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 1-10 of 19 results. Next

A088058 Duplicate of A062392.

Original entry on oeis.org

1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945
Offset: 1

Views

Author

Keywords

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)

A000584 Fifth powers: a(n) = n^5.

Original entry on oeis.org

0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049, 100000, 161051, 248832, 371293, 537824, 759375, 1048576, 1419857, 1889568, 2476099, 3200000, 4084101, 5153632, 6436343, 7962624, 9765625, 11881376, 14348907, 17210368, 20511149
Offset: 0

Views

Author

Keywords

Comments

Totally multiplicative sequence with a(p) = p^5 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A059338. The inverse binomial transform yields the (finite) 0, 1, 30, 150, 240, 120, the 5th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Equals sum of odd numbers from n^2*(n-1)+1 (A100104) to n^2*(n+1)-1 (A003777). - Bruno Berselli, Mar 14 2014
a(n) mod 10 = n mod 10. - Reinhard Zumkeller, May 10 2014
Numbers of the form a(n) + a(n+1) + ... + a(n+k) are nonprime for all n, k>=0; this can be proved by the method indicated in the comment in A256581. - Vladimir Shevelev and Peter J. C. Moses, Apr 04 2015

References

  • 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).
  • 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 give A000539.

Programs

Formula

G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (x-1)^6. [Simon Plouffe in his 1992 dissertation]
Multiplicative with a(p^e) = p^(5e). - David W. Wilson, Aug 01 2001
E.g.f.: exp(x)*(x+15*x^2+25*x^3+10*x^4+x^5). - Geoffrey Critzer, Jun 12 2013
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 120. - Ant King, Sep 23 2013
a(n) = n + Sum_{j=0..n-1}{k=1..4}binomial(5,k)*j^(5-k). - Patrick J. McNab, Mar 28 2016
From Kolosov Petro, Oct 22 2018: (Start)
a(n) = Sum_{k=1..n} A300656(n,k).
a(n) = Sum_{k=0..n-1} A300656(n,k). (End)
a(n) = Sum_{k=1..5} Eulerian(5, k)*binomial(n+5-k, 5), with Eulerian(5, k) = A008292(5, k), the numbers 1, 26, 66, 26, 1, for n >= 0. Worpitzki's identity for powers of 5. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(5) (A013663).
Sum_{n>=1} (-1)^(n+1)/a(n) = 15*zeta(5)/16 (A267316). (End)

Extensions

More terms from Henry Bottomley, Jun 21 2001

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651
Offset: 0

Views

Author

Keywords

Comments

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

Examples

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)
		

Crossrefs

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Programs

Formula

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Extensions

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A153641 Nonzero coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent, and Bernoulli numbers (triangle read by rows).

Original entry on oeis.org

1, 1, 1, -1, 1, -3, 1, -6, 5, 1, -10, 25, 1, -15, 75, -61, 1, -21, 175, -427, 1, -28, 350, -1708, 1385, 1, -36, 630, -5124, 12465, 1, -45, 1050, -12810, 62325, -50521, 1, -55, 1650, -28182, 228525, -555731, 1, -66, 2475, -56364, 685575, -3334386, 2702765, 1
Offset: 0

Views

Author

Peter Luschny, Dec 29 2008

Keywords

Comments

In the following the expression [n odd] is 1 if n is odd, 0 otherwise.
(+) W_n(0) = E_n are the Euler (or secant) numbers A122045.
(+) W_n(1) = T_n are the signed tangent numbers, see A009006.
(+) W_{n-1}(1) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli number A027641/A027642.
(+) W_n(-1) 2^{-n}(n+1) = G_n the Genocchi number A036968.
(+) W_n(1/2) 2^{n} are the signed generalized Euler (Springer) number, see A001586.
(+) | W_n([n odd]) | the number of alternating permutations A000111.
(+) | W_n([n odd]) / n! | for 0<=n the Euler zeta number A099612/A099617 (see Wikipedia on Bernoulli number). - Peter Luschny, Dec 29 2008
The diagonals in the full triangle (with zero coefficients) of the polynomials have the general form E(k)*binomial(n+k,k) (k>=0 fixed, n=0,1,...) where E(n) are the Euler numbers in the enumeration A122045. For k=2 we find the triangular numbers A000217 and for k=4 A154286. - Peter Luschny, Jan 06 2009
From Peter Bala, Jun 10 2009: (Start)
The Swiss-Knife polynomials W_n(x) may be expressed in terms of the Bernoulli polynomials B(n,x) as
... W_n(x) = 4^(n+1)/(2*n+2)*[B(n+1,(x+3)/4) - B(n+1,(x+1)/4)].
The Swiss-Knife polynomials are, apart from a multiplying factor, examples of generalized Bernoulli polynomials.
Let X be the Dirichlet character modulus 4 defined by X(4*n+1) = 1, X(4*n+3) = -1 and X(2*n) = 0. The generalized Bernoulli polynomials B(X;n,x), n = 1,2,..., associated with the character X are defined by means of the generating function
... t*exp(x*t)*(exp(t)-exp(3*t))/(exp(4*t)-1) = sum {n = 1..inf} B(X;n,x)*t^n/n!.
The first few values are B(X;1,x) = -1/2, B(X;2,x) = -x, B(X,3,x) = -3/2*(x^2-1) and B(X;4,x) = -2*(x^3-3*x).
In general, W_n(x) = -2/(n+1)*B(X;n+1,x).
For the theory of generalized Bernoulli polynomials associated to a periodic arithmetical function see [Cohen, Section 9.4].
The generalized Bernoulli polynomials may be used to evaluate twisted sums of k-th powers. For the present case the result is
sum{n = 0..4*N-1} X(n)*n^k = 1^k - 3^k + 5^k - 7^k + ... - (4*N-1)^k
= [B(X;k+1,4*N) - B(X;k+1,0)]/(k+1) = [W_k(0) - W_k(4*N)]/2.
For the proof apply [Cohen, Corollary 9.4.17 with m = 4 and x = 0].
The generalized Bernoulli polynomials and the Swiss-Knife polynomials are also related to infinite sums of powers through their Fourier series - see the formula section below. For a table of the coefficients of generalized Bernoulli polynomials attached to a Dirichlet character modulus 8 see A151751.
(End)
The Swiss-Knife polynomials provide a general formula for alternating sums of powers similar to the formula which are provided by the Bernoulli polynomials for non-alternating sums of powers (see the Luschny link). Sequences covered by this formula include A001057, A062393, A062392, A011934, A144129, A077221, A137501, A046092. - Peter Luschny, Jul 12 2009
The greatest common divisor of the nonzero coefficients of the decapitated Swiss-Knife polynomials is exp(Lambda(n)), where Lambda(n) is the von Mangoldt function for odd primes, symbolically:
gcd(coeffs(SKP_{n}(x) - x^n)) = A155457(n) (n>1). - Peter Luschny, Dec 16 2009
Another version is at A119879. - Philippe Deléham, Oct 26 2013

Examples

			1
x
x^2  -1
x^3  -3x
x^4  -6x^2   +5
x^5 -10x^3  +25x
x^6 -15x^4  +75x^2  -61
x^7 -21x^5 +175x^3 -427x
		

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag. [From Peter Bala, Jun 10 2009]

Crossrefs

W_n(k), k=0,1,...
W_0: 1, 1, 1, 1, 1, 1, ........ A000012
W_1: 0, 1, 2, 3, 4, 5, ........ A001477
W_2: -1, 0, 3, 8, 15, 24, ........ A067998
W_3: 0, -2, 2, 18, 52, 110, ........ A121670
W_4: 5, 0, -3, 32, 165, 480, ........
W_n(k), n=0,1,...
k=0: 1, 0, -1, 0, 5, 0, -61, ... A122045
k=1: 1, 1, 0, -2, 0, 16, 0, ... A155585
k=2: 1, 2, 3, 2, -3, 2, 63, ... A119880
k=3: 1, 3, 8, 18, 32, 48, 128, ... A119881
k=4: 1, 4, 15, 52, 165, 484, ........ [Peter Luschny, Jul 07 2009]

Programs

  • Maple
    w := proc(n,x) local v,k,pow,chen; pow := (a,b) -> if a = 0 and b = 0 then 1 else a^b fi; chen := proc(m) if irem(m+1,4) = 0 then RETURN(0) fi; 1/((-1)^iquo(m+1,4) *2^iquo(m,2)) end; add(add((-1)^v*binomial(k,v)*pow(v+x+1,n)*chen(k),v=0..k), k=0..n) end:
    # Coefficients with zeros:
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)*sech(t),t,16),t,i),x,i-n),n=0..i)), i=0..8);
    # Recursion
    W := proc(n,z) option remember; local k,p;
    if n = 0 then 1 else p := irem(n+1,2);
    z^n - p + add(`if`(irem(k,2)=1,0,
    W(k,0)*binomial(n,k)*(power(z,n-k)-p)),k=2..n-1) fi end:
    # Peter Luschny, edited and additions Jul 07 2009, May 13 2010, Oct 24 2011
  • Mathematica
    max = 9; rows = (Reverse[ CoefficientList[ #, x]] & ) /@ CoefficientList[ Series[ Exp[x*t]*Sech[t], {t, 0, max}], t]*Range[0, max]!; par[coefs_] := (p = Partition[ coefs, 2][[All, 1]]; If[ EvenQ[ Length[ coefs]], p, Append[ p, Last[ coefs]]]); Flatten[ par /@ rows] (* Jean-François Alcover, Oct 03 2011, after g.f. *)
    sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; Table[CoefficientList[sk[n, x], x] // Reverse // Select[#, # =!= 0 &] &, {n, 0, 13}] // Flatten (* Jean-François Alcover, May 21 2013 *)
    Flatten@Table[Binomial[n, 2k] EulerE[2k], {n, 0, 12}, {k, 0, n/2}](* Oliver Seipel, Jan 14 2025 *)
  • Sage
    def A046978(k):
        if k % 4 == 0:
            return 0
        return (-1)**(k // 4)
    def A153641_poly(n, x):
        return expand(add(2**(-(k // 2))*A046978(k+1)*add((-1)**v*binomial(k,v)*(v+x+1)**n for v in (0..k)) for k in (0..n)))
    for n in (0..7): print(A153641_poly(n, x))  # Peter Luschny, Oct 24 2011

Formula

W_n(x) = Sum_{k=0..n}{v=0..k} (-1)^v binomial(k,v)*c_k*(x+v+1)^n where c_k = frac((-1)^(floor(k/4))/2^(floor(k/2))) [4 not div k] (Iverson notation).
From Peter Bala, Jun 10 2009: (Start)
E.g.f.: 2*exp(x*t)*(exp(t)-exp(3*t))/(1-exp(4*t))= 1 + x*t + (x^2-1)*t^2/2! + (x^3-3*x)*t^3/3! + ....
W_n(x) = 1/(2*n+2)*Sum_{k=0..n+1} 1/(k+1)*Sum_{i=0..k} (-1)^i*binomial(k,i)*((x+4*i+3)^(n+1) - (x+4*i+1)^(n+1)).
Fourier series expansion for the generalized Bernoulli polynomials:
B(X;2*n,x) = (-1)^n*(2/Pi)^(2*n)*(2*n)! * {sin(Pi*x/2)/1^(2*n) - sin(3*Pi*x/2)/3^(2*n) + sin(5*Pi*x/2)/5^(2*n) - ...}, valid for 0 <= x <= 1 when n >= 1.
B(X;2*n+1,x) = (-1)^(n+1)*(2/Pi)^(2*n+1)*(2*n+1)! * {cos(Pi*x/2)/1^(2*n+1) - cos(3*Pi*x/2)/3^(2*n+1) + cos(5*Pi*x/2)/5^(2*n+1) - ...}, valid for 0 <= x <= 1 when n >= 1 and for 0 <= x < 1 when n = 0.
(End)
E.g.f.: exp(x*t) * sech(t). - Peter Luschny, Jul 07 2009
O.g.f. as a J-fraction: z/(1-x*z+z^2/(1-x*z+4*z^2/(1-x*z+9*z^2/(1-x*z+...)))) = z + x*z^2 + (x^2-1)*z^3 + (x^3-3*x)*z^4 + .... - Peter Bala, Mar 11 2012
Conjectural o.g.f.: Sum_{n >= 0} (1/2^((n-1)/2))*cos((n+1)*Pi/4)*( Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - (k + x)*t) ) = 1 + x*t + (x^2 - 1)*t^2 + (x^3 - 3*x)*t^3 + ... (checked up to O(t^13)), which leads to W_n(x) = Sum_{k = 0..n} 1/2^((k - 1)/2)*cos((k + 1)*Pi/4)*( Sum_{j = 0..k} (-1)^j*binomial(k, j)*(j + x)^n ). - Peter Bala, Oct 03 2016

A007588 Stella octangula numbers: a(n) = n*(2*n^2 - 1).

Original entry on oeis.org

0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, 2651, 3444, 4381, 5474, 6735, 8176, 9809, 11646, 13699, 15980, 18501, 21274, 24311, 27624, 31225, 35126, 39339, 43876, 48749, 53970, 59551, 65504, 71841, 78574, 85715, 93276, 101269, 109706, 118599, 127960
Offset: 0

Views

Author

Keywords

Comments

Also as a(n)=(1/6)*(12*n^3-6*n), n>0: structured hexagonal anti-diamond numbers (vertex structure 13) (Cf. A005915 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The only known square stella octangula number for n>1 is a(169) = 169*(2*169^2 - 1) = 9653449 = 3107^2. - Alexander Adamchuk, Jun 02 2008
Ljunggren proved that 9653449 = (13*239)^2 is the only square stella octangula number for n>1. See A229384 and the Wikipedia link. - Jonathan Sondow, Sep 30 2013
4*A007588 = A144138(ChebyshevU[3,n]). - Vladimir Joseph Stephan Orlovsky, Jun 30 2011
If A016813 is regarded as a regular triangle (with leading terms listed in A001844), a(n) provides the row sums of this triangle: 1, 5+9=14, 13+17+21=51 and so on. - J. M. Bergot, Jul 05 2013
Shares its digital root, A267017, with n*(n^2 + 1)/2 ("sum of the next n natural numbers" see A006003). - Peter M. Chema, Aug 28 2016

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 51.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • W. Ljunggren, Zur Theorie der Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Backwards differences give star numbers A003154: A003154(n)=a(n)-a(n-1).
1/12*t*(n^3-n)+ n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A001653 = Numbers n such that 2*n^2 - 1 is a square.
a(169) = (A229384(3)*A229384(4))^2.

Programs

Formula

G.f.: x*(1+10*x+x^2)/(1-x)^4.
a(n) = n*A056220(n).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Harvey P. Dale, Sep 16 2011
From Ilya Gutkovskiy, Jul 02 2016: (Start)
E.g.f.: x*(1 + 6*x + 2*x^2)*exp(x).
Dirichlet g.f.: 2*zeta(s-3) - zeta(s-1). (End)
a(n) = A004188(n) + A135503(n). - Miquel Cerda, Dec 25 2016
a(n) = A061317(n) - A005843(n) = A062392(n) - A062392(n-1). - J.S. Seneschal, Jul 01 2025

Extensions

In the formula given in the 1995 Encyclopedia of Integer Sequences, the second 2 should be an exponent.

A007487 Sum of 9th powers.

Original entry on oeis.org

0, 1, 513, 20196, 282340, 2235465, 12313161, 52666768, 186884496, 574304985, 1574304985, 3932252676, 9092033028, 19696532401, 40357579185, 78800938560, 147520415296, 266108291793, 464467582161, 787155279940, 1299155279940, 2093435326521, 3300704544313
Offset: 0

Views

Author

Keywords

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. 815.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 9 of array A103438.

Programs

  • Magma
    [&+[n^9: n in [0..m]]: m in [0..22]]; // Bruno Berselli, Aug 23 2011
    
  • Maple
    [seq(add(i^9,i=1..n),n=0..40)];
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^9 od: seq(a[n], n=0..22); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    lst={};s=0;Do[s=s+n^9;AppendTo[lst, s], {n, 10^2}];lst..or..Table[Sum[k^9, {k, 1, n}], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
    Accumulate[Range[0,30]^9] (* Harvey P. Dale, Oct 09 2016 *)
  • PARI
    a(n)=n^2*(n+1)^2*(n^2+n-1)*(2*n^4+4*n^3-n^2-3*n+3)/20 \\ Charles R Greathouse IV, Oct 07 2015
  • Python
    A007487_list, m = [0], [362880, -1451520, 2328480, -1905120, 834120, -186480, 18150, -510, 1, 0, 0]
    for _ in range(10**2):
        for i in range(10):
            m[i+1]+= m[i]
        A007487_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    

Formula

a(n) = n^2*(n+1)^2*(n^2+n-1)*(2*n^4+4*n^3-n^2-3*n+3)/20 (see MathWorld, Power Sum, formula 39). a(n) = n*A000542(n) - Sum_{i=0..n-1} A000542(i). - Bruno Berselli, Apr 26 2010
G.f.: x*(1 + 502*x + 14608*x^2 + 88234*x^3 + 156190*x^4 + 88234*x^5 + 14608*x^6 + 502*x^7 + x^8)/(1-x)^11. a(n) = a(-n-1). - Bruno Berselli, Aug 23 2011
a(n) = -Sum_{j=1..9} j*Stirling1(n+1,n+1-j)*Stirling2(n+9-j,n). - Mircea Merca, Jan 25 2014
a(n) = (16/5)*A000217(n)^5 - 4*A000217(n)^4 + (12/5)*A000217(n)^3 - (3/5)*A000217(n)^2. - Michael Raney, Mar 14 2016
a(n) = 11*a(n-1) - 55*a(n-2) + 165*a(n-3) - 330*a(n-4) + 462*a(n-5) - 462*a(n-6) + 330*a(n-7) - 165*a(n-8) + 55*a(n-9) - 11*a(n-10) + a(n-11) for n > 10. - Wesley Ivan Hurt, Dec 21 2016
a(n) = 288*A005585(n-1)^2 + 1728*A108679(n-3) + A062392(n)^2. - Yasser Arath Chavez Reyes, May 11 2024
a(n) = Sum_{i=1..n} J_9(i)*floor(n/i), where J_9 is A069094. - Ridouane Oudra, Jul 17 2025

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

Views

Author

Bruno Berselli, Jan 31 2014

Keywords

Comments

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

Crossrefs

Partial sums of A004188.
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).

Programs

  • Magma
    [n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];
  • Mathematica
    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 *)
  • PARI
    for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));
    

Formula

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

A152725 a(n) = n*(n+1)*(n^4 + 2*n^3 - 2*n^2 - 3*n + 3)/2.

Original entry on oeis.org

0, 1, 63, 666, 3430, 12195, 34461, 83188, 178956, 352485, 647515, 1124046, 1861938, 2964871, 4564665, 6825960, 9951256, 14186313, 19825911, 27219970, 36780030, 48986091, 64393813, 83642076, 107460900, 136679725, 172236051, 215184438
Offset: 0

Views

Author

Keywords

Crossrefs

Cf. A062392, A062393 (for 5th powers), A011934, A152726 (for 7th powers).

Programs

  • Magma
    [n*(n+1)*(n^4+2*n^3-2*n^2-3*n+3)/2: n in [0..50]]; // G. C. Greubel, Sep 01 2018
  • Mathematica
    k=0;lst={k};Do[k=n^6-k;AppendTo[lst,k],{n,1,5!}];lst
    LinearRecurrence[{7,-21,35,-35,21,-7,1}, {0,1,63,666,3430,12195,34461}, 50] (* G. C. Greubel, Sep 01 2018 *)
    CoefficientList[Series[-((x (1+56 x+246 x^2+56 x^3+x^4))/(-1+x)^7),{x,0,30}],x] (* Harvey P. Dale, Aug 03 2024 *)
  • PARI
    a(n)=n*(n+1)*(n^4+2*n^3-2*n^2-3*n+3)/2 \\ Charles R Greathouse IV, Oct 07 2015
    

Formula

a(n) = n^6 - (n-1)^6 + (n-2)^6 - ... + ((-1)^n)*0^6.
G.f.: x*(1 + 56*x + 246*x^2 + 56*x^3 + x^4) / (1-x)^7. - R. J. Mathar, Jul 08 2013
a(n) = A050492(A000217(n)). - Kelvin Voskuijl, Jun 18 2025
E.g.f.: exp(x)*x*(2 + 61*x + 160*x^2 + 95*x^3 + 18*x^4 + x^5)/2. - Stefano Spezia, Jun 19 2025

Extensions

Offset set to 0 by R. J. Mathar, Aug 15 2010

A231303 Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 16, 82, 272, 707, 1568, 3108, 5664, 9669, 15664, 24310, 36400, 52871, 74816, 103496, 140352, 187017, 245328, 317338, 405328, 511819, 639584, 791660, 971360, 1182285, 1428336, 1713726, 2042992, 2421007, 2852992, 3344528, 3901568, 4530449, 5237904
Offset: 0

Views

Author

Stanislav Sykora, Nov 07 2013

Keywords

Comments

In physics, a(n)/2^(M-1) is the trace of the spin operator |S_z|^M for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of their 4th powers is 2*82/16 = a(3)/8 (analogously for other values of M).
Partial sums of A062392. - Bruce J. Nicholson, Jun 29 2019

Examples

			a(4) = 4^4 + 2^4 = 272; a(5) = 5^4 + 3^4 + 1^4 = 707.
		

Crossrefs

Cf. A001477 (M=1), A000292 (M=2), A105636 (M=3), A231304 (M=5), A231305 (M=6), A231306 (M=7), A231307 (M=8), A231308 (M=9), A231309 (M=10).

Programs

  • GAP
    List([0..40], n-> n*(3*n^4+15*n^3+20*n^2-8)/30); # G. C. Greubel, Jul 01 2019
  • Magma
    [1/30*n*(3*n^4+15*n^3+20*n^2-8): n in [0..40]]; // Vincenzo Librandi, Dec 23 2015
    
  • Mathematica
    Table[SeriesCoefficient[x*(1+10*x+x^2)/(1-x)^6, {x, 0, n}], {n, 0, 40}] (* Michael De Vlieger, Dec 22 2015 *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 16, 82, 272, 707}, 40] (* Vincenzo Librandi, Dec 23 2015 *)
  • PARI
    nmax=40;a = vector(nmax);a[2]=1;for(i=3,#a,a[i]=a[i-2]+(i-1)^4); print(a);
    
  • PARI
    concat(0, Vec(x*(1+10*x+x^2)/(1-x)^6 + O(x^40))) \\ Colin Barker, Dec 22 2015
    
  • Sage
    [n*(3*n^4+15*n^3+20*n^2-8)/30 for n in (0..40)] # G. C. Greubel, Jul 01 2019
    

Formula

a(n) = Sum_{k=0..floor(n/2)} (n - 2*k)^4.
From Colin Barker, Dec 22 2015: (Start)
a(n) = (1/30)*n*(3*n^4 + 15*n^3 + 20*n^2 - 8).
G.f.: x*(1 + 10*x + x^2) / (1-x)^6.
(End)
E.g.f.: x*(30 + 210*x + 185*x^2 + 45*x^3 + 3*x^4)*exp(x)/30. - G. C. Greubel, Apr 24 2016
From Bruce J. Nicholson, Jun 29 2019: (Start)
a(n) = 12*A000389(n+3) + A000292(n);
a(n) = (12*A000579(n+4)+A000332(n+3)) - (12*A000579(n+3)+A000332(n+2));
a(n) - a(n-2) = A000583(n). (End)
Showing 1-10 of 19 results. Next