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.

Previous Showing 11-16 of 16 results.

A161743 Fourth left hand column of the RSEG2 triangle A161739.

Original entry on oeis.org

1, 10, 73, 425, 1561, -2856, -73520, 380160, 15376416, -117209664, -7506967104, 72162155520, 7045087741056, -80246202992640, -11448278791372800, 149576169325363200, 30017051616972275712, -440857664887810867200
Offset: 3

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Crossrefs

Equals fourth left hand column of A161739 (RSEG2 triangle).
Other left hand columns are A129825 and A161742.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.
A001710 (n!/2!), A001715 (n!/3!), A001720 (n!/4!), A001725 (n!/5!), A001730 (n!/6!), A049388 (n!/7!), A049389 (n!/8!), A049398 (n!/9!), A051431 (n!/10!) appear in Maple program.

Programs

  • Maple
    nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 3 to nmax do a(n) := sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n,k+3)* A008955(k+2,k), k=0..n-3) od: seq(a(n),n=3..nmax);

Formula

a(n) = sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n, k+3)*A008955(k+2, k), k = 0..n-3).

A162995 A scaled version of triangle A162990.

Original entry on oeis.org

1, 3, 1, 12, 4, 1, 60, 20, 5, 1, 360, 120, 30, 6, 1, 2520, 840, 210, 42, 7, 1, 20160, 6720, 1680, 336, 56, 8, 1, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Johannes W. Meijer, Jul 27 2009

Keywords

Comments

We get this scaled version of triangle A162990 by dividing the coefficients in the left hand columns by their 'top-values' and then taking the square root.
T(n,k) = A173333(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Feb 19 2010
T(n,k) = A094587(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

Examples

			The first few rows of the triangle are:
[1]
[3, 1]
[12, 4, 1]
[60, 20, 5, 1]
		

Crossrefs

Cf. A094587.
A056542(n) equals the row sums for n>=1.
A001710, A001715, A001720, A001725, A001730, A049388, A049389, A049398, A051431 are related to the left hand columns.
A000012, A009056, A002378, A007531, A052762, A052787, A053625 and A159083 are related to the right hand columns.

Programs

  • Haskell
    a162995 n k = a162995_tabl !! (n-1) !! (k-1)
    a162995_row n = a162995_tabl !! (n-1)
    a162995_tabl = map fst $ iterate f ([1], 3)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
  • Maple
    a := proc(n, m): (n+1)!/(m+1)! end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 23 2012
  • Mathematica
    Table[(n+1)!/(m+1)!, {n, 10}, {m, n}] (* Paolo Xausa, Mar 31 2024 *)

Formula

a(n,m) = (n+1)!/(m+1)! for n = 1, 2, 3, ..., and m = 1, 2, ..., n.

A094646 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -2, 1, 2, -3, 1, 0, 2, -3, 1, 0, 2, -1, -2, 1, 0, 4, 0, -5, 0, 1, 0, 12, 4, -15, -5, 3, 1, 0, 48, 28, -56, -35, 7, 7, 1, 0, 240, 188, -252, -231, 0, 42, 12, 1, 0, 1440, 1368, -1324, -1638, -231, 252, 114, 18, 1, 0, 10080, 11016, -7900, -12790, -3255, 1533, 1050, 240, 25, 1
Offset: 0

Views

Author

Vladeta Jovovic, May 17 2004

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows, given by [ -2, 1, -1, 2, 0, 3, 1, 4, 2, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 23 2006
From Wolfdieter Lang, Jun 23 2011: (Start)
The row polynomials s(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy risefac(x-2,n)=s(n,x), with the rising factorials risefac(x-2,n):=Product_{j=0..n-1} (x-2+j), n >= 1, risefac(x-2,0)=1. Compare this with the formula risefac(x,n)=|S1|(n,x), with the row polynomials |S1|(n,x) of A132393 (unsigned Stirling1).
This is the third triangle of an a-family of Sheffer arrays, call them |S1|(a), with e.g.f. of the row polynomials |S1|(a;x;z) = ((1-z)^a)*exp(-x*log(1-z)). In the notation showing the column e.g.f.s this is Sheffer ((1-z)^a,-log(1-z)). In the umbral notation (see the Roman reference, given under A094645) this is called Sheffer for (exp(a*t),1-exp(-t)). For a=0 this becomes the unsigned Stirling1 triangle |S1|(0) = A132393 with row polynomials |S1|(0;n,x) =: s1(n,x).
E.g.f. column number k (with leading zeros): ((1-x)^a)*((-log(1-x))^k)/k!, k >= 0.
E.g.f. for row sums is (1-x)^(a-1), and the e.g.f. for the alternating row sums is (1-x)^(a+1).
Row polynomial recurrence:
|S1|(a;n,x)=(x+(n-1-a))*|S1|(a;n-1,m), |S1|(a;0,x)=1.
Meixner identity (see the reference under A060338):
|S1|(a;n,x) - |S1|(a;n,x-1) = n*|S1|(a;n-1,x), n >= 1,
Also (from the corollary 3.7.2 on p. 50 of the Roman reference): |S1|(a;n,x) = (x-a)*|S1|(a;n-1,x+1), n >= 1.
Recurrence: |S1|(a;n,k) = |S1|(a;n-1,k-1) + (n-(a+1))*|S1|(a;n-1,k); |S1|(a;n,k)=0 if n < m, |S1|(a;n,-1)=0, |S1|(a;0,0)=1.
Connection to |Stirling1|=|S1|(0):
|S1|(a;n,k) = Sum_{p=0..a} |S1|(a;a,p)*abs(Stirling1(n-a,k-p)), n >= a.
The exponential convolution identity is
|S1|(a;n,x+y) = Sum_{k=0..n} binomial(n,k)*|S1|(a;k,y)*s1(n-k,x), n >= 0, with symmetry x <-> y.
The Sheffer a- and z-sequences are (see the W. Lang link under A006232): Sha(a;n)=A164555(n)/A027642(n) (independent of a) with e.g.f. x/(1-exp(-x)), and the z-sequence has e.g.f. (exp(a*x)-1)/(exp(-x)-1).
The inverse Sheffer matrix has e.g.f. exp(a*z)*exp(x*(1-exp(-z))), in short notation (exp(a*z),1-exp(-z)),
(or in umbral notation ((1-t)^a,-log(1-t))).
(End)

Examples

			Triangle begins
   1;
  -2,  1;
   2, -3,  1;
   0,  2, -3,  1;
   0,  2, -1, -2,  1;
   0,  4,  0, -5,  0,  1;
   ...
risefac(x-2,3) = (x-2)*(x-1)*x = 2*x-3*x^2+x^3.
-1 = T(4,2) = T(3,1) + 1*T(3,2) =  2 + (-3).
T(4,3) = 2*abs(S1(2,3)) - 3*abs(S1(2,2)) + 1*abs(S1(2,1)) = 2*0 - 3*1 + 1*1 = -2.
		

Crossrefs

Programs

  • Maple
    A094646_row := n -> seq((-1)^(n-k)*coeff(expand(pochhammer(x-n+3, n)), x, k), k=0..n): seq(print(A094646_row(n)), n = 0..6); # Peter Luschny, May 16 2013
  • Mathematica
    Flatten[ Table[ CoefficientList[ Pochhammer[x-2, n], x], {n, 0, 10}]] (* Jean-François Alcover, Sep 26 2011 *)

Formula

E.g.f.: (1-y)^(2-x).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 respectively. - Philippe Deléham, Nov 13 2007
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then |T(n,i)| = |f(n,i,-2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Wolfdieter Lang, Jun 23 2011: (Start)
risefac(x-2,n) = Sum_{k=0..n} T(n,k)*x^k, n >= 0, with the rising factorials (see a comment above).
Recurrence: T(n,k) = T(n-1,k-1) + (n-3)*T(n-1,k); T(n,k)=0 if n < m, T(n,-1)=0, T(0,0)=1.
T(n,k) = 2*abs(S1(n-2,k)) - 3*abs(S1(n-2,k-1)) + abs(S1(n-2,k-2)), n >= 2, with S1(n,k) = Stirling1(n,k) = A048994(n,k).
E.g.f. column number k (with leading zeros):
((1-x)^2)*((-log(1-x))^k)/k!, k >= 0.
E.g.f. for row sums is 1-x, i.e., [1,-1,0,0,...],
and the e.g.f. for the alternating row sums is (1-x)^3. i.e., [1,-3,3,1,0,0,...]. (End)

A176736 a(n) = (n+9)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 10, 111, 1352, 17909, 256134, 3931555, 64441684, 1123029513, 20730064706, 403978495031, 8286870547680, 178468044946621, 4025739435397822, 94912091598455979, 2334250550458513004, 59779945135439664785, 1591626582328767492474, 43990176790179196598143, 1260374228606935319612536
Offset: 0

Views

Author

Wolfdieter Lang, Jul 14 2010

Keywords

Comments

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=10 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049398(n) = (n+9)!/9!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010).

Examples

			Necklaces and 10 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c10(1), (binomial(4,2)*! 2)*c10(2), and 1*c10(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c10(n):=A049398(n) numbers for the pure 10-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=10: 1/(1-x)^10). This adds up as 9 + 4*2*10 + (6*1)*110 + 17160 = 17909 = a(4).
		

Crossrefs

Cf. A176735 (necklaces and k=9 cords).

Formula

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^10) = exp(-x)/(1-x)^11, equivalent to the given recurrence.
a(n) = A086764(n+10,10).

A051565 Third unsigned column of triangle A051523.

Original entry on oeis.org

0, 0, 1, 33, 791, 17100, 358024, 7491484, 159168428, 3463513704, 77559615576, 1792139785920, 42789106278720, 1056302350122240, 26964471256888320, 711643650545422080, 19410244660543737600, 546854985563699289600
Offset: 0

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Author

Keywords

Comments

From Johannes W. Meijer, Oct 20 2009: (Start)
The asymptotic expansion of the higher order exponential integral E(x,m=3,n=10) ~ exp(-x)/x^3*(1 - 33/x + 791/x^2 - 17100/x^3 + 358024/x^4 + ...) leads to the sequence given above. See A163931 and A163932 for more information.
(End)

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051523.

Crossrefs

Cf. A049398 (m=0), A051564 (m=1) unsigned columns.

Formula

a(n) = A051523(n, 2)*(-1)^n; e.g.f.: ((log(1-x))^2)/(2*(1-x)^10).
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then a(n) = |f(n,2,10)|, for n>=1. - Milan Janjic, Dec 21 2008

A373168 Triangle read by rows: the exponential almost-Riordan array ( exp(x/(1-x)) | 1/(1-x), x ).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 13, 2, 2, 1, 73, 6, 6, 3, 1, 501, 24, 24, 12, 4, 1, 4051, 120, 120, 60, 20, 5, 1, 37633, 720, 720, 360, 120, 30, 6, 1, 394353, 5040, 5040, 2520, 840, 210, 42, 7, 1, 4596553, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 58941091, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1
Offset: 0

Views

Author

Stefano Spezia, May 26 2024

Keywords

Examples

			The triangle begins:
     1;
     1,   1;
     3,   1,   1;
    13,   2,   2,  1;
    73,   6,   6,  3,  1;
   501,  24,  24, 12,  4, 1;
  4051, 120, 120, 60, 20, 5, 1;
  ...
		

Crossrefs

Programs

  • Mathematica
    T[n_,0]:=n!SeriesCoefficient[Exp[x/(1-x)],{x,0,n}]; T[n_,k_]:=(n-1)!/(k-1)!SeriesCoefficient[1/(1-x)*x^(k-1),{x,0,n-1}]; Table[T[n,k],{n,0,10},{k,0,n}]//Flatten

Formula

T(n,0) = n! * [x^n] exp(x/(1-x)); T(n,k) = (n-1)!/(k-1)! * [x^(n-1)] 1/(1-x)*x^(k-1).
T(n,3) = A001710(n-1) for n > 2.
T(n,4) = A001715(n-1) for n > 3.
T(n,5) = A001720(n-1) for n > 4.
T(n,6) = A001725(n-1) for n > 5.
T(n,7) = A001730(n-1) for n > 6.
T(n,8) = A049388(n-8) for n > 7.
T(n,9) = A049389(n-9) for n > 8.
T(n,10) = A049398(n-10) for n > 9.
T(n,11) = A051431(n-11) for n > 10.
Previous Showing 11-16 of 16 results.