing Lambert series. (End)
a(n) = 6*A000332(n+4) - 12*A000292(n+1) + 7*A000217(n+1) - n - 1. - Adam Mohamed, Sep 05 2024

Extensions

Edited by M. F. Hasler, May 02 2015

A007775 Numbers not divisible by 2, 3 or 5.

Original entry on oeis.org

1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 91, 97, 101, 103, 107, 109, 113, 119, 121, 127, 131, 133, 137, 139, 143, 149, 151, 157, 161, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 203, 209
Offset: 1

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Author

Keywords

Comments

Also numbers n such that the sum of the 4th powers of the first n positive integers is divisible by n, or A000538(n) = n*(n+1)(2*n+1)(3*n^2+3*n-1)/30 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Also the 7-rough numbers: positive integers that have no prime factors less than 7. - Michael B. Porter, Oct 09 2009
a(n) mod 3 has period 8, repeating [1,1,2,1,2,1,2,2] = (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. floor(a(n)/3) is the set of numbers k such that k is congruent to {0,2,3,4,5,6,7,9} mod 10 = floor((5*n-2)/4)-floor((n mod 8)/6). - Gary Detlefs, Jan 08 2012
Numbers k such that C(k+3,3)==1 (mod k) and C(k+5,5)==1 (mod k). - Gary Detlefs, Sep 15 2013
a(n) mod 30 has period 8 repeating [1, 7, 11, 13, 17, 19, 23, 29]. The mean of these 8 numbers is 120/8 = 15. (a(n)-15) mod 30 has period 8 repeating [-14, -8, -4, -2, 2, 4, 8, 14]. One half of the absolute value produces the symmetric sequence [7, 4, 2, 1, 1, 2, 4, 7] = A061501(((n-1) + 16) mod 8). - Gary Detlefs, Sep 24 2013
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)*(n + 3*m)(n + 4*m)/120 is integral. Cf. A007310. - Peter Bala, Nov 13 2015
The asymptotic density of this sequence is 4/15. - Amiram Eldar, Sep 30 2020
If a(n) + a(n+1) = 0 (mod 30), then a(n-j) + a(n+j+1) = a(n) + a(n+1) for each j in [1, n-1]. - Alexandre Herrera, Jun 27 2023

Crossrefs

Cf. A000538, A054403, A145011 (first differences).
For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063.
Complement is A080671.
For digital root of Fibonacci numbers indexed by this sequence, see A227896.

Programs

  • Haskell
    a007775 n = a007775_list !! (n-1)
    a007775_list = 1 : filter ((> 5) . a020639) [1..]
    -- Reinhard Zumkeller, Jan 06 2013
    
  • Magma
    I:=[1, 7, 11, 13, 17, 19, 23, 29, 31]; [n le 9 select I[n] else Self(n-1) +Self(n-8) - Self(n-9): n in [1..80]]; // G. C. Greubel, Oct 22 2018
    
  • Maple
    for i from 1 to 500 do if gcd(i,30) = 1 then print(i); fi; od;
    for k from 1 to 300 do if ((k^2 mod 48=1) or (k^2 mod 48=25)) and ((k^2 mod 120=1) or (k^2 mod 120=49)) then print(k) fi od. # Gary Detlefs, Dec 30 2011
  • Mathematica
    Select[ Range[ 300 ], GCD[ #1, 30 ]==1& ]
    Select[Range[250], Mod[#, 2]>0&&Mod[#, 3]>0&&Mod[#, 5]>0&] (* Vincenzo Librandi, Feb 08 2014 *)
    a[ n_] := Quotient[ n, 8, 1] 30 + {1, 7, 11, 13, 17, 19, 23, 29}[[Mod[n, 8, 1]]]; (* Michael Somos, Jun 02 2014 *)
    LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 7, 11, 13, 17, 19, 23, 29, 31}, 100] (* Mikk Heidemaa, Dec 07 2017 *)
    Cases[Range@1000, x_ /; NoneTrue[Array[Prime, 3], Divisible[x, #] &]] (* Mikk Heidemaa, Dec 07 2017 *)
    CoefficientList[ Series[(x^8 + 6x^7 + 4x^6 + 2x^5 + 4x^4 + 2x^3 + 4x^2 + 6x + 1)/((x - 1)^2 (x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)), {x, 0, 55}], x] (* Robert G. Wilson v, Dec 07 2017 *)
  • PARI
    isA007775(n) = gcd(n,30)==1 \\ Michael B. Porter, Oct 09 2009
    
  • PARI
    {a(n) = n\8 * 30 + [ -1, 1, 7, 11, 13, 17, 19, 23][n%8 + 1]} /* Michael Somos, Feb 05 2011 */
    
  • PARI
    {a(n) = n\8 * 6 + 9 + 3 * (n+1)\2 * 2 - max(5, (n-2)%8) * 2} /* Michael Somos, Jun 02 2014 */
    
  • PARI
    Vec(x*(1+6*x+4*x^2+2*x^3+4*x^4+2*x^5+4*x^6+6*x^7+x^8)/((1+x)*(x^2+1)*(x^4+1)*( x-1)^2) + O(x^100)) \\ Altug Alkan, Nov 16 2015
    
  • Python
    def A007775(n): return ((m:=n-1)<<2|1)-(m>>2&-2)+(2,0,-2,0)[m-1>>1&3] # Chai Wah Wu, Feb 02 2025
  • Sage
    a = lambda n: ((((n-1)<< 2)-((n-1)>>2))|1) + ((((n-1)<<1)-((n-1)>> 1)) & 2)
    print([a(n) for n in (1..56)]) # after Andrew Lelechenko, Peter Luschny, Jul 08 2017
    

Formula

A141256(a(n)) = n+1. - Reinhard Zumkeller, Jun 17 2008
From R. J. Mathar, Feb 27 2009: (Start)
a(n+8) = a(n) + 30.
a(n) = a(n-1) + a(n-8) - a(n-9).
G.f.: x*(1 + 6*x + 4*x^2 + 2*x^3 + 4*x^4 + 2*x^5 + 4*x^6 + 6*x^7 + x^8)/((1 + x)*(x^2 + 1)*(x^4 + 1)*(x-1)^2). (End)
a(n) = 4*n - 3 - 2*floor((n-1)/8) + (1 + (-1)^floor((n-2)/2))*(-1)^floor((n-2)/4), n >= 1. - Timothy Hopper, Mar 14 2010
a(1 - n) = -a(n). - Michael Somos, Feb 05 2011
Numbers k such that ((k^2 mod 48=1) or (k^2 mod 48=25)) and ((k^2 mod 120=1) or (k^2 mod 120=49)). - Gary Detlefs, Dec 30 2011
Numbers k such that k^2 mod 30 is 1 or 19. - Gary Detlefs, Dec 31 2011
a(n) = 3*(floor((5*n-2)/4) - floor((n mod 8)/6)) + (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. - Gary Detlefs, Jan 08 2012
a(n) = 4*n - 3 + 2*(floor((n+6)/8) - floor((n+4)/8) - floor((n+2)/8) + floor(n/8) - floor((n-1)/8)), n >= 1. From the o.g.f. given above by R. J. Mathar (with the denominator written as (1-x^8)*(1-x)), and a two-step reduction of the floor functions. Compare with Hopper's and Detlefs's formulas above. - Wolfdieter Lang, Jan 26 2012
a(n) = (6*f(n) - 3 + (-1)^f(n))/2, where f(n)= n + floor(n/4)+ floor(((n+4) mod 8)/6). - Gary Detlefs, Sep 15 2013
a(n) = 30*floor((n-1)/8) + 15 + 2*f((n-1) mod 8 + 16)*(-1)^floor(((n+3) mod 8)/4), where f(n) = (n*(n+1)/2+1) mod 10. - Gary Detlefs, Sep 24 2013
a(n) = 3*n + 6*floor(n/8) + (n mod 2) - 2*floor(((n-2) mod 8)/6) - 2*floor(((n-2) mod 8)/7) + 1. - Gary Detlefs, Jun 01 2014
a(n+1) = ((n << 2 - n >> 2) || 1) + ((n << 1 - n >> 1) && 2), where << and >> are bitwise left and right shifts, || and && are bitwise "or" and "and". - Andrew Lelechenko, Jul 08 2017
a(n) = 2*n + 2*floor(1/2 + (7*n)/8) + 2*(91 mod (2 - ((3*n)/4 + n^2/4) mod 2)) - 3 (n > 0). - Mikk Heidemaa, Dec 06 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(23 + sqrt(5) - sqrt(6*(5 + sqrt(5))))*Pi/15. - Amiram Eldar, Dec 13 2021

A000539 Sum of 5th powers: 0^5 + 1^5 + 2^5 + ... + n^5.

Original entry on oeis.org

0, 1, 33, 276, 1300, 4425, 12201, 29008, 61776, 120825, 220825, 381876, 630708, 1002001, 1539825, 2299200, 3347776, 4767633, 6657201, 9133300, 12333300, 16417401, 21571033, 28007376, 35970000, 45735625, 57617001, 71965908, 89176276, 109687425, 133987425, 162616576
Offset: 0

Views

Author

Keywords

Comments

This sequence is related to A000538 by a(n) = n*A000538(n) - Sum_{i=0..n-1} A000538(i). - Bruno Berselli, Apr 26 2010
See comment in A008292 for a formula for r-th successive summation of Sum_{k=1..n} k^j. - Gary Detlefs, Jan 02 2014

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
  • 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).

Crossrefs

Partial sums of A000584. Row 5 of array A103438.

Programs

  • Magma
    [n^2*(n+1)^2*(2*n^2+2*n-1)/12: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015
    
  • Maple
    A000539:=-(1+26*z+66*z**2+26*z**3+z**4)/(z-1)**7; # Simon Plouffe in his 1992 dissertation
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^5 od: seq(a[n], n=0..30); # Zerinvary Lajos, Feb 22 2008
    a:=n->sum(j^5,j=0..n): seq(a(n), n=0..30); # Zerinvary Lajos, Jun 05 2008
  • Mathematica
    Accumulate[Range[0, 40]^5]
    LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 1, 33, 276, 1300, 4425, 12201}, 41] (* Jean-François Alcover, Feb 09 2016 *)
  • Maxima
    A000539(n):=n^2*(n+1)^2*(2*n^2+2*n-1)/12$ makelist(A000539(n),n,0,30); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    a(n)=n^2*(n+1)^2*(2*n^2+2*n-1)/12 \\ Charles R Greathouse IV, Jul 15 2011
    
  • PARI
    concat(0, Vec(x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)^7 + O(x^100))) \\ Altug Alkan, Dec 07 2015
    
  • Python
    A000539_list, m = [0], [120, -240, 150, -30, 1, 0, 0]
    for _ in range(10**2):
        for i in range(6):
            m[i+1] += m[i]
        A000539_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Python
    def A000539(n): return n**2*(n**2*(n*(n+3<<1)+5)-1)//12 # Chai Wah Wu, Oct 03 2024

Formula

a(n) = n^2*(n+1)^2*(2*n^2+2*n-1)/12.
a(n) = sqrt(Sum_{j=1..n}Sum_{i=1..n}(i*j)^5). - Alexander Adamchuk, Oct 26 2004
a(n) = Sum_{i = 1..n} J_5(i)*floor(n/i), where J_5 is A059378. - Enrique Pérez Herrero, Feb 26 2012
a(n) = 6*a(n-1) - 15* a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + 120. - Ant King, Sep 23 2013
a(n) = 120*C(n+3,6) + 30*C(n+2,4) + C(n+1,2) (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..5} j*Stirling1(n+1,n+1-j)*Stirling2(n+5-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} 1/a(n) = 60 - 4*Pi^2 + 8*sqrt(3)*Pi * tan(sqrt(3)*Pi/2). - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)^2*n^2*(n + 1/2 + sqrt(3/4))*(n + 1/2 - sqrt(3/4))/6. See the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015
G.f.: x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)^7. - Robert Israel, Dec 07 2015
a(n) = (4/3)*A000217(n)^3 - (1/3)*A000217(n)^2. - Michael Raney, Feb 19 2016
a(n) = (binomial(n+1,4) + 6*binomial(n+2,4) + binomial(n+3,4))*(binomial(n+2,3) - binomial(n+1,3)). - Tony Foster III, Oct 21 2018
a(n) = 24*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, May 04 2024
E.g.f.: exp(x)*x*(12 + 186*x + 360*x^2 + 195*x^3 + 36*x^4 + 2*x^5)/12. - Stefano Spezia, May 04 2024

A103438 Square array T(m,n) read by antidiagonals: Sum_{k=1..n} k^m.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 5, 6, 4, 0, 1, 9, 14, 10, 5, 0, 1, 17, 36, 30, 15, 6, 0, 1, 33, 98, 100, 55, 21, 7, 0, 1, 65, 276, 354, 225, 91, 28, 8, 0, 1, 129, 794, 1300, 979, 441, 140, 36, 9, 0, 1, 257, 2316, 4890, 4425, 2275, 784, 204, 45, 10
Offset: 0

Views

Author

Ralf Stephan, Feb 11 2005

Keywords

Comments

For the o.g.f.s of the column sequences for this array, see A196837 and the link given there. - Wolfdieter Lang, Oct 15 2011
T(m,n)/n is the m-th moment of the discrete uniform distribution on {1,2,...,n}. - Geoffrey Critzer, Dec 31 2018
T(1,n) divides T(m,n) for odd m. - Franz Vrabec, Dec 23 2020

Examples

			Square array begins:
  0, 1,  2,   3,    4,     5,     6,      7,      8,      9, ... A001477;
  0, 1,  3,   6,   10,    15,    21,     28,     36,     45, ... A000217;
  0, 1,  5,  14,   30,    55,    91,    140,    204,    285, ... A000330;
  0, 1,  9,  36,  100,   225,   441,    784,   1296,   2025, ... A000537;
  0, 1, 17,  98,  354,   979,  2275,   4676,   8772,  15333, ... A000538;
  0, 1, 33, 276, 1300,  4425, 12201,  29008,  61776, 120825, ... A000539;
  0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, ... A000540;
Antidiagonal triangle begins as:
  0;
  0, 1;
  0, 1,  2;
  0, 1,  3,  3;
  0, 1,  5,  6,  4;
  0, 1,  9, 14, 10,  5;
  0, 1, 17, 36, 30, 15, 6;
		

References

  • J. Faulhaber, Academia Algebrae, Darinnen die miraculosische inventiones zu den höchsten Cossen weiters continuirt und profitirt werden, Augspurg, bey Johann Ulrich Schönigs, 1631.

Crossrefs

Diagonals include A076015 and A031971.
Antidiagonal sums are in A103439.
Antidiagonals are the rows of triangle A192001.

Programs

  • Magma
    T:= func< n,k | n eq 0 select k else (&+[j^n: j in [0..k]]) >;
    [T(n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 22 2021
    
  • Maple
    seq(print(seq(Zeta(0,-k,1)-Zeta(0,-k,n+1),n=0..9)),k=0..6);
    # (Produces the square array from the example.) Peter Luschny, Nov 16 2008
    # alternative
    A103438 := proc(m,n)
        (bernoulli(m+1,n+1)-bernoulli(m+1))/(m+1) ;
        if m = 0 then
            %-1 ;
        else
            % ;
        end if;
    end proc: # R. J. Mathar, May 10 2013
    # simpler:
    A103438 := proc(m,n)
        (bernoulli(m+1,n+1)-bernoulli(m+1,1))/(m+1) ;
    end proc: # Peter Luschny, Mar 20 2024
  • Mathematica
    T[m_, n_]:= HarmonicNumber[m, -n]; Flatten[Table[T[m-n, n], {m, 0, 11}, {n, m, 0, -1}]] (* Jean-François Alcover, May 11 2012 *)
  • PARI
    T(m,n)=sum(k=0,n,k^m)
    
  • Python
    from itertools import count, islice
    from math import comb
    from fractions import Fraction
    from sympy import bernoulli
    def A103438_T(m,n): return sum(k**m for k in range(1,n+1)) if n<=m else int(sum(comb(m+1,i)*(bernoulli(i) if i!=1 else Fraction(1,2))*n**(m-i+1) for i in range(m+1))/(m+1))
    def A103438_gen(): # generator of terms
        for m in count(0):
            for n in range(m+1):
                yield A103438_T(m-n,n)
    A103438_list = list(islice(A103438_gen(),100)) # Chai Wah Wu, Oct 23 2024
  • SageMath
    def T(n,k): return (bernoulli_polynomial(k+1, n+1) - bernoulli_polynomial(1, n+1)) /(n+1)
    flatten([[T(n-k,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Dec 22 2021
    

Formula

E.g.f.: e^x*(e^(x*y)-1)/(e^x-1).
T(m, n) = Zeta(-n, 1) - Zeta(-n, m + 1), for m >= 0 and n >= 0, where Zeta(z, v) is the Hurwitz zeta function. - Peter Luschny, Nov 16 2008
T(m, n) = HarmonicNumber(m, -n). - Jean-François Alcover, May 11 2012
T(m, n) = (Bernoulli(m + 1, n + 1) - Bernoulli(m + 1, 1)) / (m + 1). - Peter Luschny, Mar 20 2024
T(m, n) = Sum_{k=0...m-n} B(k)*(-1)^k*binomial(m-n,k)*n^(m-n-k+1)/(m-n-k+1), where B(k) = Bernoulli number A027641(k) / A027642(k). - Robert B Fowler, Aug 20 2024
T(m, n) = Sum_{i=1..n} J_m(i)*floor(n/i), where J_m is the m-th Jordan totient function. - Ridouane Oudra, Jul 19 2025

A005585 5-dimensional pyramidal numbers: a(n) = n*(n+1)*(n+2)*(n+3)*(2n+3)/5!.

Original entry on oeis.org

1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289, 5005, 7371, 10556, 14756, 20196, 27132, 35853, 46683, 59983, 76153, 95634, 118910, 146510, 179010, 217035, 261261, 312417, 371287, 438712, 515592, 602888, 701624, 812889, 937839, 1077699, 1233765, 1407406
Offset: 1

Views

Author

Keywords

Comments

Convolution of triangular numbers (A000217) and squares (A000290) (n>=1). - Graeme McRae, Jun 07 2006
p^k divides a(p^k-3), a(p^k-2), a(p^k-1) and a(p^k) for prime p > 5 and integer k > 0. p^k divides a((p^k-3)/2) for prime p > 5 and integer k > 0. - Alexander Adamchuk, May 08 2007
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-5) is the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
5-dimensional square numbers, fourth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+4, i+4)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Antidiagonal sums of the convolution array A213550. - Clark Kimberling, Jun 17 2012
Binomial transform of (1, 6, 14, 16, 9, 2, 0, 0, 0, ...). - Gary W. Adamson, Jul 28 2015
2*a(n) is number of ways to place 4 queens on an (n+3) X (n+3) chessboard so that they diagonally attack each other exactly 6 times. The maximal possible attack number, p=binomial(k,2)=6 for k=4 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015
While adjusting for offsets, add A000389 to find the next in series A000389, A005585, A051836, A034263, A027800, A051843, A051877, A051878, A051879, A051880, A056118, A271567. (See Bruno Berselli's comments in A271567.) - Bruce J. Nicholson, Jun 21 2018
Coefficients in the terminating series identity 1 - 7*n/(n + 6) + 27*n*(n - 1)/((n + 6)*(n + 7)) - 77*n*(n - 1)*(n - 2)/((n + 6)*(n + 7)*(n + 8)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A040977. - Peter Bala, Feb 18 2019

Examples

			G.f. = x + 7*x^2 + 27*x^3 + 77*x^4 + 182*x^5 + 378*x^6 + 714*x^7 + 1254*x^8 + ... - _Michael Somos_, Jun 24 2018
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 797.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = ((-1)^(n+1))*A053120(2*n+3, 5)/16, (1/16 of sixth unsigned column of Chebyshev T-triangle, zeros omitted).
Partial sums of A002415.
Cf. A006542, A040977, A047819, A111125 (third column).
Cf. a(n) = ((-1)^(n+1))*A084960(n+1, 2)/16 (compare with the first line). - Wolfdieter Lang, Aug 04 2014

Programs

  • Magma
    I:=[1, 7, 27, 77, 182, 378]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Jun 09 2013
    
  • Maple
    [seq(binomial(n+2,6)-binomial(n,6), n=4..45)]; # Zerinvary Lajos, Jul 21 2006
    A005585:=(1+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    With[{c=5!},Table[n(n+1)(n+2)(n+3)(2n+3)/c,{n,40}]] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{1,7,27,77,182,378},40] (* Harvey P. Dale, Oct 04 2011 *)
    CoefficientList[Series[(1 + x) / (1 - x)^6, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
  • PARI
    a(n)=binomial(n+3,4)*(2*n+3)/5 \\ Charles R Greathouse IV, Jul 28 2015

Formula

G.f.: x*(1+x)/(1-x)^6.
a(n) = 2*C(n+4, 5) - C(n+3, 4). - Paul Barry, Mar 04 2003
a(n) = C(n+3, 5) + C(n+4, 5). - Paul Barry, Mar 17 2003
a(n) = C(n+2, 6) - C(n, 6), n >= 4. - Zerinvary Lajos, Jul 21 2006
a(n) = Sum_{k=1..n} T(k)*T(k+1)/3, where T(n) = n(n+1)/2 is a triangular number. - Alexander Adamchuk, May 08 2007
a(n-1) = (1/4)*Sum_{1 <= x_1, x_2 <= n} |x_1*x_2*det V(x_1,x_2)| = (1/4)*Sum_{1 <= i,j <= n} i*j*|i-j|, where V(x_1,x_2) is the Vandermonde matrix of order 2. First differences of A040977. - Peter Bala, Sep 21 2007
a(n) = C(n+4,4) + 2*C(n+4,5). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), a(1)=1, a(2)=7, a(3)=27, a(4)=77, a(5)=182, a(6)=378. - Harvey P. Dale, Oct 04 2011
a(n) = (1/6)*Sum_{i=1..n+1} (i*Sum_{k=1..i} (i-1)*k). - Wesley Ivan Hurt, Nov 19 2014
E.g.f.: x*(2*x^4 + 35*x^3 + 180*x^2 + 300*x + 120)*exp(x)/120. - Robert Israel, Nov 19 2014
a(n) = A000389(n+3) + A000389(n+4). - Bruce J. Nicholson, Jun 21 2018
a(n) = -a(-3-n) for all n in Z. - Michael Somos, Jun 24 2018
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 40*(16*log(2) - 11)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 20*(8*Pi - 25)/3. (End)
a(n) = A004302(n+1) - A207361(n+1). - J. M. Bergot, May 20 2022
a(n) = Sum_{i=0..n+1} Sum_{j=i..n+1} i*j*(j-i)/2. - Darío Clavijo, Oct 11 2023
a(n) = (A000538(n+1) - A000330(n+1))/12. - Yasser Arath Chavez Reyes, Feb 21 2024

A254640 Third partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 67, 927, 6677, 32942, 126378, 404634, 1129854, 2833479, 6515509, 13947505, 28115451, 53846156, 98669156, 173975076, 296541132, 490504893, 789878583, 1241708083, 1909993393, 2880500634, 4266609710, 6216356510, 8920844010, 12624212835, 17635378761
Offset: 1

Views

Author

Luciano Ancora, Feb 04 2015

Keywords

Comments

This is one of a sequence of arrays that are the convolutions of the zero-padded sequences binomial(2n-1+k,k) with the Eulerian polynomials E(n,x) of A008292, represented by E(n,x) (1-x)^(-2n), which generate increasing partial sums of powers of integers:
n= 2) (1 + 4*x + x^2)/(1-x)^4 is the o.g.f. of A000578, the convolution of (1,4,1) with A000292, giving the powers of m^3.
n= 3) (1 + 11*x + 11*x^2 + x^3)/(1-x)^6 is the o.g.f. of A000538, convolution of (1,11,11,1) with A000389, giving the partial sums of m^4.
n= 4) (1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1-x)^8, the o.g.f. of A101092, convolution of (1,26,66,26,1) with A000580, the second partial sums of m^5
n= 5) (1 + 57*x + 302*x^2 + 302*x^3 + 57*x^4 + x^5)/(1-x)^10, the o.g.f. of A254460, convolution of (1,57,302,302,57,1) with A000582, giving the third partial sums of m^6. - Tom Copeland, Dec 07 2015

Crossrefs

Programs

  • GAP
    List([1..30], n-> Binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210); # G. C. Greubel, Aug 28 2019
    
  • Magma
    [n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+ 5*n^4)/5040: n in [1..30]]; // Vincenzo Librandi, Feb 05 2015
    
  • Maple
    seq(binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210, n=1..30); # G. C. Greubel, Aug 28 2019
  • Mathematica
    Table[n(1+n)(2+n)(3+n)(3+2n)(2 -30n +35n^2 +30n^3 +5n^4)/5040, {n, 30}] (* or *) CoefficientList[Series[(x+1)(x^4 +56x^3 +246x^2 +56x +1)/(x - 1)^10, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 05 2015 *)
  • PARI
    vector(30, n, n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+5*n^4)/5040) \\ Colin Barker, Feb 04 2015
    
  • Python
    def A254640(n): return n*(n*(n*(n*(n*(n*(n*(n*(10*n + 135) + 720) + 1890) + 2394) + 945) - 640) - 450) + 36)//5040 # Chai Wah Wu, Dec 07 2021
  • Sage
    [binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210 for n in (1..30)] # G. C. Greubel, Aug 28 2019
    

Formula

a(n) = n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2 -30*n +35*n^2 +30*n^3 +5*n^4)/5040.
G.f.: x*(1+x)*(1 +56*x +246*x^2 +56*x^3 +x^4)/(1-x)^10. - Colin Barker, Feb 04 2015

A062392 a(n) = n^4 - (n-1)^4 + (n-2)^4 - ... 0^4.

Original entry on oeis.org

0, 1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945, 28680, 36856, 46665, 58311, 72010, 87990, 106491, 127765, 152076, 179700, 210925, 246051, 285390, 329266, 378015, 431985, 491536, 557040, 628881, 707455, 793170, 886446, 987715
Offset: 0

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Author

Henry Bottomley, Jun 21 2001

Keywords

Comments

Number of edges in the join of two complete graphs of order n^2 and n, K_n^2 * K_n. - Roberto E. Martinez II, Jan 07 2002
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-5)(P(4,1)-(-1)^k P(4,2k+1))|. - Peter Luschny, Jul 12 2009
Define an infinite symmetric array by T(n,m) = n*(n-1) + m for 0 <= m <= n and T(n,m) = T(m,n), n >= 0. Then a(n) is the sum of terms in the top left (n+1) X (n+1) subarray: a(n) = Sum_{r=0..n} Sum_{c=0..n} T(r,c). - J. M. Bergot, Jul 05 2013
a(n) is the sum of all positive numbers less than A002378(n). - J. M. Bergot, Aug 30 2013
Except the first term, these are triangular numbers that remain triangular when divided by their index, e.g., 66 divided by 11 gives 6. - Waldemar Puszkarz, Sep 14 2017
a(n) is the semiperimeter of the unique primitive Pythagorean triple such that (a-b+c)/2 = T(n) = A000217(n). Its long leg and hypotenuse are consecutive natural numbers and the triple is (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

Examples

			From _Bruno Berselli_, Oct 30 2017: (Start)
After 0:
1   =                 -(1) + (2);
15  =             -(1 + 2) + (3 + 4 + 5 + 2*3);
66  =         -(1 + 2 + 3) + (4 + 5 + 6 + 7 + ... + 11 + 3*4);
190 =     -(1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + ... + 19 + 4*5);
435 = -(1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + ... + 29 + 5*6), etc. (End)
		

References

  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Crossrefs

Cf. A000538, A000583. A062393 provides the result for 5th powers, A011934 for cubes, A000217 for squares, A001057 (unsigned) for nonnegative integers, A000035 (offset) for 0th powers.
Cf. A236770 (see crossrefs).

Programs

  • Maple
    a := n -> (2*n^2+n^3-1)*n/2; # Peter Luschny, Jul 12 2009
  • Mathematica
    Table[n (n + 1) (n^2 + n - 1)/2, {n, 0, 40}] (* Harvey P. Dale, Oct 19 2011 *)
  • PARI
    { a=0; for (n=0, 1000, write("b062392.txt", n, " ", a=n^4 - a) ) } \\ Harry J. Smith, Aug 07 2009

Formula

a(n) = n*(n+1)*(n^2 + n - 1)/2 = n^4 - a(n-1) = A000583(n) - a(n-1) = A000217(A028387(n-1)) = A000217(n)*A028387(n-1).
a(n) = Sum_{i=0..n} A007588(i) for n > 0. - Jonathan Vos Post, Mar 15 2006
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Harvey P. Dale, Oct 19 2011
G.f.: x*(x*(x + 10) + 1)/(1 - x)^5. - Harvey P. Dale, Oct 19 2011
a(n) = A000384(A000217(n)). - Bruno Berselli, Jan 31 2014
a(n) = A110450(n) - A002378(n). - Gionata Neri, May 13 2015
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*2*Pi/sqrt(5). - Amiram Eldar, Jan 22 2024
a(n) = sqrt(144*A288876(n-2) + 72*A006542(n+2) + A000537(n)). - Yasser Arath Chavez Reyes, Jul 22 2024
E.g.f.: exp(x)*x*(2 + 13*x + 8*x^2 + x^3)/2. - Stefano Spezia, Apr 27 2025
a(n) = A000217(n)*(2*A000217(n)-1). - Miguel-Ángel Pérez García-Ortega, May 27 2025
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