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 1-8 of 8 results.

A002414 Octagonal pyramidal numbers: a(n) = n*(n+1)*(2*n-1)/2.

Original entry on oeis.org

1, 9, 30, 70, 135, 231, 364, 540, 765, 1045, 1386, 1794, 2275, 2835, 3480, 4216, 5049, 5985, 7030, 8190, 9471, 10879, 12420, 14100, 15925, 17901, 20034, 22330, 24795, 27435, 30256, 33264, 36465, 39865, 43470, 47286, 51319, 55575, 60060, 64780
Offset: 1

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Author

Keywords

Comments

Number of ways of covering a 2n X 2n lattice with 2n^2 dominoes with exactly 2 horizontal dominoes. - Yong Kong (ykong@curagen.com), May 06 2000
Equals binomial transform of [0, 1, 7, 6, 0, 0, 0, ...]. - Gary W. Adamson, Jun 14 2008, corrected Oct 25 2012
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF3 denominators of A156927. See A157704 for background information. - Johannes W. Meijer, Mar 07 2009
This sequence is related to A000326 by a(n) = n*A000326(n) - Sum_{i=0..n-1} A000326(i) and this is the case d=3 in the identity n*(n*(d*n-d+2)/2)-Sum_{k=0..n-1} k*(d*k-d+2)/2 = n*(n+1)*(2*d*n-2*d+3)/6. - Bruno Berselli, Apr 21 2010
2*a(n) gives the principal diagonal of the convolution array A213819. - Clark Kimberling, Jul 04 2012
Partial sums of the figurate octagonal numbers A000567. For each sequence with a linear recurrence with constant coefficients, the values reduced modulo some constant m generate a periodic sequence. For this sequence, these Pisano periods have length 1, 4, 3, 8, 5, 12, 7, 16, 9, 20, 11, 24, 13, 28, 15, 32, 17, ... for m >= 1. - Ant King, Oct 26 2012
Partial sums of the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 773", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
On a square grid of side length n+1, the number of embedded rectangles (where each side is greater than 1). For example, in a 2 X 2 square there is one rectangle, in a 3 X 3 square there are nine rectangles, etc. - Peter Woodward, Nov 26 2017
a(n) is the sum of the numbers in the n X n square array A204154(n). - Ali Sada, Jun 21 2019
Sum of all multiples of n and n+1 that are <= n^2. - Wesley Ivan Hurt, May 25 2023

Examples

			a(2) = 9 since there are 9 ways to cover a 4 X 4 lattice with 8 dominoes, 2 of which is horizontal and the other 6 are vertical. - Yong Kong (ykong@curagen.com), May 06 2000
G.f. = x + 9*x^2 + 30*x^3 + 70*x^4 + 135*x^5 + 231*x^6 + 364*x^7 + 540*x^8 + 765*x^9 + ...
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • 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. 2.
  • 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

Cf. A093563 (( 6, 1) Pascal, column m=3). A000567 (differences).
Cf. A156927, A157704. - Johannes W. Meijer, Mar 07 2009
Cf. A000326.
Cf. similar sequences listed in A237616.
Cf. A260234 (largest prime factor of a(n+1)).

Programs

  • Magma
    [n*(n+1)*(2*n-1)/2: n in [1..50]]; // Vincenzo Librandi, May 24 2016
  • Maple
    A002414 := n-> 1/2*n*(n+1)*(2*n-1): seq(A002414(n), n=1..100);
  • Mathematica
    LinearRecurrence[{4,-6,4,-1},{1,9,30,70},40] (* Harvey P. Dale, Apr 12 2013 *)
  • PARI
    {a(n) = (2*n - 1) * n * (n + 1) / 2} \\ Michael Somos, Mar 17 2011
    

Formula

a(n) = odd numbers * triangular numbers = (2*n-1)* binomial(n+1,2). - Xavier Acloque, Oct 27 2003
G.f.: x*(1+5*x)/(1-x)^4. [Conjectured by Simon Plouffe in his 1992 dissertation.]
a(n) = A000578(n) + A000217(n-1). - Kieren MacMillan, Mar 19 2007
a(-n) = -A160378(n). - Michael Somos, Mar 17 2011
From Ant King, Oct 26 2012: (Start)
a(n) = a(n-1) + n*(3*n-2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = n*A000326(n) - A002411(n-1), see Berselli's comment.
a(n) = (n+1)*(2*A000567(n)+n)/6.
a(n) = A000292(n) + 5*A000292(n-1) = binomial(n+2,3)+5*binomial(n+1,3).
a(n) = A002413(n) + A000292(n-1).
a(n) = A000217(n) + 6*A000292(n-1).
Sum_{n>=1} 1/a(n) = 2*(4*log(2)-1)/3 = 1.1817258148265...
(End)
a(n) = Sum_{i=0..n-1} (n-i)*(6*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014
E.g.f.: x*(2 + 7*x + 2*x^2)*exp(x)/2. - Ilya Gutkovskiy, May 23 2016
a(n) = A080851(6,n-1). - R. J. Mathar, Jul 28 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(Pi + 1 - 4*log(2))/3. - Amiram Eldar, Jul 02 2020

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 09 2000
Incorrect formula deleted by Ant King, Oct 04 2012

A156927 FP3 polynomials related to the generating functions of the columns of the A156921 matrix.

Original entry on oeis.org

1, 1, 1, -6, 29, 31, -283, 245, 298, -286, -108, 119, -3106, 29469, -104585, -220481, 3601363, -15487305, 34949165, -39821950, 4356011, 46881744, -51274736, 9005908, 14663472, -5205168, -1456704, -20736
Offset: 0

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Author

Johannes W. Meijer, Feb 20 2009

Keywords

Comments

For the matrix of the FP1 polynomials see A156921. The coefficients in the columns of this matrix are the powers of z^m, m=0, 1, 2, ... . The columns are numbered 1, 2, 3... .
The GF3(z;m) generate the sequences of the z^m coefficients. The general structure of the GF3(z;m) is given below.
The FP3(z,m) in the numerator of the GF3(z;m) is a polynomial of a certain degree, let's say k3. The (k3+1) coefficients of this polynomial can be determined one by one by comparing the series expansion of the FP3(z,m) with the values of the powers of z^m in column (m+1). These values can be generated with the GF1 formulas, see A156921.
An appropriate name for the polynomials FP3(z;m) in the numerators of the GF(3;m) seems to be the flower polynomials of the third kind, the FP3, because the zero patterns of these polynomials look like flowers. The zero patterns of the FP3 and the FP4, see A156933, resemble each other closely and look like the zero patterns of the FP1 and FP2.
The sequence of the (k3+1) number of terms of the FP3(z;m) polynomials for m from 0 to 11 is 1, 2, 8, 17, 29, 45, 63, 84, 109, 137, 167, 200.

Examples

			The first few rows of the "triangle" of the FP3(z,m) coefficients are:
  [1]
  [1, 1]
  [-6, 29, 31, -283, 245, 298, -286, -108]
The first few FP3 polynomials are:
  FP3(z; m=0) = 1
  FP3(z; m=1) = (1+z)
  FP3(z; m=2) = (-6+29*z+31*z^2-283*z^3+245*z^4+298*z^5-286*z^6-108*z^7)
Some GF3(z;m) are:
  GF3(z;m=1) = z^2*(1+z)/((1-z)^4*(1-2*z))
  GF3(z;m=2) = z^2*(-6+29*z+31*z^2-283*z^3+245*z^4+298*z^5-286*z^6-108*z^7)/((1-z)^7*(1-2*z)^4*(1-3*z))
		

Crossrefs

For the first few GF3's see A156928, A156929, A156930, A156931.
Row sums A156932.
For the polynomials in the denominators of the GF3(z;m) see A157704.

Formula

G.f.: GF3(z;m):= z^p*FP3(z;m)/Product_{k=0..m} (1-(k+1)*z)^(1+3*k).

A064350 a(n) = (3*n)!/n!.

Original entry on oeis.org

1, 6, 360, 60480, 19958400, 10897286400, 8892185702400, 10137091700736000, 15388105201717248000, 30006805143348633600000, 73096577329197271449600000, 217535414131691079834009600000, 776601428450137155007414272000000, 3275704825202678519821273399296000000
Offset: 0

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Author

Karol A. Penson, Sep 18 2001

Keywords

Comments

Also a(n) = (((n)!)^2)*A006480(n). [corrected by Johannes W. Meijer, Mar 02 2009]
a(n) is the number of ways to partition the set {1,2,...,3n} into n blocks of size 3 and then linearly order the elements within each block. - Geoffrey Critzer, Dec 30 2012

Crossrefs

From Johannes W. Meijer, Mar 07 2009: (Start)
Equals A001525*3!
Equals row sums of A157704 and A157705. (End)

Programs

  • Mathematica
    Table[(3n)!/n!,{n,0,20}]  (* Geoffrey Critzer, Dec 30 2012 *)
  • PARI
    { t=f=1; for (n=0, 70, if (n, t*=3*n*(3*n - 1)*(3*n - 2); f*=n); write("b064350.txt", n, " ", t/f) ) } \\ Harry J. Smith, Sep 12 2009

Formula

Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*BesselK(1/3, 2*sqrt(x/27))/(3*Pi*sqrt(x)) dx, n >= 0.
A recursive formula: a(n) = (27 * (n - 1)^2 + 27 * (n - 1) + 6) * a(n - 1) with a(0) = 1. An explicit formula following from the recursion equation: a(n) = (3/2)*27^n*GAMMA(n+2/3)*GAMMA(n+1/3)/(Pi*3^(1/2)). - Thomas Wieder, Nov 15 2004
E.g.f.: (of aerated sequence) 2*cos(arcsin((3*sqrt(3)*x/2)/3))/sqrt(4-27*x^2). - Paul Barry, Jul 27 2010
E.g.f.: (with interpolated zeros) exp(x^3). - Geoffrey Critzer, Dec 30 2012
Sum_{n>=1} 1/a(n) = A248759. - Amiram Eldar, Nov 15 2020

Extensions

a(11) from Harry J. Smith, Sep 12 2009

A157705 G.f.s of the z^p coefficients of the polynomials in the GF4 denominators of A156933.

Original entry on oeis.org

1, 1, 3, 2, 18, 128, 171, 42, 1, 22, 1348, 11738, 26293, 17693, 3271, 115, 13, 6122, 228986, 2070813, 6324083, 7397855, 3361536, 544247, 24590, 155, 3, 17248, 2413434, 67035224, 612026240, 2274148882
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF4(z;n) polynomials in the GF4 denominators of A156933 can be found below.
The general structure of the GFKT4(z;p) that generate the z^p coefficients of the PDGF4(z;n) polynomials can also be found below. The KT4(z;p) polynomials in the numerators of the GFKT4(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT4(z;p) polynomials is 1, 3, 5, 7, 10, 13, 15, 18, 20, 23, 26, 29, 32, 34, 37, 40, 42. The differences of this sequence and that of the number of terms of the KT3(z;p), see A157704, follow a simple pattern.
A Maple algorithm that generates relevant GFKT4(z;p) information can be found below.

Examples

			Some PDGF4 (z;n) are:
  PDGF4(z; n=3) = (1-7*z)*(1-5*z)^4*(1-3*z)^7*(1-z)^10
  PDGF4(z; n=4) = (1-9*z)*(1-7*z)^4*(1-5*z)^7*(1-3*z)^10*(1-z)^13
The first few GFKT4's are:
  GFKT4(z;p=0) = 1/(1-z)
  GFKT4(z;p=1) = -(1+3*z+2*z^2)/(1-z)^4
  GFKT4(z;p=2) = z*(18+128*z+171*z^2+42*z^3+z^4)/(1-z)^7
Some KT4(z,p) polynomials are:
  KT4(z;p=2) = 18+128*z+171*z^2+42*z^3+z^4
  KT4(z;p=3) = 22+1348*z+11738*z^2+26293*z^3+17693*z^4+3271*z^5+115*z^6
		

Crossrefs

Originator sequence A156933.
See A081436 for the z^1 coefficients and A157708 for the z^2 coefficients.
Row sums equal A064350 and those of A157704.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(2*n2+1-(2*k))*z)^(3*k+1), k=0..n2): a(n2):= coeff(fz(n2),z,p): end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)= sort (simplify(fk)); GFKT4(p):=sum((fk)*z^k,k=0..infinity); q4:=ldegree((numer (GFKT4(p)))): KT4(p):=sort((-1)^(p)*simplify((GFKT4(p)*(1-z)^(3*p+1))/z^q4),z, ascending);

Formula

PDGF4(z;n) = Product_{k=0..n} (1-(2*n+1-2*k)*z)^(3*k+1) with n = 1, 2, 3, ...
GFKT4(z;p) = (-1)^(p)*(z^q4)*KT4(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A157702 G.f.s of the z^p coefficients of the polynomials in the GF1 denominators of A156921.

Original entry on oeis.org

1, 1, 1, 7, 26, 7, 3, 166, 951, 951, 166, 3, 263, 8999, 59637, 108602, 59637, 8999, 263, 174, 33124, 848555, 6062651, 15477896, 15477896, 6062651, 848555, 33124, 174, 45, 66963, 5856626, 122966782, 920090513
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF1(z;n) polynomials in the GF1 denominators of A156921 can be found below.
The general structure of the GFKT1(z;p) that generate the z^p coefficients of the PDGF1(z; n) polynomials can also be found below. The KT1(z;p) polynomials in the numerators of the GFKT1(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT1(z;p) polynomials is: 1, 2, 3, 6, 7, 10, 13, 14, 17, 20, 23, 24, 27, 30, 33, 36, 37, 40. The first differences follow a simple pattern. The positions of the 1's follow the Lazy Caterer's sequence A000124.
A Maple algorithm that generates relevant GFKT1(z;p) information can be found below.

Examples

			Some PDGF1 (z;n) are:
  PDGF1(z;n=3) = (1-5*z)*(1-3*z)^2*(1-z)^3
  PDGF1(z;n=4) = ((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)
The first few GFKT1's are:
  GFKT1(z;p=0) = 1/(1-z)
  GFKT1(z;p=1) = -z*(1+z)/(1-z)^4
  GFKT1(z;p=2) = z^2*(7+26*z+7*z^2)/(1-z)^7
Some KT1(z;p) polynomials are:
  KT1(z;p=2) = 7+26*z+7*z^2
  KT1(z;p=3) = 3+166*z+951*z^2+951*z^3+166*z^4+3*z^5
  KT1(z;p=4) = 263+8999*z+59637*z^2+108602*z^3+59637*z^4+8999*z^5+263*z^6
		

Crossrefs

Originator sequence A156921.
See A000330 for the z^1 coefficients and A157706 for the z^2 coefficients.
Row sums equal A052502.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn, a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(2*m-1)*z)^(n2+1-m),m=1..n2); a(n2):= coeff(fz(n2),z,p); end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT1(p):=(sum(fk*z^k,k=0..infinity)); q1:=ldegree((numer(GFKT1(p)))): KT1(p):=sort((-1)^p*simplify((GFKT1(p))*(1-z)^(3*p+1)/z^q1),z, ascending);

Formula

PDGF1(z;n) = Product_{m=1..n} (1-(2*m-1)*z)^(n+1-m) with n = 1, 2, 3, ... .
GFKT1(z;p) = (-1)^(p)*(z^q1)*KT1(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ... .
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A157703 G.f.s of the z^p coefficients of the polynomials in the GF2 denominators of A156925.

Original entry on oeis.org

1, 1, 5, 5, 2, 62, 152, 62, 2, 91, 1652, 5957, 5957, 1652, 91, 52, 5240, 77630, 342188, 551180, 342188, 77630, 5240, 52, 12, 8549, 424921, 5629615, 28123559, 61108544, 61108544, 28123559, 5629615, 424921, 8549, 12
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF2(z;n) polynomials in the GF2 denominators of A156925 can be found below.
The general structure of the GFKT2(z;p) that generate the z^p coefficients of the PDGF2(z; n) polynomials can also be found below. The KT2(z;p) polynomials in the numerators of the GFKT2(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT2(z;p) polynomials is: 1, 1, 2, 5, 6, 9, 12, 13, 16, 19, 22, 23, 26. The first differences follow a simple pattern. The positions of the 1's follow the Lazy Caterer's sequence A000124 with one exception, here a(0) = 0.
A Maple algorithm that generates relevant GFKT2(z;p) information can be found below.

Examples

			Some PDGF2 (z;n) are:
  PDGF2(z;n=3) = (1-z)^3*(1-2*z)^2*(1-3*z)
  PDGF2(z;n=4) = (1-z)^4*(1-2*z)^3*(1-3*z)^2*(1-4*z)
The first few GFKT2's are:
  GFKT2(z;p=0) = 1/(1-z)
  GFKT2(z;p=1) = -z/(z-1)^4
  GFKT2(z;p=2) = z^2*(5+5*z)/(1-z)^7
Some KT2(z,p) polynomials are:
  KT2(z;p=2) = 5+5*z
  KT2(z;p=3) = 2+62*z+152*z^2+62*z^3+2*z^4
  KT2(z;p=4) = 91+1652*z+5957*z^2+5957*z^3+1652*z^4+91*z^5
		

Crossrefs

Originator sequence A156925.
See A000292 for the z^1 coefficients and A040977 for the z^2 coefficients divided by 5.
Row sums equal A025035.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-m*z)^(n2+1-m),m=1..n2): a(n2):= coeff(fz(n2),z,p): end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT2(p):=sum((fk)*z^k,k=0..infinity); q2:=ldegree((numer(GFKT2(p)))): KT2(p):=sort((-1)^p*simplify((GFKT2(p)*(1-z)^(3*p+1))/z^q2),z, ascending);

Formula

PDGF2(z;n) = Product_{m=1..n} (1-m*z)^(n+1-m) with n = 1, 2, 3, ...
GFKT2(z;p) = (-1)^(p)*(z^q2)*KT2(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A001525 a(n) = (3n)!/(3!n!).

Original entry on oeis.org

1, 60, 10080, 3326400, 1816214400, 1482030950400, 1689515283456000, 2564684200286208000, 5001134190558105600000, 12182762888199545241600000, 36255902355281846639001600000, 129433571408356192501235712000000, 545950804200446419970212233216000000
Offset: 1

Views

Author

Keywords

References

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

Crossrefs

Equals A064350/3!, row sums of A157704/3! and row sums of A157705/3!.

Programs

  • Mathematica
    Table[(3 n)!/(3! n!), {n, 20}] (* Harvey P. Dale, May 25 2011 *)

Extensions

More terms from Harvey P. Dale, May 25 2011

A157707 The z^2 coefficients of the polynomials in the GF3 denominators of A156927 divided by 2.

Original entry on oeis.org

16, 205, 1165, 4415, 13055, 32606, 72030, 144930, 270930, 477235, 800371, 1288105, 2001545, 3017420, 4430540, 6356436, 8934180, 12329385, 16737385, 22386595, 29542051, 38509130, 49637450, 63324950
Offset: 1

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

See A157704 for background information.

Crossrefs

Programs

  • Maple
    nmax:=24; for n from 0 to nmax do fz(n):=product((1-(k+1)*z)^(1+3*k),k=0..n); c(n):= coeff(fz(n),z,2)/2; end do: a:=n-> c(n): seq(a(n), n=1..nmax);

Formula

a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7)
a(n) = 1/4*n^6+7/4*n^5+37/8*n^4+34/6*n^3+25/8*n^2+7/12*n
G.f.: (16 + 93*z + 66*z^2 + 5*z^3)/(1-z)^7
Showing 1-8 of 8 results.