A000914 -id:A000914 - cp's OEIS frontend

A346415 Triangle T(n,k), n>=0, 0<=k<=n, read by rows, where column k is (1/k!) times the k-fold exponential convolution of Fibonacci numbers with themselves.

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 3, 11, 6, 1, 0, 5, 35, 35, 10, 1, 0, 8, 115, 180, 85, 15, 1, 0, 13, 371, 910, 630, 175, 21, 1, 0, 21, 1203, 4494, 4445, 1750, 322, 28, 1, 0, 34, 3891, 22049, 30282, 16275, 4158, 546, 36, 1, 0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1

Offset: 0

Alois P. Heinz, Jul 15 2021

The sequence of column k>0 satisfies a linear recurrence with constant coefficients of order k+1.

			Triangle T(n,k) begins:
  1;
  0,  1;
  0,  1,     1;
  0,  2,     3,      1;
  0,  3,    11,      6,      1;
  0,  5,    35,     35,     10,      1;
  0,  8,   115,    180,     85,     15,     1;
  0, 13,   371,    910,    630,    175,    21,    1;
  0, 21,  1203,   4494,   4445,   1750,   322,   28,   1;
  0, 34,  3891,  22049,  30282,  16275,  4158,  546,  36,  1;
  0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-4 give: A000007, A000045, A014335, A014337, A014341.
T(n+j,n) for j=0-2 give: A000012, A000217, A000914.
Row sums give A256180.

b:= proc(n) option remember; `if`(n=0, 1, add(expand(x*b(n-j)
      *binomial(n-1, j-1)*(<<0|1>, <1|1>>^j)[1, 2]), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..12);
# second Maple program:
b:= proc(n, k) option remember; `if`(k=0, 0^n, `if`(k=1,
       combinat[fibonacci](n), (q-> add(binomial(n, j)*
       b(j, q)*b(n-j, k-q), j=0..n))(iquo(k, 2))))
    end:
T:= (n, k)-> b(n, k)/k!:
seq(seq(T(n, k), k=0..n), n=0..12);

b[n_, k_] := b[n, k] = If[k == 0, 0^n, If[k == 1, Fibonacci[n], With[{q = Quotient[k, 2]}, Sum[Binomial[n, j] b[j, q] b[n-j, k-q], {j, 0, n}]]]];
T[n_, k_] := b[n, k]/k!;
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Nov 06 2021, after 2nd Maple program *)

A347107 a(n) = Sum_{1 <= i < j <= n} j^3*i^3.

Original entry on oeis.org

0, 0, 8, 251, 2555, 15055, 63655, 214918, 616326, 1561110, 3586110, 7612385, 15139553, 28506101, 51229165, 88438540, 147420940, 238291788, 374813076, 575377095, 864177095, 1272587195, 1840775123, 2619572626, 3672629650, 5078879650, 6935344650, 9360309933

Offset: 0

Roudy El Haddad, Jan 27 2022

a(n) is the sum of all products of two distinct cubes of positive integers up to n, i.e., the sum of all products of two distinct elements from the set of cubes {1^3, ..., n^3}.

			For n=3, a(3) = (2*1)^3+(3*1)^3+(3*2)^3 = 251.

Roudy El Haddad, Multiple Sums and Partition Identities, arXiv:2102.00821 [math.CO], 2021.
Roudy El Haddad, A generalization of multiple zeta value. Part 2: Multiple sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 200-233, DOI: 10.7546/nntdm.2022.28.2.200-233.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000537, A000578.
Cf. A346642 (for nondistinct cubes).
Cf. A000217 (for power 0), A000914 (for power 1), A000596 (for squares).

CoefficientList[Series[-(x^5 + 64 x^4 + 424 x^3 + 584 x^2 + 179 x + 8) x^2/(x - 1)^9, {x, 0, 27}], x] (* Michael De Vlieger, Feb 04 2022 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,0,8,251,2555,15055,63655,214918,616326},30] (* Harvey P. Dale, Jul 07 2025 *)

a(n) = sum(i=2, n, sum(j=1, i-1, i^3*j^3));

{a(n) = n*(n+1)*(n-1)*(21*n^5+36*n^4-21*n^3-48*n^2+8)/672};

def A347107(n): return n*(n**2*(n*(n*(n*(n*(21*n + 36) - 42) - 84) + 21) + 56) - 8)//672 # Chai Wah Wu, Feb 17 2022

a(n) = Sum_{j=2..n} Sum_{i=1..j-1} j^3*i^3.
a(n) = n*(n+1)*(n-1)*(21*n^5+36*n^4-21*n^3-48*n^2+8)/672 (from the generalized form of Faulhaber's formula).
From Alois P. Heinz, Jan 27 2022: (Start)
a(n) = Sum_{i=1..n} A000578(i)*A000537(i-1) = Sum_{i=1..n} i^3*(i*(i-1)/2)^2.
G.f.: -(x^5+64*x^4+424*x^3+584*x^2+179*x+8)*x^2/(x-1)^9. (End)

A103115 a(n) = 6n(n-1) - 1.

Original entry on oeis.org

-1, -1, 11, 35, 71, 119, 179, 251, 335, 431, 539, 659, 791, 935, 1091, 1259, 1439, 1631, 1835, 2051, 2279, 2519, 2771, 3035, 3311, 3599, 3899, 4211, 4535, 4871, 5219, 5579, 5951, 6335, 6731, 7139, 7559, 7991, 8435, 8891, 9359, 9839, 10331, 10835, 11351, 11879, 12419

Offset: 0

Jacob Landon (jacoblandon(AT)aol.com), May 09 2009

Star numbers A003154 minus 2.
What A163433 does for a triangle, this sequence is doing for a square but giving one-half the results. Take a square with vertices n, n+1, n+2, and n+3 and find the sum of the four products of each four vertices times the sum of the other three; at n you have n((n+1)+(n+2)+(n+3)) and so on for the other three vertices. The result of all four is 12*n^2 + 36*n + 22; half this is 6*n^2 + 18*n + 11 and gives the numbers in this sequence starting with n = 0. - J. M. Bergot, May 23 2012
Multiplying a(n) by 16 gives the sum of the convolution with itself of each of the 24 permutations of four consecutive numbers. - J. M. Bergot, May 15 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000914, A003154, A163433.

[6*n*(n-1)-1: n in [0..50]]; // Vincenzo Librandi, May 16 2017

Table[6n(n-1)-1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{-1,-1,11},50] (* Harvey P. Dale, Nov 14 2011 *)
CoefficientList[Series[(1 - 2 x - 11 x^2) / (x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, May 16 2017 *)

a(n)=6*n*(n-1)-1 \\ Charles R Greathouse IV, Jun 16 2017

a(n) = A003154(n) - 2.
G.f.: (1-2*x-11*x^2)/(x-1)^3. - R. J. Mathar, May 11 2009 (adapted by Vincenzo Librandi, May 16 2017).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=-1, a(1)=-1, a(2)=11. - Harvey P. Dale, Nov 14 2011
a(n) = (n-2)*(n-1 + n + n+1) + (n-1)*(n + n+1) + n*(n+1), which is applying A000914 to four consecutive numbers. - J. M. Bergot, May 15 2017
Sum_{n>=1} 1/a(n) = tan(sqrt(5/3)*Pi/2)*Pi/(2*sqrt(15)). Amiram Eldar, Aug 20 2022
E.g.f.: exp(x)*(6*x^2 - 1). - Elmo R. Oliveira, Jan 16 2025

Edited and extended by R. J. Mathar, May 11 2009

Entries rechecked by N. J. A. Sloane, Jul 18 2009

A110426 The r-th term of the n-th row of the following array contains the sum of r successively decreasing integers beginning from n, 0 < r <= n (see Example).

Original entry on oeis.org

1, 3, 3, -5, -30, -84, -182, -342, -585, -935, -1419, -2067, -2912, -3990, -5340, -7004, -9027, -11457, -14345, -17745, -21714, -26312, -31602, -37650, -44525, -52299, -61047, -70847, -81780, -93930, -107384, -122232, -138567, -156485, -176085, -197469, -220742, -246012, -273390, -302990

Offset: 1

Amarnath Murthy, Aug 01 2005

			The r-th term of the n-th row of the following array contains the sum of r successively decreasing integers beginning from n, 0 < r <=n.
E.g., the row corresponding to 4 contains 4, (3+2),{(1) +(0)+(-1)}, {(-2)+(-3)+(-4)+(-5)} ----> 4,5,0,-14
  1
  2 1
  3 3 -3
  4 5 0 -14
  5 7 3 -10 -35
  6 9 6 -6 -30 -69
  ...
Sequence contains the row sums.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000330, A000914, A110425, A110427.

A110426[n_] := n*(n + 1)*(2 - (n - 3)*n)/8; Array[A110426, 50] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 3, 3, -5, -30}, 50] (* Paolo Xausa, Aug 26 2025 *)

Vec(x*(1 - 2*x - 2*x^2)/(1 - x)^5 + O(x^50)) \\ Colin Barker, May 27 2017

a(n) = Sum_{i=(2-n)*(n+1)/2..n} i = (-n^4 + 2*n^3 + 5*n^2 + 2*n)/8. - Theresa Guinard, Nov 15 2013
a(n) = A000330(n) - A000914(n-1). - J. M. Bergot, May 27 2017
From Colin Barker, May 27 2017: (Start)
G.f.: x*(1 - 2*x - 2*x^2)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5. (End)
E.g.f.: -x*(-8 - 4*x + 4*x^2 + x^3)*exp(x)/8. - Elmo R. Oliveira, Aug 24 2025

More terms from Joshua Zucker, May 10 2006

A259460 From higher-order arithmetic progressions.

Original entry on oeis.org

4, 220, 10500, 535500, 30870000, 2044828800, 156029328000, 13673998800000, 1369285948800000, 155756276676000000, 20005336176265440000, 2884501462544301600000, 464334381775424160000000, 83021688624014300160000000, 16408769917253890176000000000, 3569104362241728159962112000000, 850861011640079911341911040000000

Offset: 0

N. J. A. Sloane, Jun 30 2015

nonn

The expression "2 over n!" in the article is A006472(n+1). It is used in C_1, C_2, C_3 (A259459, A259460, A378234) on page 13. A_2 is A000914. - Georg Fischer, Dec 06 2024

Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]

Cf. A000914, A006472, A259459, A378234.

rV := proc(n,a,d)
    n*(n+1)/2*a+(n-1)*n*(n+1)/6*d;
end proc:
A259460 := proc(n)
    mul(rV(i,a,d),i=1..n+2) ;
    coeftayl(%,d=0,2) ;
    coeftayl(%,a=0,n) ;
end proc:
seq(A259460(n),n=1..18) ; # R. J. Mathar, Jul 14 2015

rV[n_, a_, d_] := n (n + 1)/2*a + (n - 1) n (n + 1)/6*d;
A259460[n_] :=
   Product[rV[i, a, d], {i, 1, n + 3}] //
   SeriesCoefficient[#, {d, 0, 2}] & //
   SeriesCoefficient[#, {a, 0, n + 1}] & ;
Table[A259460[n], {n, 0, 16}] (* Jean-François Alcover, Apr 27 2023, after R. J. Mathar *)

From Georg Fischer, Dec 06 2024: (Start)
a(n) = (n+4)!*(n+3)!/2^(n+3)/9 * (n+2)*(n+1)*(n+3)*(3*n+8)/24.
D-finite with recurrence: 2*n*(3*n+5)*a(n) - (n+3)^2*(n+4)*(3*n+8)*a(n-1) = 0. (End)

Typos in data corrected by Jean-François Alcover, Apr 27 2023

A259463 From higher-order arithmetic progressions.

Original entry on oeis.org

5, 550, 61250, 8330000, 1440600000, 318084480000, 88994505600000, 31196975040000000, 13537335651840000000, 7186069008518400000000, 4614893517270516480000000, 3548831033950800998400000000, 3237226357799023349760000000000, 3472842105575052965314560000000000

Offset: 0

N. J. A. Sloane, Jun 30 2015

nonn

"3 over n!" on page 15 in the Dienger article is A087047; A_2 is A000914. - Georg Fischer, Dec 16 2024

Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]

Cf. A000914, A087047, A259462, A259464.

rXI := proc(n, a, d)
        n*(n+1)*(n+2)/6*a+(n+2)*(n+1)*n*(n-1)/24*d;
end proc:
A259463 := proc(n)
        mul(rXI(i, a, d), i=1..n+2) ;
        coeftayl(%, d=0, 2) ;
        coeftayl(%, a=0, n) ;
end proc:
seq(A259463(n), n=1..25) ; # R. J. Mathar, Jul 15 2015

rXI[n_, a_, d_] := n(n+1)(n+2)/6 a + (n+2)(n+1)n(n-1)/24 d;
A259463[n_] := Product[rXI[i, a, d], {i, 1, n+3}]//
   SeriesCoefficient[#, {d, 0, 2}]&//
   SeriesCoefficient[#, {a, 0, n+1}]&;
Table[A259463[n], {n, 0, 13}] (* Jean-François Alcover, May 02 2023, after R. J. Mathar *)

D-finite with recurrence: -6*n*(3*n+5)*a(n) +(n+5)*(n+4)*(3*n+8)*(n+3)^2*a(n-1)=0. - R. J. Mathar, Jul 15 2015

a(n) = 2^(-n-4)*3^(-n-3)*(n+3)!*(n+4)!*(n+5)!*(n+2)*(n+1)*(n+3)*(3*n+8)/384. - Georg Fischer, Dec 16 2024

Corrected by Jean-François Alcover, May 02 2023

A191685 Eighth diagonal a(n) = s(n,n-7) of the unsigned Stirling numbers of the first kind with n>7.

Original entry on oeis.org

5040, 109584, 1172700, 8409500, 45995730, 206070150, 790943153, 2681453775, 8207628000, 23057159840, 60202693980, 147560703732, 342252511900, 756111184500, 1599718388730, 3256091103430, 6400590336096, 12191224980000, 22563937825000, 40681506808800

Offset: 8

Paul Weisenhorn, Jun 11 2011

The Maple programs under I generate the sequence. The Maple program under II generates explicit formulas for a(n+1) = s(n+1,n+1-c) with c>=1 and n>=c.

			c=1; a(n+1) = binomial(n+1,2)
c=2; a(n+1) = binomial(n+1,3)*(2+3*n)/4
c=3; a(n+1) = binomial(n+1,4)*(n+n^2)/2
c=4; a(n+1) = binomial(n+1,5)*(-8-10*n+15*n^2 +15*n^3)/48
c=5; a(n+1) = binomial(n+1,6)*(-6*n-7*n^2+2*n^3+ 3*n^4)/16
c=6; a(n+1) = binomial(n+1,7)*(96+140*n-224*n^2-315*n^3+63*n^5)/576
c=7; a(n+1) = binomial(n+1,8)*(80*n+114*n^2-23*n^3-75*n^4-9*n^5+9*n^6)/144
c=8; a(n+1) = binomial(n+1,9)*(-1152-1936*n+2820*n^2+
  5320*n^3+735*n^4-1575*n^5-315*n^6+135*n^7)/3840
c=9; a(n+1) = binomial(n+1,10)*(-1008*n-1676*n^2 +100*n^3+1295*n^4+392*n^5-210*n^6-60*n^7 +15*n^8)/768

K. Seidel, Variation der Binomialkoeffizienten, Bild
der Wissenschaft, 6 (1980), pp. 127-128.

T. D. Noe, Table of n, a(n) for n = 8..1000

Cf. A130534, A000012 (c=0; 1st diagonal), A000217 (c=1; 2nd diagonal), A000914 (c=2; 3rd diagonal), A001303 (c=3; 4th diagonal), A000915 (c=4; 5th diagonal), A053567 (c=5; 6th diagonal), A112002 (c=6; 7th diagonal), A191685 (c=7; 8th diagonal).

I: programs generate the sequence:
with(combinat): c:=7; a:= proc(n) a(n):=abs(stirling1(n,n-c)); end: seq(a(n), n=c+1..28);
for n from 7 to 27 do a(n+1) := binomial(n+1,8)*(80*n+ 114*n^2- 23*n^3- 75*n^4- 9*n^5+ 9*n^6)/144 end do: seq(a(n),n=8..28);
II: program generates explicit formulas for a(n+1) =  s(n+1,n+1-c):
k[1,0]:=1: v:=1:
for c from 2 to 10 do
  c1:=c-1: c2:=c-2: p0:=0:
  for j from 0 to c2 do p0:=p0+k[c1,j]*m^j: end do:
  f:=expand(2*c*(m+1)*p0/v):
  p1:=0: p2:=0:
  for j from 0 to c1 do
    p1:=p1+k[c,j]*(m+1)^j:
    p2:=p2+k[c,j]*m^j:
  end do:
  g:=collect((m+2)*p1-(m-c1)*p2-f,m):
  kh[0]:=rem(g,m,m): Mk:=[kh[0]]: Mv:=[k[c,0]]:
  for j from 1 to c1 do
    kh[j]:=coeff(g,m^j):
    Mk:=[op(Mk),kh[j]]: Mv:=[k[c,j],op(Mv)]:
  end do:
  sol:=solve(Mk,Mv):
  v:=1:
  for j from 1 to c do
    k[c,c-j]:=eval(k[c,c-j],sol[1,j]):
    nen[j]:=denom(k[c,c-j]):
    v:=ilcm(v,nen[j]):
  end do:
  for j from 0 to c1 do k[c,j]:=k[c,j]*v:
    printf("%8d",k[c,j]): end do:
  p3:=0:
  for j from 0 to c1 do p3:=p3+k[c,j]*n^j: end do:
  s[n+1,n+1-c]:=binomial(n+1,c+1)*(c+1)*p3/(2^c*k[c,c1]):
end do:
for c from 2 to 10 do print("%a\n",s[n+1,n+1-c]):
end do:

a(n+1) = A130534(T(n,n-7)) = s(n+1,n+1-7)

a(n+1) = binomial(n+1,8)*(80*n+114*n^2-23*n^3-75*n^4-9*n^5+9*n^6)/144

A253171 a(n) = number of permutations of (1,2,...,n) producible by an ordered triple of distinct transpositions.

Original entry on oeis.org

3, 12, 60, 240, 756, 1988, 4572, 9495, 18205, 32736, 55848, 91182, 143430, 218520, 323816, 468333, 662967, 920740, 1257060, 1689996, 2240568, 2933052, 3795300, 4859075, 6160401, 7739928, 9643312, 11921610, 14631690, 17836656, 21606288, 26017497, 31154795

Offset: 3

Andrew Woods, Dec 28 2014

			For n=4, the 12 permutations are 0132, 0213, 0321, 1023, 1230, 1302, 2031, 2103, 2310, 3012, 3120, and 3201. For example, 0123 is permuted into 0132 by ((b,d),(c,d), (b,c)).

Colin Barker, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000914, for two transpositions.

Vec(-x^3*(x^6-7*x^5+21*x^4-33*x^3+39*x^2-9*x+3)/(x-1)^7 + O(x^100)) \\ Colin Barker, Dec 30 2014

a(n) = n * (n^5 - 7*n^4 + 17*n^3 - 17*n^2 + 30*n - 24) / 48 for n>=3.
a(n) = C(n,2)*(C(n-2,2)*C(n-4,2)/6 + 1) + 2*C(n,3)*C(n-3,2) + 6*C(n,4) for n>=3.
G.f.: -x^3*(x^6-7*x^5+21*x^4-33*x^3+39*x^2-9*x+3) / (x-1)^7. - Colin Barker, Dec 30 2014

More terms from Colin Barker, Dec 30 2014

A259457 From higher-order arithmetic progressions.

Original entry on oeis.org

3, 66, 1050, 15300, 220500, 3245760, 49533120, 789264000, 13172544000, 230519520000, 4229703878400, 81315551116800, 1636227552960000, 34417989365760000, 755835784704000000, 17305616126582784000, 412559358036553728000, 10227311816872550400000, 263309943217447526400000, 7032029553158658048000000

Offset: 0

N. J. A. Sloane, Jun 30 2015

nonn

Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]

Cf. A000914, A259248.

rX := proc(n, a, d)
        n*a+(n-1)*n/2*d;
end proc:
A259457 := proc(n)
        mul(rX(i, a, d), i=1..n+2) ;
        coeftayl(%, d=0, 2) ;
        coeftayl(%, a=0, n) ;
end proc:
seq(A259457(n), n=1..25) ; # R. J. Mathar, Jul 15 2015

rX[n_, a_, d_] := n*a + (n-1)*n/2*d;
A259457[n_] :=
   Product[rX[i, a, d], {i, 1, n+3}]//
   SeriesCoefficient[#, {d, 0, 2}]&//
   SeriesCoefficient[#, {a, 0, n+1}]&;
Table[A259457[n], {n, 0, 17}] (* Jean-François Alcover, Apr 27 2023, after R. J. Mathar *)

Conjecture: 3*n*a(n) +(-3*n^2-19*n-44)*a(n-1) -2*(n+2)^2*a(n-2)=0. - R. J. Mathar, Jul 15 2015
From Georg Fischer, Dec 06 2024: (Start)
a(n) = (n+3)!*(n+2)*(n+1)*(n+3)*(3*n+8)/96.
D-finite with recurrence: -n*(3*n+5)*a(n) + (n+3)^2*(3*n+8)*a(n-1) = 0. (End)

A024169 Integer part of ((2nd elementary symmetric function of 1,2,...,n)/(1+2+...+n)).

Original entry on oeis.org

0, 0, 1, 3, 5, 8, 11, 15, 19, 24, 29, 34, 41, 47, 54, 62, 70, 79, 88, 98, 108, 119, 130, 141, 154, 166, 179, 193, 207, 222, 237, 253, 269, 286, 303, 320, 339, 357, 376, 396, 416, 437, 458, 480, 502

Offset: 1

Clark Kimberling

nonn

Ivan Neretin, Table of n, a(n) for n = 1..10000

List([1..50],n->Int((1/12)*(n-1)*(3*n+2))); # Muniru A Asiru, May 19 2018

seq(floor((1/12)*(n-1)*(3*n+2)),n=1..50); # Muniru A Asiru, May 19 2018

Table[Floor[1/12 (n - 1) (3 n + 2)], {n, 45}] (* Ivan Neretin, May 19 2018 *)

From R. J. Mathar, Sep 15 2009: (Start)
a(n) = floor( A000914(n-1)/A000217(n)).
G.f.: x^3*(-1-x-x^5-2*x^7+x^4+x^8)/((x^2+1) * (1+x+x^2) * (x^4-x^2+1) * (x-1)^3). (End)
a(n) = floor((1/12)*(n - 1)*(3*n + 2)). - Ivan Neretin, May 19 2018

cp's OEIS Frontend

A346415 Triangle T(n,k), n>=0, 0<=k<=n, read by rows, where column k is (1/k!) times the k-fold exponential convolution of Fibonacci numbers with themselves.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A347107 a(n) = Sum_{1 <= i < j <= n} j^3*i^3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

A103115 a(n) = 6*n*(n-1) - 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A110426 The r-th term of the n-th row of the following array contains the sum of r successively decreasing integers beginning from n, 0 < r <= n (see Example).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A259460 From higher-order arithmetic progressions.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A259463 From higher-order arithmetic progressions.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A191685 Eighth diagonal a(n) = s(n,n-7) of the unsigned Stirling numbers of the first kind with n>7.

Views

Author

Keywords

Comments

A103115 a(n) = 6n(n-1) - 1.