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 1-9 of 9 results.

A286718 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
Offset: 0

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Author

Wolfdieter Lang, May 18 2017

Keywords

Comments

This is a generalization of the unsigned Stirling1 triangle A132393.
In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018

Examples

			The triangle T(n, k) begins:
n\k        0        1        2       3      4     5    6  7 8 ...
O:         1
1:         1        1
2:         4        5        1
3:        28       39       12       1
4:       280      418      159      22      1
5:      3640     5714     2485     445     35     1
6:     58240    95064    45474    9605   1005    51    1
7:   1106560  1864456   959070  227969  28700  1974   70  1
8:  24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
From _Wolfdieter Lang_, Aug 09 2017: (Start)
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Crossrefs

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
Column sequences for k = 0..4: A007559, A024216(n-1), A286721(n-2), A382984, A382985.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).

Programs

  • Mathematica
    T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[, -1] = 0; T[0, 0] = 1; T[n, k_] /; nJean-François Alcover, Jun 20 2018 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0,a_1,...,a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017

A225468 Triangle read by rows, S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 1, 4, 7, 1, 8, 39, 15, 1, 16, 203, 159, 26, 1, 32, 1031, 1475, 445, 40, 1, 64, 5187, 12831, 6370, 1005, 57, 1, 128, 25999, 107835, 82901, 20440, 1974, 77, 1, 256, 130123, 888679, 1019746, 369061, 53998, 3514, 100, 1
Offset: 0

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Author

Peter Luschny, May 16 2013

Keywords

Comments

The definition of the Stirling-Frobenius subset numbers: S_m(n, k) = (Sum_{j=0..n} binomial(j, n-k)*A_m(n, j)) / (m^k*k!) where A_m(n, j) are the generalized Eulerian numbers. For m = 1 this gives the classical Stirling set numbers A048993. (See the links for details.)
From Peter Bala, Jan 27 2015: (Start)
Exponential Riordan array [ exp(2*z), 1/3*(exp(3*z) - 1)].
Triangle equals P * A111577 = P^(-1) * A075498, where P is Pascal's triangle A007318.
Triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*3^n*binomial((x - 2)/3,n) }n>=0. An example is given below.
This triangle is the particular case a = 3, b = 0, c = 2 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)

Examples

			[n\k][ 0,    1,     2,    3,    4,  5,  6]
[0]    1,
[1]    2,    1,
[2]    4,    7,     1,
[3]    8,   39,    15,    1,
[4]   16,  203,   159,   26,    1,
[5]   32, 1031,  1475,  445,   40,  1,
[6]   64, 5187, 12831, 6370, 1005, 57,  1.
Connection constants: Row 3: [8, 39, 15, 1] so
x^3 = 8 + 39*(x - 2) + 15*(x - 2)*(x - 5) + (x - 2)*(x - 5)*(x - 8). - _Peter Bala_, Jan 27 2015
		

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225469 (m=4).

Programs

  • Maple
    SF_S := proc(n, k, m) option remember;
    if n = 0 and k = 0 then return(1) fi;
    if k > n or k < 0 then return(0) fi;
    SF_S(n-1, k-1, m) + (m*(k+1)-1)*SF_S(n-1, k, m) end:
    seq(print(seq(SF_S(n, k, 3), k=0..n)), n = 0..5);
  • Mathematica
    EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m) + (m*k+1)*EulerianNumber(n-1,k,m)
    def SF_S(n, k, m):
        return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/ (factorial(k)*m^k)
    for n in (0..6): [SF_S(n, k, 3) for k in (0..n)]

Formula

T(n, k) = (Sum_{j=0..n} binomial(j, n-k)*A_3(n, j)) / (3^k*k!) with A_3(n,j) = A225117.
For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A016127; T(n, 2) ~ A016297; T(n, 3) ~ A025999;
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024212.
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i)*3^(i-k)*binomial(n,i)*Stirling2(i+1,k+1).
E.g.f.: exp(2*z)*exp(x/3*(exp(3*z) - 1)) = 1 + (2 + x)*z + (4 + 7*x + x^2)*z^2/2! + ....
T(n,k) = 1/(3^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(3*j + 2)^n.
O.g.f. for n-th diagonal: exp(-2*x/3)*Sum_{k >= 0} (3*k + 2)^(k + n - 1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k: 1/( (1 - 2*x)*(1 - 5*x)...(1 - (3*k + 2)*x) ). (End)
E.g.f. column k: exp(2*x)*((exp(3*x) - 1)/3)^k, k >= 0. See the Bala link for the S(3,0,2) exponential Riordan aka Sheffer triangle. - Wolfdieter Lang, Apr 10 2017

A024216 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 1 mod 3.

Original entry on oeis.org

1, 5, 39, 418, 5714, 95064, 1864456, 42124592, 1077459120, 30777463360, 971142388160, 33547112941440, 1259204418129280, 51032742579123200, 2220990565060377600, 103308619261574809600, 5114702794181847910400
Offset: 0

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Author

Keywords

Comments

Comment by R. J. Mathar, Oct 01 2016: (Start)
The k-th elementary symmetric functions of the integers 1+j*3, j=0..n-1, form a triangle T(n,k), 0 <= k <= n, n >= 0:
1
1 1
1 5 4
1 12 39 28
1 22 159 418 280
1 35 445 2485 5714 3640
1 51 1005 9605 45474 95064 58240
1 70 1974 28700 227969 959070 1864456 1106560
1 92 3514 72128 859369 5974388 22963996 42124592 24344320
This here is the first subdiagonal. The diagonal seems to be A007559. The first columns are A000012, A000326, A024212, A024213, A024214. (End)

Examples

			From _Gheorghe Coserea_, Dec 24 2015: (Start)
For n = 1 we have a(1) = 1*4*(1/1 + 1/4) = 5.
For n = 2 we have a(2) = 1*4*7*(1/1 + 1/4 + 1/7) = 39.
For n = 3 we have a(3) = 1*4*7*10*(1/1 + 1/4 + 1/7 + 1/10) = 418.
(End)
		

Crossrefs

Cf. A024395, A024382, A286718 (first column).

Programs

  • Magma
    I:=[5,39]; [1] cat [n le 2 select I[n] else (6*n-1) * Self(n-1) - (3*n-2)^2 * Self(n-2) : n in [1..30]]; // Vincenzo Librandi, Aug 30 2015
  • Maple
    f:= gfun:-rectoproc({-(3*n+1)^2*a(n-1)+(6*n+5)*a(n)-a(n+1), a(0) = 1, a(1) = 5, a(2) = 39}, a(n), remember):
    map(f, [$0..30]); # Robert Israel, Aug 30 2015
  • Mathematica
    Rest[CoefficientList[Series[-(1/3)*Log[1-3*x]/(1-3*x)^(1/3), {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Oct 07 2013 *)
  • PARI
    n = 33; a = vector(n); a[1] = 5; a[2] = 39;
    for (k = 2, n-1, a[k+1] = (6*k+5) * a[k] - (3*k+1)^2 * a[k-1]);
    print(concat(1,a));  \\ Gheorghe Coserea, Aug 29 2015
    

Formula

E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(1/3). - Vladeta Jovovic, Sep 26 2003
For n >= 1, a(n-1) = 3^(n-1)*n!*Sum_{k=0..n-1} binomial(k-2/3, k)/(n-k). - Milan Janjic, Dec 14 2008, corrected by Peter Bala, Oct 08 2013
a(n) ~ (n+1)! * GAMMA(2/3) * 3^(n+3/2) * (log(n) + gamma + Pi*sqrt(3)/6 + 3*log(3)/2) / (6*Pi*n^(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 07 2013
a(n+1) = (6*n+5) * a(n) - (3*n+1)^2 * a(n-1). - Gheorghe Coserea, Aug 29 2015
E.g.f.: (3 - log(1-3*x))/(3*(1-3*x)^(4/3)). - Robert Israel, Aug 30 2015
a(n) = A286718(n+1, 1), n >= 0.
Boas-Buck type recurrence: a(0) = 1 and for n >= 1: a(n) = ((n+1)!/n) * Sum_{p=1..n} 3^(n-p)*(1 + 3*beta(n-p))*a(p-1)/p!, with beta(k) = A002208(k+1) / A002209(k+1). Proof from a(n) = A286718(n+1, 1). - Wolfdieter Lang, Aug 09 2017

Extensions

More terms from Vladeta Jovovic, Sep 26 2003

A096038 Triangle T(n,m) = (3*n^2-3*m^2+5*m-4+n)/2 read by rows.

Original entry on oeis.org

1, 6, 4, 14, 12, 7, 25, 23, 18, 10, 39, 37, 32, 24, 13, 56, 54, 49, 41, 30, 16, 76, 74, 69, 61, 50, 36, 19, 99, 97, 92, 84, 73, 59, 42, 22, 125, 123, 118, 110, 99, 85, 68, 48, 25, 154, 152, 147, 139, 128, 114, 97, 77, 54, 28, 186, 184, 179, 171, 160, 146, 129, 109, 86, 60, 31
Offset: 1

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Author

Gary W. Adamson, Jun 17 2004

Keywords

Comments

The triangle is obtained by subtracting the triangle A094930 from
its square root (also described in A094930) and then dividing each element of column m through 3*m-1.
For the first three rows n=1 to 3 this yields for example:
4;.................2;............2......................1;
14,25;......minus..2,5;.......=..12,20;......->.divide..6,4;
30,65,64;..........2,5,8;........28,60,56;..............14;12,7;

Crossrefs

Programs

  • Python
    def A096038(n,m):
        return (3*n**2-3*m**2+5*m-4+n)//2
    print( [A096038(n,m) for n in range(20) for m in range(1,n+1)] )
    # R. J. Mathar, Oct 11 2009

Formula

T(n,1) = A095794(n).
T(n,n) = 3*n-2.
T(n,m) = A094930(n,m)/(3*m-1)-1.

Extensions

Edited, T(3,2) corrected, and extended by R. J. Mathar, Oct 11 2009

A116689 Partial sums of dodecahedral numbers (A006566).

Original entry on oeis.org

0, 1, 21, 105, 325, 780, 1596, 2926, 4950, 7875, 11935, 17391, 24531, 33670, 45150, 59340, 76636, 97461, 122265, 151525, 185745, 225456, 271216, 323610, 383250, 450775, 526851, 612171, 707455, 813450, 930930, 1060696, 1203576, 1360425
Offset: 0

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Author

Jonathan Vos Post, Mar 15 2006

Keywords

Comments

Geometrically, the partial sums of dodecahedral numbers may be interpreted as 4-dimensional dodecahedral hyperpyramidal numbers. It is somewhat surprising that this is (with proper offset) the triangular number of the "second pentagonal numbers, minus 1."
Also, the sequence is related to A004188 by a(n) = n*A004188(n)-sum(A004188(i), i=0..n-1). [Bruno Berselli, Apr 05 2012]

Crossrefs

Programs

Formula

a(n) = Sum_{i = 0..n} A006566(i).
a(n) = Sum_{i = 0..n} i*(3*i-1)*(3*i-2)/2.
a(n+1) = A000217(A095794(n)).
a(n+1) = A000217(A005449(n) - 1).
a(n+1) = A000217(n*(3n+1)/2-1).
a(n+1) = A000217(A001844(n) - A000217(n+1) - 1).
a(n) = n*(9*n^3+6*n^2-5*n-2)/8. G.f.: x*(1+16*x+10*x^2)/(1-x)^5. [Colin Barker, Apr 04 2012]
a(n) = binomial(A005449(n), 2). - Wesley Ivan Hurt, Oct 06 2013
From Peter Bala, Sep 03 2023: (Start)
a(n) = n*(n + 1)*(3*n + 1)*(3*n - 2)/8.
a(n) = Sum_{0 <= i <= j <= n-1} (3*i + 1)*(3*j + 1). Cf. A024212 (End)

A094930 Triangle T(n,m) read by rows, defined by squaring a matrix with row entries 2+3*(m-1).

Original entry on oeis.org

4, 14, 25, 30, 65, 64, 52, 120, 152, 121, 80, 190, 264, 275, 196, 114, 275, 400, 462, 434, 289, 154, 375, 560, 682, 714, 629, 400, 200, 490, 744, 935, 1036, 1020, 860, 529, 252, 620, 952, 1221, 1400, 1462, 1380, 1127, 676, 310, 765, 1184, 1540, 1806, 1955, 1960
Offset: 1

Views

Author

Gary W. Adamson, Jun 17 2004

Keywords

Comments

Matrix square of the matrix B(n,m) = 2+3*(m-1), B containing the first terms of A016789
in its row n, n>0, 1<=m<=n.

Examples

			The matrix B starts as
  2 ;
  2,5 ;
  2,5,8 ;
  2,5,8,11 ;
  2,5,8,11,14 ;
and interpreting this as a lower triangular matrix, its square T = B^2 starts
  4;
  14,25;
  30,65,64;
  52,120,152,121;
		

Crossrefs

Programs

  • Maple
    A094930 := proc(n,m) (3*m-1)*(3*m+3*n-2)*(n+1-m)/2 ; end: seq(seq(A094930(n,m),m=1..n),n=1..20) ; # R. J. Mathar, Oct 09 2009

Formula

T(n,m) = sum_{k=m..n} B(n,k)*B(k,m) = (3*m-1)*(3*m+3*n-2)*(n+1-m)/2.
Row sums: sum_{m=1..n} T(n,m) = A024212(n).
G.f. as triangle: x*y*(4+2*x+13*x*y-16*x^2*y+x^2*y^2-4*x^3*y^2)/((1-x)*(1-x*y))^3. - Robert Israel, May 06 2019

Extensions

Edited and extended by R. J. Mathar, Oct 09 2009

A096037 Triangle T(n,m) = (3*n+3*m-2)*(n+1-m)/2 read by rows.

Original entry on oeis.org

2, 7, 5, 15, 13, 8, 26, 24, 19, 11, 40, 38, 33, 25, 14, 57, 55, 50, 42, 31, 17, 77, 75, 70, 62, 51, 37, 20, 100, 98, 93, 85, 74, 60, 43, 23, 126, 124, 119, 111, 100, 86, 69, 49, 26, 155, 153, 148, 140, 129, 115, 98, 78, 55, 29, 187, 185, 180, 172, 161, 147, 130, 110, 87, 61, 32
Offset: 1

Views

Author

Gary W. Adamson, Jun 17 2004

Keywords

Examples

			The triangle starts in row n=1 as
2;
7,5;
15,13,8;
26,24,19,11;
		

Crossrefs

Programs

  • Python
    def A096037(n,m):
        return (3*n+3*m-2)*(n+1-m)//2
    print( [A096037(n,m) for n in range(20) for m in range(1,n+1)] )
    # R. J. Mathar, Oct 11 2009

Formula

T(n,m) = (3*n+3*m-2)*(n+1-m)/2 .
T(n,m) = A094930(n,m)/(3*m-1).
T(n,1) = A005449(n).
T(n,n) = A016768(n-1).
Row sums: sum_{m=1..n} T(n,m) = n^2*(n+1) = A011379(n).

Extensions

Edited and extended, A-numbers corrected by R. J. Mathar, Oct 11 2009

A024219 a(n) = floor( (2nd elementary symmetric function of S(n))/(first elementary symmetric function of S(n)) ), where S(n) = {first n+1 positive integers congruent to 1 mod 3}.

Original entry on oeis.org

0, 3, 7, 12, 19, 28, 38, 49, 62, 77, 93, 110, 129, 150, 172, 195, 220, 247, 275, 304, 335, 368, 402, 437, 474, 513, 553, 594, 637, 682, 728, 775, 824, 875, 927, 980, 1035, 1092, 1150, 1209, 1270, 1333, 1397, 1462, 1529, 1598, 1668, 1739, 1812, 1887, 1963, 2040
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3,-4,4,-3,1},{0,3,7,12,19},60] (* Harvey P. Dale, May 20 2019 *)
  • PARI
    a(n)=floor(sum(j=0, n, sum(k=j+1, n, (3*j+1)*(3*k+1)))/sum(i=0, n, (3*i+1))) \\ Andrew Howroyd, Aug 12 2018
    
  • PARI
    a(n) = floor(n*(9*n^2+9*n-2)/(4*(3*n+2))); \\ Andrew Howroyd, Aug 12 2018

Formula

From R. J. Mathar, Oct 08 2011: (Start)
Conjecture: a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5);
g.f.: x^2*(-3+2*x-3*x^2+x^3) / ( (x^2+1)*(x-1)^3 ). (End)
From Andrew Howroyd, Aug 12 2018: (Start)
The above conjectures are true.
a(n) = floor(A024212(n) / A000326(n+1)).
a(n) = floor(n*(9*n^2 + 9*n - 2)/(4*(3*n + 2))).
(End)

Extensions

More terms from Joshua Zucker, May 20 2006

A024223 [ (4th elementary symmetric function of S(n))/(2nd elementary symmetric function of S(n)) ], where S(n) = {first n+3 positive integers congruent to 1 mod 3}.

Original entry on oeis.org

1, 12, 45, 115, 244, 457, 785, 1262, 1926, 2822, 3997, 5504, 7400, 9746, 12608, 16057, 20168, 25021, 30699, 37290, 44888, 53591, 63500, 74722, 87368, 101553, 117397, 135026, 154567, 176154, 199926, 226024, 254597, 285795, 319774, 356694, 396722, 440026
Offset: 1

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Formula

a(n) = floor(A024214(n) / A024212(n+2)). - Sean A. Irvine, Jun 25 2019
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