Search: a092182 -id:a092182
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1, 121, 1068, 4720, 14705, 36981, 80416, 157368, 284265, 482185, 777436, 1202136, 1794793, 2600885, 3673440, 5073616, 6871281, 9145593, 11985580, 15490720, 19771521, 24950101, 31160768, 38550600, 47280025, 57523401, 69469596, 83322568, 99301945
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OFFSET
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1,2
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COMMENTS
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Geometrically, the partial sums of A092182 may be interpreted as 5-dimensional hexacosichoronal hyperpyramidal numbers. The hexacosichoron is a convex regular 4-D polytope with Schlaefli symbol {3,3,5}.
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LINKS
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FORMULA
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G.f.: x*(1 + 115*x + 357*x^2 + 107*x^3) / (1 - x)^6.
a(n) = (n*(12 + n - 64*n^2 + 5*n^3 + 58*n^4)) / 12.
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) for n>6.
(End)
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MATHEMATICA
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Accumulate[LinearRecurrence[{5, -10, 10, -5, 1}, {1, 120, 947, 3652, 9985}, 30]] (* or *) LinearRecurrence[ {6, -15, 20, -15, 6, -1}, {1, 121, 1068, 4720, 14705, 36981}, 30] (* Harvey P. Dale, May 04 2024 *)
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PROG
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(PARI) Vec(x*(1 + 115*x + 357*x^2 + 107*x^3) / (1 - x)^6 + O(x^40)) \\ Colin Barker, Aug 15 2018
(PARI) a(n) = (n*(12 + n - 64*n^2 + 5*n^3 + 58*n^4)) / 12 \\ Colin Barker, Aug 15 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A000583
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Fourth powers: a(n) = n^4.
(Formerly M5004 N2154)
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+10
399
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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
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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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).
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LINKS
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FORMULA
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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).
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
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) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 24. - Ant King, Sep 23 2013
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)
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MAPLE
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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
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MATHEMATICA
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PROG
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(Haskell)
a000583 = (^ 4)
a000583_list = scanl (+) 0 a005917_list
(Maxima) makelist(n^4, n, 0, 30); /* Martin Ettl, Nov 12 2012 */
(Python)
def a(n): return n**4
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CROSSREFS
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KEYWORD
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nonn,core,easy,nice,mult
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AUTHOR
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STATUS
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approved
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A000332
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Binomial coefficient binomial(n,4) = n*(n-1)*(n-2)*(n-3)/24.
(Formerly M3853 N1578)
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+10
381
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0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410
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OFFSET
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0,6
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COMMENTS
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Number of intersection points of diagonals of convex n-gon where no more than two diagonals intersect at any point in the interior.
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation. - Ignacio Larrosa Cañestro, Apr 09 2002. [See Les Reid link for proof. - N. J. A. Sloane, Apr 02 2016]
Start from cubane and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink on chemistry. - Robert G. Wilson v, Aug 02 2002
For n>0, a(n) = (-1/8)*(coefficient of x in Zagier's polynomial P_(2n,n)). (Zagier's polynomials are used by PARI/GP for acceleration of alternating or positive series.)
Figurate numbers based on the 4-dimensional regular convex polytope called the regular 4-simplex, pentachoron, 5-cell, pentatope or 4-hypertetrahedron with Schlaefli symbol {3,3,3}. a(n)=((n*(n-1)*(n-2)*(n-3))/4!). - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 07 2009
Maximal number of crossings that can be created by connecting n vertices with straight lines. - Cameron Redsell-Montgomerie (credsell(AT)uoguelph.ca), Jan 30 2007
If X is an n-set and Y a fixed (n-1)-subset of X then a(n) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Product of four consecutive numbers divided by 24. - Artur Jasinski, Dec 02 2007
For strings consisting entirely of 0's and 1's, the number of distinct arrangements of four 1's such that 1's are not adjacent. The shortest possible string is 7 characters, of which there is only one solution: 1010101, corresponding to a(5). An eight-character string has 5 solutions, nine has 15, ten has 35 and so on, congruent to A000332. - Gil Broussard, Mar 19 2008
Except for the 4 initial 0's, is equivalent to the partial sums of the tetrahedral numbers A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
If the first 3 zeros are disregarded, that is, if one looks at binomial(n+3, 4) with n>=0, then it becomes a 'Matryoshka doll' sequence with alpha=0: seq(add(add(add(i,i=alpha..k),k=alpha..n),n=alpha..m),m=alpha..50). - Peter Luschny, Jul 14 2009
For n>=1, a(n) is the number of n-digit numbers the binary expansion of which contains two runs of 0's. - Vladimir Shevelev, Jul 30 2010
For n>0, a(n) is the number of crossing set partitions of {1,2,..,n} into n-2 blocks. - Peter Luschny, Apr 29 2011
For n > 3, a(n) is the hyper-Wiener index of the path graph on n-2 vertices. - Emeric Deutsch, Feb 15 2012
Except for the four initial zeros, number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Aug 31 2012
a(n+3) is the number of different ways to color the faces (or the vertices) of a regular tetrahedron with n colors if we count mirror images as the same.
a(n) = fallfac(n,4)/4! is also the number of independent components of an antisymmetric tensor of rank 4 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Number of chiral pairs of colorings of the vertices (or faces) of a regular tetrahedron with n available colors. Chiral colorings come in pairs, each the reflection of the other. - Robert A. Russell, Jan 22 2020
a(n+3) is the number of lattice rectangles (squares included) in a staircase of order n; this is obtained by stacking n rows of consecutive unit lattice squares, aligned either to the left or to the right, which consist of 1, 2, 3, ..., n squares and which are stacked either in the increasing or in the decreasing order of their lengths. Below, there is a staircase or order 4 which contains a(7) = 35 rectangles. [See the Teofil Bogdan and Mircea Dan Rus link, problem 3, under A004320]
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(End)
a(n+4) is the number of strings of length n on an ordered alphabet of 5 letters where the characters in the word are in nondecreasing order. E.g., number of length-2 words is 15: aa,ab,ac,ad,ae,bb,bc,bd,be,cc,cd,ce,dd,de,ee. - Jim Nastos, Jan 18 2021
Aside from the zeros, this is the fifth diagonal of the Pascal matrix A007318, the only nonvanishing diagonal (fifth) of the matrix representation IM = (A132440)^4/4! of the differential operator D^4/4!, when acting on the row vector of coefficients of an o.g.f., or power series.
M = e^{IM} is the matrix of coefficients of the Appell sequence p_n(x) = e^{D^4/4!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^4/4!}, which is that for A025036 aerated with triple zeros, the first column of M.
See A099174 and A000292 for analogous relationships for the third and fourth diagonals of the Pascal matrix. (End)
For integer m and positive integer r >= 3, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (3 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 8.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 7.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
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).
Charles W. Trigg: Mathematical Quickies. New York: Dover Publications, Inc., 1985, p. 53, #191
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Eric Weisstein's World of Mathematics, Pentatope.
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FORMULA
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a(n) = n*(n-1)*(n-2)*(n-3)/24.
a(n) = Sum_{k=1..n-3} Sum_{i=1..k} i*(i+1)/2. - Benoit Cloitre, Jun 15 2003
Convolution of natural numbers {1, 2, 3, 4, ...} and A000217, the triangular numbers {1, 3, 6, 10, ...}. - Jon Perry, Jun 25 2003
a(n+1) = ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24 - (n^5-(n-1)^5-1)/30; a(n) = A006322(n-2)-A006325(n-1). - Xavier Acloque, Oct 20 2003; R. J. Mathar, Jul 07 2009
a(4*n+2) = Pyr(n+4, 4*n+2) where the polygonal pyramidal numbers are defined for integers A>2 and B>=0 by Pyr(A, B) = B-th A-gonal pyramid number = ((A-2)*B^3 + 3*B^2 - (A-5)*B)/6; For all positive integers i and the pentagonal number function P(x) = x*(3*x-1)/2: a(3*i-2) = P(P(i)) and a(3*i-1) = P(P(i) + i); 1 + 24*a(n) = (n^2 + 3*n + 1)^2. - Jonathan Vos Post, Nov 15 2004
For n > 3, the sum of the first n-2 tetrahedral numbers (A000292). - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005 [Corrected by Doug Bell, Jun 25 2017]
Starting (1, 5, 15, 35, ...), = binomial transform of [1, 4, 6, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
Sum_{n>=4} 1/a(n) = 4/3, from the Taylor expansion of (1-x)^3*log(1-x) in the limit x->1. - R. J. Mathar, Jan 27 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1. - Harvey P. Dale, Aug 22 2011
a(n) = (binomial(n-1,2)^2 - binomial(n-1,2))/6. - Gary Detlefs, Nov 20 2011
Sum_{n>=0} a(n)/n! = e/24. Sum_{n>=3} a(n)/(n-3)! = 73*e/24. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
a(n+3) = C(n,1) + 3*C(n,2) + 3*C(n,3) + C(n,4). Each term indicates the number of ways to use n colors to color a tetrahedron with exactly 1, 2, 3, or 4 colors.
G.f.: Starting (1, 5, 14, ...), x/(1-x)^5 can be written
as (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1+x)^5;
as (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1+x+x^2)^5;
as (x * r(x) * r(x^4) * r(x^16) * r(x^64) * ...) where r(x) = (1+x+x^2+x^3)^5;
... (as a conjectured infinite set). (End)
Product_{n>=5} (1 - 1/a(n)) = cosh(sqrt(15)*Pi/2)/(100*Pi). - Amiram Eldar, Jan 21 2021
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EXAMPLE
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a(5) = 5 from the five independent components of an antisymmetric tensor A of rank 4 and dimension 5, namely A(1,2,3,4), A(1,2,3,5), A(1,2,4,5), A(1,3,4,5) and A(2,3,4,5). See the Dec 10 2015 comment. - Wolfdieter Lang, Dec 10 2015
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MAPLE
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A000332 := n->binomial(n, 4); [seq(binomial(n, 4), n=0..100)];
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MATHEMATICA
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Table[ Binomial[n, 4], {n, 0, 45} ] (* corrected by Harvey P. Dale, Aug 22 2011 *)
Table[(n-4)(n-3)(n-2)(n-1)/24, {n, 100}] (* Artur Jasinski, Dec 02 2007 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 0, 0, 1}, 45] (* Harvey P. Dale, Aug 22 2011 *)
CoefficientList[Series[x^4 / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 23 2014 *)
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PROG
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(PARI) a(n)=binomial(n, 4);
(Python)
# Starts at a(3), i.e. computes n*(n+1)*(n+2)*(n+3)/24
x, y, z, u = 1, 1, 1, 1
yield 0
while True:
yield x
x, y, z, u = x + y + z + u + 1, y + z + u + 1, z + u + 1, u + 1
(Python)
print([n*(n-1)*(n-2)*(n-3)//24 for n in range(50)])
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CROSSREFS
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Cf. A006008 (Number of ways to color the faces (or vertices) of a regular tetrahedron with n colors when mirror images are counted as two).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Some formulas that referred to another offset corrected by R. J. Mathar, Jul 07 2009
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STATUS
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approved
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A014820
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a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.
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+10
30
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1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481
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OFFSET
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0,2
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COMMENTS
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a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009
If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008
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REFERENCES
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T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
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LINKS
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Eric Weisstein's World of Mathematics, 16-Cell
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FORMULA
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Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009
G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013
E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024
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MAPLE
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al:=proc(s, n) binomial(n+s-1, s); end; be:=proc(d, n) local r; add( (-1)^r*binomial(d-1, r)*2^(d-1-r)*al(d-r, n), r=0..d-1); end; [seq(be(4, n), n=0..100)];
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MATHEMATICA
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LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)
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PROG
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(R)
a <- c(1, 8, 33, 96, 225)
for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
(Sage) [((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019
(GAP) List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019
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CROSSREFS
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Cf. A000332, A000583, A005900, A069038, A069039, A070212, A081277, A092181, A092182, A092183, A099175, A099193, A099195, A099196, A099197, A142978.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A092181
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Figurate numbers based on the 24-cell (4-D polytope with Schlaefli symbol {3,4,3}).
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+10
9
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1, 24, 153, 544, 1425, 3096, 5929, 10368, 16929, 26200, 38841, 55584, 77233, 104664, 138825, 180736, 231489, 292248, 364249, 448800, 547281, 661144, 791913, 941184, 1110625, 1301976, 1517049, 1757728, 2025969, 2323800, 2653321, 3016704
(list;
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refs;
listen;
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internal format)
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OFFSET
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1,2
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COMMENTS
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This is the 4-dimensional regular convex polytope called the 24-cell, hyperdiamond or icositetrachoron.
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LINKS
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Eric Weisstein's World of Mathematics, 24-Cell
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FORMULA
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a(n) = n^2*((3*n^2)-(4*n)+2).
a(n) = C(n+3,4) + 19 C(n+2,4) + 43 C(n+1,4) + 9 C(n,4).
a(n) = +5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5). G.f.: x*(1+19*x+43*x^2+9*x^3)/(1-x)^5. [R. J. Mathar, Jun 21 2010]
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EXAMPLE
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a(3)= 3^2*((3*3^2)-(4*3)+2) = 9*(27-12+2) = 9*17 = 153
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MATHEMATICA
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Table[SeriesCoefficient[x (1 + 19 x + 43 x^2 + 9 x^3)/(1 - x)^5, {x, 0, n}], {n, 32}] (* Michael De Vlieger, Dec 14 2015 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 24, 153, 544, 1425}, 40] (* Harvey P. Dale, May 25 2022 *)
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PROG
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Michael J. Welch (mjw1(AT)ntlworld.com), Mar 31 2004
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STATUS
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approved
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A092183
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Figurate numbers based on the 120-cell (4-D polytope with Schlaefli symbol {5,3,3}).
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+10
8
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1, 600, 4983, 19468, 53505, 119676, 233695, 414408, 683793, 1066960, 1592151, 2290740, 3197233, 4349268, 5787615, 7556176, 9701985, 12275208, 15329143, 18920220, 23108001, 27955180, 33527583, 39894168, 47127025, 55301376
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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This is the 4-dimensional regular convex polytope called the 120-cell, hecatonicosachoron or hyperdodecahedron.
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LINKS
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Eric Weisstein's World of Mathematics, 120-Cell
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FORMULA
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a(n) = n*((261*n^3)-(504*n^2)+(283*n)-38)/2.
a(n) = C(n+3,4) + 595 C(n+2,4) + 1993 C(n+1,4) + 543 C(n,4).
a(n) = +5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5). G.f.: x*(1+595*x+1993*x^2+543*x^3)/(1-x)^5. [R. J. Mathar, Jun 21 2010]
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EXAMPLE
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a(3) = 3*((261*3^3)-(504*3^2)+(283*3)-38)/2 = 3*(7047-4536+849-38)/2 = 1.5*3322 = 4983
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MATHEMATICA
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Table[SeriesCoefficient[x (1 + 595 x + 1993 x^2 + 543 x^3)/(1 - x)^5, {x, 0, n}], {n, 26}] (* Michael De Vlieger, Dec 14 2015 *)
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PROG
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(Magma) [n*((261*n^3)-(504*n^2)+(283*n)-38)/2: n in [1..40]]; // Vincenzo Librandi, May 22 2011
(PARI) a(n) = n*(261*n^3 - 504*n^2 + 283*n - 38)/2; \\ Michel Marcus, Dec 14 2015
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Michael J. Welch (mjw1(AT)ntlworld.com), Mar 31 2004
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STATUS
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approved
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A195163
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1000-gonal numbers: a(n) = n*(499*n - 498).
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+10
4
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0, 1, 1000, 2997, 5992, 9985, 14976, 20965, 27952, 35937, 44920, 54901, 65880, 77857, 90832, 104805, 119776, 135745, 152712, 170677, 189640, 209601, 230560, 252517, 275472, 299425, 324376, 350325, 377272, 405217, 434160, 464101, 495040, 526977, 559912, 593845, 628776
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,3
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COMMENTS
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LINKS
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FORMULA
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a(n) = 998*n*(n-1)/2 + n, according to the common formula for s-gonal numbers, a(n) = (s-2)*n*(n-1)/2 + n. - Sergey Pavlov, Aug 14 2015
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 4. - Muniru A Asiru, Sep 12 2017
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MAPLE
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MATHEMATICA
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PROG
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(PARI) x='x+O('x^99); concat(0, Vec(x*(1+997*x)/(1-x)^3)) \\ Altug Alkan, Apr 10 2016
(JavaScript) function a(n){return 998*n*(n-1)/2+n} // Sergey Pavlov, Aug 14 2015
(GAP)
a:=[0, 1, 1000];; for n in [4..10^2] do a[n]:=3*a[n-1]-3*a[n-2]+*a[n-3]; od; a; # Muniru A Asiru, Sep 12 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A100012
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Let h(k) = a(k)*((145*a(k)^3)-(280*a(k)^2)+(179*a(k))-38)/6, then a(n) = h(a(n-1)) for n >= 1 and a(0) =2.
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+10
0
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OFFSET
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0,1
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COMMENTS
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The next term has 163 digits.
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LINKS
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MATHEMATICA
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NestList[#/6*(145#^3-280#^2+179#-38)&, 2, 3] (* Harvey P. Dale, Apr 09 2015 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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