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 11-20 of 123 results. Next

A242091 a(n) = r * (n-1)! where r is the rational number that satisfies the equation Sum_{k>=n} (-1)^(k + n)/C(k,n) = n*2^(n-1)*log(2) - r.

Original entry on oeis.org

0, 2, 15, 128, 1310, 15864, 222936, 3572736, 64354608, 1287495360, 28328889600, 679936896000, 17678878214400, 495015296025600, 14850552286080000, 475219068007219200, 16157470542709708800, 581669316147767500800, 22103440771676298854400
Offset: 1

Views

Author

Richard R. Forberg, Aug 14 2014

Keywords

Comments

The sum of the terms of the inverse of the binomial coefficients, 1/C(k,n), with alternating signs, equals an irrational number which is expressed as m * log(2) - r, where m is the integer n*2^(n-1) = A001787(n), n>=1, and r is rational. a(n) = r * (n-1)!.

Examples

			Sum_{k>=1} (-1)^(k + 1)/C(k,1) = Sum_{k>=1} (-1)^(k + 1)/k = log(2) where m = 1 and r = 0. (See A002162.)
Sum_{k>=2} (-1)^(k + 2)/C(k,2) = 4*log(2) - 2. (See A000217.)
Sum_{k>=3} (-1)^(k + 3)/C(k,3) = 12*log(2) - 15/2. (See A000292.)
Sum_{k>=4} (-1)^(k + 4)/C(k,4) = 30*log(2) - 64/3. (See A000332.)
Sum_{k>=5} (-1)^(k + 5)/C(k,5) = 80*log(2) - 655/12. (See A000389.)
		

Crossrefs

Programs

  • Magma
    [n le 1 select 0 else 2*(n)*Self(n-1)+(Factorial(n) div (n-1)): n in [1..20]]; // Vincenzo Librandi, Sep 22 2015
    
  • Maple
    seq(add(2^(n-j-1)*n!/j, j=1..n-1), n=1..100); # Robert Israel, Aug 14 2014
  • Mathematica
    Table[Sum[2^(n - j - 1)*n!/j, {j, n - 1}], {n, 20}] (* Wesley Ivan Hurt, Aug 14 2014 *)
    FullSimplify[Table[-1/2*n!*(LerchPhi[1/2, 1, n] - 2^n*Log[2]),{n, 1, 20}]] (* Vaclav Kotesovec, Aug 15 2014 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(serlaplace(x*log(1-x)/(2*x-1)))) \\ G. C. Greubel, Nov 25 2017

Formula

From Robert Israel, Aug 14 2014: (Start)
a(n) = n * A068102(n-1).
a(n) = n! * Sum_{j=1..(n-1)} 2^(n-j-1)/j.
a(n) = n! * (2^(n-1)*log(2)-(1/2)*LerchPhi(1/2, 1, n)).
a(n+1) = 2*(n+1)*a(n) + (n+1)!/n.
E.g.f.: x*log(1-x)/(2*x-1).
(End)
Recurrence: (n-1)*a(n) = n*(3*n-4)*a(n-1) - 2*(n-2)*(n-1)*n*a(n-2). - Vaclav Kotesovec, Aug 15 2014

A002412 Hexagonal pyramidal numbers, or greengrocer's numbers.

Original entry on oeis.org

0, 1, 7, 22, 50, 95, 161, 252, 372, 525, 715, 946, 1222, 1547, 1925, 2360, 2856, 3417, 4047, 4750, 5530, 6391, 7337, 8372, 9500, 10725, 12051, 13482, 15022, 16675, 18445, 20336, 22352, 24497, 26775, 29190, 31746, 34447, 37297, 40300
Offset: 0

Views

Author

Keywords

Comments

Binomial transform of (1, 6, 9, 4, 0, 0, 0, ...). - Gary W. Adamson, Oct 16 2007
a(n) is the sum of the maximum(m,n) over {(m,n): m,n in positive integers, m<=n}. - Geoffrey Critzer, Oct 11 2009
We obtain these numbers for d=2 in the identity n*(n*(d*n-d+2)/2)-sum(k*(d*k-d+2)/2, k=0..n-1) = n*(n+1)*(2*d*n-2*d+3)/6 (see Klaus Strassburger in Formula lines). - Bruno Berselli, Apr 21 2010, Nov 16 2010
q^a(n) is the Hankel transform of the q-Catalan numbers. - Paul Barry, Dec 15 2010
Row 1 of the convolution array A213835. - Clark Kimberling, Jul 04 2012
From Ant King, Oct 24 2012: (Start)
For n>0, the digital roots of this sequence A010888(A002412(n)) form the purely periodic 27-cycle {1,7,4,5,5,8,9,3,3,4,1,7,8,8,2,3,6,6,7,4,1,2,2,5,6,9,9}.
For n>0, the units' digits of this sequence A010879(A002412(n)) form the purely periodic 20-cycle {1,7,2,0,5,1,2,2,5,5,6,2,7,5,0,6,7,7,0,0}.
(End)
Partial sums of A000384. - Omar E. Pol, Jan 12 2013
Row sums of A094728. - J. M. Bergot, Jun 14 2013
Number of orbits of Aut(Z^7) as function of the infinity norm (n+1) of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 40320. - Philippe A.J.G. Chevalier, Dec 28 2015
Coefficients in the hypergeometric series identity 1 - 7*(x - 1)/(3*x + 1) + 22*(x - 1)*(x - 2)/((3*x + 1)*(3*x + 2)) - 50*(x - 1)*(x - 2)*(x - 3)/((3*x + 1)*(3*x + 2)*(3*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A000326 and A002418. Column 3 of A103450. - Peter Bala, Mar 14 2019

Examples

			Let n=5, 2*n=10. Since 10 = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5, a(5) = 1*9 + 2*8 + 3*7 + 4*6 + 5*5 = 95. - _Vladimir Shevelev_, May 11 2012
		

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.
  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
  • I. Siap, Linear codes over F_2 + u*F_2 and their complete weight enumerators, in Codes and Designs (Ohio State, May 18, 2000), pp. 259-271. De Gruyter, 2002.
  • 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

Bisection of A002623. Equals A000578(n) - A000330(n-1).
a(n) = A093561(n+2, 3), (4, 1)-Pascal column.
Cf. A220084 for a list of numbers of the form n*P(k,n)-(n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number (see Adamson's formula).
Cf. similar sequences listed in A237616.
Orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008585, A005843, A001477, A000217.

Programs

  • GAP
    List([0..40],n->n*(n+1)*(4*n-1)/6); # Muniru A Asiru, Mar 18 2019
    
  • Magma
    [n*(n+1)*(4*n-1)/6: n in [0..40]]; // Vincenzo Librandi, Nov 28 2015
    
  • Maple
    seq(sum(i*(2*k-i), i=1..k), k=0..100); # Wesley Ivan Hurt, Sep 25 2013
  • Mathematica
    Figurate[ ngon_, rank_, dim_] := Binomial[rank + dim - 2, dim - 1] ((rank - 1)*(ngon - 2) + dim)/dim; Table[ Figurate[6, r, 3], {r, 0, 40}] (* Robert G. Wilson v, Aug 22 2010 *)
    Table[n(n+1)(4n-1)/6, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,7,22}, 40] (* Harvey P. Dale, Jul 16 2011 *)
  • Maxima
    A002412(n):=n*(n+1)*(4*n-1)/6$ makelist(A002412(n),n,0,20); /* Martin Ettl, Dec 12 2012 */
    
  • PARI
    v=vector(40,i,(i*(i+1))\2); s=0; print1(s","); forstep(i=1,40,2,s+=v[i]; print1(s","))
    
  • Python
    print([n*(n+1)*(4*n-1)//6 for n in range(40)]) # Michael S. Branicky, Mar 28 2022

Formula

a(n) = n(n + 1)(4n - 1)/6.
G.f.: x*(1+3*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation.
a(n) = n^3 - Sum_{i=1..n-1} i^2. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)
Partial sums of n odd-indexed triangular numbers, e.g., a(3) = t(1)+t(3)+t(5) = 1+6+15 = 22. - Jon Perry, Jul 23 2003
a(n) = Sum_{i=0..n-1} (n - i)*(n + i). - Jon Perry, Sep 26 2004
a(n) = n*A000292(n) - (n-1)*A000292(n-1) = n*binomial((n+2),3) - (n-1)*binomial((n+1),3); e.g., a(5) = 95 = 5*35 - 4*20. - Gary W. Adamson, Dec 28 2007
a(n) = Sum_{i=0..n} (2i^2 + 3i + 1), for n >= 0 (Omits the leading 0). - William A. Tedeschi, Aug 25 2010
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4), with a(0)=0, a(1)=1, a(2)=7, a(3)=22. - Harvey P. Dale, Jul 16 2011
a(n) = sum a*b, where the summing is over all unordered partitions 2*n = a+b. - Vladimir Shevelev, May 11 2012
From Ant King, Oct 24 2012: (Start)
a(n) = a(n-1) + n*(2*n-1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 4.
a(n) = (n+1)*(2*A000384(n) + n)/6 = (4*n-1)*A000217(n)/3.
a(n) = A000292(n) + 3*A000292(n-1) = A002411(n) + A000292(n-1).
a(n) = binomial(n+2,3) + 3*binomial(n+1,3) = (4*n-1)*binomial(n+1,2)/3.
Sum_{n>=1} 1/a(n) = 6*(12*log(2)-2*Pi-1)/5 = 1.2414...
(End)
a(n) = Sum_{i=1..n} Sum_{j=1..n} max(i,j) = Sum_{i=1..n} i*(2*n-i). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A005900(n+1) - A000326(n+1) = Octahedral - Pentagonal Numbers. - Richard R. Forberg, Aug 07 2013
a(n) = n*A000217(n) + Sum_{i=0..n-1} A000217(i). - Bruno Berselli, Dec 18 2013
a(n) = 2n * A000217(n) - A000330(n). - J. M. Bergot, Apr 05 2014
a(n) = A080851(4,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: x*(6 + 15*x + 4*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(1 + 2*sqrt(2)*Pi - 2*(3+sqrt(2))*log(2) + 4*sqrt(2)*log(2-sqrt(2)))/5. - Amiram Eldar, Jan 04 2022

A000580 a(n) = binomial coefficient C(n,7).

Original entry on oeis.org

1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 11440, 19448, 31824, 50388, 77520, 116280, 170544, 245157, 346104, 480700, 657800, 888030, 1184040, 1560780, 2035800, 2629575, 3365856, 4272048, 5379616, 6724520, 8347680, 10295472
Offset: 7

Views

Author

Keywords

Comments

Figurate numbers based on 7-dimensional regular simplex. According to Hyun Kwang Kim, it appears that every nonnegative integer can be represented as the sum of g = 15 of these numbers. - Jonathan Vos Post, Nov 28 2004
a(n) is the number of terms in the expansion of (Sum_{i=1..8} a_i)^n. - Sergio Falcon, Feb 12 2007
Product of seven consecutive numbers divided by 7!. - Artur Jasinski, Dec 02 2007
In this sequence there are no primes. - Artur Jasinski, Dec 02 2007
For a set of integers {1,2,...,n}, a(n) is the sum of the 2 smallest elements of each subset with 6 elements, which is 3*C(n+1,7) (for n>=6), hence a(n) = 3*C(n+1,7) = 3*A000580(n+1). - Serhat Bulut, Mar 13 2015
Partial sums of A000579. In general, the iterated sums S(m, n) = Sum_{j=1..n} S(m-1, j) with input S(1, n) = A000217(n) = 1 + 2 + ... + n are S(m, n) = risefac(n, m+1)/(m+1)! = binomial(n+m, m+1) = Sum_{k = 1..n} risefac(k, m)/m!, with the rising factorials risefac(x, m):= Product_{j=0..m-1} (x+j), for m >= 1. Such iterated sums of arithmetic progressions have been considered by Narayana Pandit (see The MacTutor History of Mathematics archive link, and the Gottwald et al. reference, p. 338, where the name Narayana Daivajna is also used). - Wolfdieter Lang, Mar 20 2015
a(n) = fallfac(n,7)/7! = binomial(n, 7) is also the number of independent components of an antisymmetric tensor of rank 7 and dimension n >= 7 (for n=1..6 this becomes 0). Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
From Juergen Will, Jan 02 2016: (Start)
Number of compositions (ordered partitions) of n+1 into exactly 8 parts.
Number of weak compositions (ordered weak partitions) of n-7 into exactly 8 parts. (End)

Examples

			For A={1,2,3,4,5,6,7}, subsets with 6 elements are {1,2,3,4,5,6}, {1,2,3,4,5,7}, {1,2,3,4,6,7}, {1,2,3,5,6,7}, {1,2,4,5,6,7}, {1,3,4,5,6,7}, {2,3,4,5,6,7}.
Sum of 2 smallest elements of each subset: a(7) = (1+2)+(1+2)+(1+2)+(1+2)+(1+2)+(1+3)+(2+3) = 24 = 3*C(7+1,7) = 3*A000580(7+1). - _Serhat Bulut_, Mar 13 2015
		

References

  • 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.
  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • 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.
  • S. Gottwald, H.-J. Ilgauds and K.-H. Schlote (eds.), Lexikon bedeutender Mathematiker (in German), Bibliographisches Institut Leipzig, 1990.
  • 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).

Crossrefs

Programs

  • Magma
    [Binomial(n,7): n in [7..40]]; // Vincenzo Librandi, Mar 21 2015
    
  • Maple
    ZL := [S, {S=Prod(B,B,B,B,B,B,B,B), B=Set(Z, 1 <= card)}, unlabeled]: seq(combstruct[count](ZL, size=n), n=8..38); # Zerinvary Lajos, Mar 13 2007
    A000580:=1/(z-1)**8; # Simon Plouffe in his 1992 dissertation, offset 0.
    seq(binomial(n+7,7)*1^n,n=0..30); # Zerinvary Lajos, Jun 23 2008
    G(x):=x^7*exp(x): f[0]:=G(x): for n from 1 to 38 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n]/7!,n=7..37); # Zerinvary Lajos, Apr 05 2009
  • Mathematica
    Table[n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)/7!, {n, 1, 100}] (* Artur Jasinski, Dec 02 2007 *)
    Binomial[Range[7,40],7] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{1,8,36,120,330,792,1716,3432},40] (* Harvey P. Dale, Nov 28 2011 *)
    CoefficientList[Series[1 / (1-x)^8, {x, 0, 33}], x] (* Vincenzo Librandi, Mar 21 2015 *)
  • PARI
    a(n)=binomial(n,7) \\ Charles R Greathouse IV, Sep 24 2015

Formula

G.f.: x^7/(1-x)^8.
a(n) = (n^7 - 21*n^6 + 175*n^5 - 735*n^4 + 1624*n^3 - 1764*n^2 + 720*n)/5040.
a(n) = -A110555(n+1,7). - Reinhard Zumkeller, Jul 27 2005
Convolution of the nonnegative numbers (A001477) with the sequence A000579. Also convolution of the triangular numbers (A000217) with the sequence A000332. Also convolution of the sequence {1,1,1,1,...} (A000012) with the sequence A000579. Also self-convolution of the tetrahedral numbers (A000292). - Sergio Falcon, Feb 12 2007
a(n+4) = (1/3!)*(d^3/dx^3)S(n,x)|A049310.%20-%20_Wolfdieter%20Lang">{x=2}, n >= 3. One sixth of third derivative of Chebyshev S-polynomials evaluated at x=2. See A049310. - _Wolfdieter Lang, Apr 04 2007
a(n) = n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)/7!. - Artur Jasinski, Dec 02 2007, R. J. Mathar, Jul 07 2009
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) with a(7)=1, a(8)=8, a(9)=36, a(10)=120, a(11)=330, a(12)=792, a(13)=1716, a(14)=3432. - Harvey P. Dale, Nov 28 2011
a(n) = 3*C(n+1,7) = 3*A000580(n+1). - Serhat Bulut, Mar 13 2015
From Wolfdieter Lang, Mar 21 2015: (Start)
a(n) = A104712(n, 7), n >= 7.
a(n+6) = sum(A000579(j+5), j = 1..n), n >= 1. See the Mar 20 2015 comment above. (End)
Sum_{k >= 7} 1/a(k) = 7/6. - Tom Edgar, Sep 10 2015
Sum_{n>=7} (-1)^(n+1)/a(n) = A001787(7)*log(2) - A242091(7)/6! = 448*log(2) - 9289/30 = 0.8966035575... - Amiram Eldar, Dec 10 2020

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Mar 17 2000
Some formulas that referred to other offsets corrected by R. J. Mathar, Jul 07 2009

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11
Offset: 0

Views

Author

Wouter Meeussen, Oct 01 2002

Keywords

Comments

This sequence counts the "almost triangular" partitions of n. A partition is triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An "almost triangular" partition is a triangular partition with at most 1 added to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4. Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of "Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A function f is acyclic if for every subset B of the domain the image of B under f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)}, f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. - Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) = 1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2, 1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2, 1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688. Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) = RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . - Paul Curtz, Jun 21 2010
A054143 is the fission of the polynomial sequence (p(n,x)) given by p(n,x) = x^n + x^(n-1) + ... + x + 1 by the polynomial sequence ((x+1)^n). See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Reversal of A135278. - Philippe Deléham, Feb 11 2012
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From A238363, the operator equation d/d(:xD:)f(xD)={exp[d/d(xD)]-1}f(xD) = f(xD+1)-f(xD) follows. Choosing f(x) = x^n and using :xD:^n/n! = binomial(xD,n) and (xD)^n = Bell(n,:xD:), the Bell polynomials of A008277, it follows that the lower triangular matrix [padded A074909]
A) = [St2]*[dP]*[St1] = A048993*A132440*[padded A008275]
B) = [St2]*[dP]*[St2]^(-1)
C) = [St1]^(-1)*[dP]*[St1],
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 whereas [padded A074909]=A007318-I with I=identity matrix. - Tom Copeland, Apr 25 2014
T(n,k) generated by m-gon expansions in the case of odd m with "vertex to side" version or even m with "vertex to vertes" version. Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as a triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to side" version or (ii) m is even and expansion is "vertex to vertex" version. m*Sum_{i=1..k} T(n,k) gives the total label value at the n-th iteration. See also A247976. Vertex to vertex: A061777, A247618, A247619, A247620. Vertex to side: A101946, A247903, A247904, A247905. - Kival Ngaokrajang Sep 28 2014
From Tom Copeland, Nov 12 2014: (Start)
With P(n,x) = [(x+1)^(n+1)-x^(n+1)], the row polynomials of this entry, Up(n,x) = P(n,x)/(n+1) form an Appell sequence of polynomials that are the umbral compositional inverses of the Bernoulli polynomials B(n,x), i.e., B[n,Up(.,x)] = x^n = Up[n,B(.,x)] under umbral substitution, e.g., B(.,x)^n = B(n,x).
The e.g.f. for the Bernoulli polynomials is [t/(e^t - 1)] e^(x*t), and for Up(n,x) it's exp[Up(.,x)t] = [(e^t - 1)/t] e^(x*t).
Another g.f. is G(t,x) = log[(1-x*t)/(1-(1+x)*t)] = log[1 + t /(1 + -(1+x)t)] = t/(1-t*Up(.,x)) = Up(0,x)*t + Up(1,x)*t^2 + Up(2,x)*t^3 + ... = t + (1+2x)/2 t^2 + (1+3x+3x^2)/3 t^3 + (1+4x+6x^2+4x^3)/4 t^4 + ... = -log(1-t*P(.,x)), expressed umbrally.
The inverse, Ginv(t,x), in t of the g.f. may be found in A008292 from Copeland's list of formulas (Sep 2014) with a=(1+x) and b=x. This relates these two sets of polynomials to algebraic geometry, e.g., elliptic curves, trigonometric expansions, Chebyshev polynomials, and the combinatorics of permutahedra and their duals.
Ginv(t,x) = [e^((1+x)t) - e^(xt)] / [(1+x) * e^((1+x)t) - x * e^(xt)] = [e^(t/2) - e^(-t/2)] / [(1+x)e^(t/2) - x*e^(-t/2)] = (e^t - 1) / [1 + (1+x) (e^t - 1)] = t - (1 + 2 x) t^2/2! + (1 + 6 x + 6 x^2) t^3/3! - (1 + 14 x + 36 x^2 + 24 x^3) t^4/4! + ... = -exp[-Perm(.,x)t], where Perm(n,x) are the reverse face polynomials, or reverse f-vectors, for the permutahedra, i.e., the face polynomials for the duals of the permutahedra. Cf. A090582, A019538, A049019, A133314, A135278.
With L(t,x) = t/(1+t*x) with inverse L(t,-x) in t, and Cinv(t) = e^t - 1 with inverse C(t) = log(1 + t). Then Ginv(t,x) = L[Cinv(t),(1+x)] and G(t,x) = C[L[t,-(1+x)]]. Note L is the special linear fractional (Mobius) transformation.
Connections among the combinatorics of the permutahedra, simplices (cf. A135278), and the associahedra can be made through the Lagrange inversion formula (LIF) of A133437 applied to G(t,x) (cf. A111785 and the Schroeder paths A126216 also), and similarly for the LIF A134685 applied to Ginv(t,x) involving the simplicial Whitehouse complex, phylogenetic trees, and other structures. (See also the LIFs A145271 and A133932). (End)
R = x - exp[-[B(n+1)/(n+1)]D] = x - exp[zeta(-n)D] is the raising operator for this normalized sequence UP(n,x) = P(n,x) / (n+1), that is, R UP(n,x) = UP(n+1,x), where D = d/dx, zeta(-n) is the value of the Riemann zeta function evaluated at -n, and B(n) is the n-th Bernoulli number, or constant B(n,0) of the Bernoulli polynomials. The raising operator for the Bernoulli polynomials is then x + exp[-[B(n+1)/(n+1)]D]. [Note added Nov 25 2014: exp[zeta(-n)D] is abbreviation of exp(a.D) with (a.)^n = a_n = zeta(-n)]. - Tom Copeland, Nov 17 2014
The diagonals T(n, n-m), for n >= m, give the m-th iterated partial sum of the positive integers; that is A000027(n+1), A000217(n), A000292(n-1), A000332(n+1), A000389(n+1), A000579(n+1), A000580(n+1), A000581(n+1), A000582(n+1), ... . - Wolfdieter Lang, May 21 2015
The transpose gives the numerical coefficients of the Maurer-Cartan form matrix for the general linear group GL(n,1) (cf. Olver, but note that the formula at the bottom of p. 6 has an error--the 12 should be a 15). - Tom Copeland, Nov 05 2015
The left invariant Maurer-Cartan form polynomial on p. 7 of the Olver paper for the group GL^n(1) is essentially a binomial convolution of the row polynomials of this entry with those of A133314, or equivalently the row polynomials generated by the product of the e.g.f. of this entry with that of A133314, with some reindexing. - Tom Copeland, Jul 03 2018
From Tom Copeland, Jul 10 2018: (Start)
The first column of the inverse matrix is the sequence of Bernoulli numbers, which follows from the umbral definition of the Bernoulli polynomials (B.(0) + x)^n = B_n(x) evaluated at x = 1 and the relation B_n(0) = B_n(1) for n > 1 and -B_1(0) = 1/2 = B_1(1), so the Bernoulli numbers can be calculated using Cramer's rule acting on this entry's matrix and, therefore, from the ratios of volumes of parallelepipeds determined by the columns of this entry's square submatrices. - Tom Copeland, Jul 10 2018
Umbrally composing the row polynomials with B_n(x), the Bernoulli polynomials, gives (B.(x)+1)^(n+1) - (B.(x))^(n+1) = d[x^(n+1)]/dx = (n+1)*x^n, so multiplying this entry as a lower triangular matrix (LTM) by the LTM of the coefficients of the Bernoulli polynomials gives the diagonal matrix of the natural numbers. Then the inverse matrix of this entry has the elements B_(n,k)/(k+1), where B_(n,k) is the coefficient of x^k for B_n(x), and the e.g.f. (1/x) (e^(xt)-1)/(e^t-1). (End)

Examples

			T(4,2) = 0+0+1+3+6 = 10 = binomial(5, 2).
Triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7   8   9 10 11
0:  1
1:  1  2
2:  1  3  3
3:  1  4  6   4
4:  1  5 10  10   5
5:  1  6 15  20  15   6
6:  1  7 21  35  35  21   7
7:  1  8 28  56  70  56  28   8
8:  1  9 36  84 126 126  84  36  9
9:  1 10 45 120 210 252 210 120 45   10
10: 1 11 55 165 330 462 462 330 165  55 11
11: 1 12 66 220 495 792 924 792 495 220 66 12
... Reformatted. - _Wolfdieter Lang_, Nov 04 2014
.
Can be seen as the square array A(n, k) = binomial(n + k + 1, n) read by descending antidiagonals. A(n, k) is the number of monotone nondecreasing functions f: {1,2,..,k} -> {1,2,..,n}. - _Peter Luschny_, Aug 25 2019
[0]  1,  1,   1,   1,    1,    1,     1,     1,     1, ... A000012
[1]  2,  3,   4,   5,    6,    7,     8,     9,    10, ... A000027
[2]  3,  6,  10,  15,   21,   28,    36,    45,    55, ... A000217
[3]  4, 10,  20,  35,   56,   84,   120,   165,   220, ... A000292
[4]  5, 15,  35,  70,  126,  210,   330,   495,   715, ... A000332
[5]  6, 21,  56, 126,  252,  462,   792,  1287,  2002, ... A000389
[6]  7, 28,  84, 210,  462,  924,  1716,  3003,  5005, ... A000579
[7]  8, 36, 120, 330,  792, 1716,  3432,  6435, 11440, ... A000580
[8]  9, 45, 165, 495, 1287, 3003,  6435, 12870, 24310, ... A000581
[9] 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, ... A000582
		

Crossrefs

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n+1,k)))); # Muniru A Asiru, Jul 10 2018
    
  • Haskell
    a074909 n k = a074909_tabl !! n !! k
    a074909_row n = a074909_tabl !! n
    a074909_tabl = iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Feb 25 2012
    
  • Magma
    /* As triangle */ [[Binomial(n+1,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2018
    
  • Maple
    A074909 := proc(n,k)
        if k > n or k < 0 then
            0;
        else
            binomial(n+1,k) ;
        end if;
    end proc: # Zerinvary Lajos, Nov 09 2006
  • Mathematica
    Flatten[Join[{1}, Table[Sum[Binomial[k, m], {k, 0, n}], {n, 0, 12}, {m, 0, n}] ]] (* or *) Flatten[Join[{1}, Table[Binomial[n, m], {n, 12}, {m, n}]]]
  • PARI
    print1(1);for(n=1,10,for(k=1,n,print1(", "binomial(n,k)))) \\ Charles R Greathouse IV, Mar 26 2013
    
  • Python
    from math import comb, isqrt
    def A074909(n): return comb(r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),n-comb(r,2)) # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)*e^t - x) e^(x*t). The row polynomials p_n(x) satisfy dp_n(x)/dx = (n+1)*p_(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. - Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)*y^(n-k), where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the n-th row polynomial of this entry. dUp(n,x)/dx = n * Up(n-1,x) and dP(n,x)/dx = (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-t*x)*(1-(1+t)x)) = x + (1+2t)*x^2 + (1+3t+3t^2)*x^3 + ... has the inverse GFinv(x,t) = (1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f., i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...) associated with the moments defined by GFinv, and, in fact, these free cumulants generate these moments through the noncrossing partitions of A134264. The associated e.g.f. and relations to Grassmannians are described in A248727, whose polynomials are the basis for an Appell sequence of polynomials that are umbral compositional inverses of the Appell sequence formed from this entry's polynomials (distinct from the one described in the comments above, without the normalizing reciprocal). - Tom Copeland, Jan 07 2015
T(n, k) = (1/k!) * Sum_{i=0..k} Stirling1(k,i)*(n+1)^i, for 0<=k<=n. - Ridouane Oudra, Oct 23 2022

Extensions

I added an initial 1 at the suggestion of Paul Barry, which makes the triangle a little nicer but may mean that some of the formulas will now need adjusting. - N. J. A. Sloane, Feb 11 2003
Formula section edited, checked and corrected by Wolfdieter Lang, Nov 04 2014

A000581 a(n) = binomial coefficient C(n,8).

Original entry on oeis.org

1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 43758, 75582, 125970, 203490, 319770, 490314, 735471, 1081575, 1562275, 2220075, 3108105, 4292145, 5852925, 7888725, 10518300, 13884156, 18156204, 23535820, 30260340, 38608020, 48903492, 61523748, 76904685
Offset: 8

Views

Author

Keywords

Comments

Figurate numbers based on 8-dimensional regular simplex. - Jonathan Vos Post, Nov 28 2004
Just as A005712 and A000574 are described as the coefficients of x^4 and x^5 in the expansion of (1+x+x^2)^n, so should this sequence be described as the coefficients of x^3 therein. - R. K. Guy, Oct 19 2007
Product of 8 consecutive numbers divided by 8!. - Artur Jasinski, Dec 02 2007
In this sequence there are no primes. - Artur Jasinski, Dec 02 2007
a(n) = number of (n-8)-digit numbers with nondescending digits. E.g., a(9) = 9 = {1,2,3,..,9}, a(10) = 45 = {11-19, 22-29, 33-39, ..., 99} [0 is counted as a zero-digit number rather than a 1-digit number]. - Toby Gottfried, Feb 14 2012
a(n) =fallfac(n, 8)/8! = binomial(n, 8) is also the number of independent components of an antisymmetric tensor of rank 8 and dimension n >= 8 (for n = 1..7 this becomes 0). Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Number of compositions (ordered partitions) of n+1 into exactly 9 parts. - Juergen Will, Jan 02 2016
Number of weak compositions (ordered weak partitions) of n-8 into exactly 9 parts. - Juergen Will, Jan 02 2016
Partial sums of A000580. - Art Baker, Mar 26 2019

References

  • 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.
  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • 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).

Crossrefs

Programs

Formula

G.f.: x^8/(1-x)^9.
a(n) = A110555(n+1,8). - Reinhard Zumkeller, Jul 27 2005
a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)/8!. - Artur Jasinski, Dec 02 2007
Sum_{k>=8} 1/a(k) = 8/7. - Tom Edgar, Sep 10 2015
Sum_{n>=8} (-1)^n/a(n) = A001787(8)*log(2) - A242091(8)/7! = 1024*log(2) - 74432/105 = 0.9065224171... - Amiram Eldar, Dec 10 2020

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Mar 17 2000
Some formulas referring to other offsets rewritten by R. J. Mathar, Jul 07 2009
3 more terms from William Boyles, Aug 06 2015

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

Original entry on oeis.org

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422
Offset: 0

Views

Author

Keywords

Comments

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

Examples

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014
		

References

  • Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

Crossrefs

Programs

  • Haskell
    a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
    -- Reinhard Zumkeller, Oct 14 2001
    
  • Magma
    [n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017
    
  • Maple
    A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017
  • Mathematica
    Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
    Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)
  • PARI
    a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008
    
  • PARI
    for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

Formula

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A135278 Triangle read by rows, giving the numbers T(n,m) = binomial(n+1, m+1); or, Pascal's triangle A007318 with its left-hand edge removed.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 12, 66, 220, 495, 792, 924, 792
Offset: 0

Views

Author

Zerinvary Lajos, Dec 02 2007

Keywords

Comments

T(n,m) is the number of m-faces of a regular n-simplex.
An n-simplex is the n-dimensional analog of a triangle. Specifically, a simplex is the convex hull of a set of (n + 1) affinely independent points in some Euclidean space of dimension n or higher, i.e., a set of points such that no m-plane contains more than (m + 1) of them. Such points are said to be in general position.
Reversing the rows gives A074909, which as a linear sequence is essentially the same as this.
From Tom Copeland, Dec 07 2007: (Start)
T(n,k) * (k+1)! = A068424. The comment on permuted words in A068424 shows that T is related to combinations of letters defined by connectivity of regular polytope simplexes.
If T is the diagonally-shifted Pascal matrix, binomial(n+m, k+m), for m=1, then T is a fundamental type of matrix that is discussed in A133314 and the following hold.
The infinitesimal matrix generator is given by A132681, so T = LM(1) of A132681 with inverse LM(-1).
With a(k) = (-x)^k / k!, T * a = [ Laguerre(n,x,1) ], a vector array with index n for the Laguerre polynomials of order 1. Other formulas for the action of T are given in A132681.
T(n,k) = (1/n!) (D_x)^n (D_t)^k Gf(x,t) evaluated at x=t=0 with Gf(x,t) = exp[ t * x/(1-x) ] / (1-x)^2.
[O.g.f. for T ] = 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 }. [ O.g.f. for row sums ] = 1 / { (1-x) * (1-2x) }, giving A000225 (without a leading zero) for the row sums. Alternating sign row sums are all 1. [Sign correction noted by Vincent J. Matsko, Jul 19 2015]
O.g.f. for row polynomials = [ (1+q)**(n+1) - 1 ] / [ (1+q) -1 ] = A(1,n+1,q) on page 15 of reference on Grassmann cells in A008292. (End)
Given matrices A and B with A(n,k) = T(n,k)*a(n-k) and B(n,k) = T(n,k)*b(n-k), then A*B = C where C(n,k) = T(n,k)*[a(.)+b(.)]^(n-k), umbrally. The e.g.f. for the row polynomials of A is {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. - Tom Copeland, Aug 21 2008
A007318*A097806 as infinite lower triangular matrices. - Philippe Deléham, Feb 08 2009
Riordan array (1/(1-x)^2, x/(1-x)). - Philippe Deléham, Feb 22 2012
The elements of the matrix inverse are T^(-1)(n,k)=(-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Relation to K-theory: T acting on the column vector (-0,d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168 and also A104712, A111492, and A238363. - Tom Copeland, Apr 11 2014
Number of walks of length p>0 between any two distinct vertices of the complete graph K_(n+2) is W(n+2,p)=(-1)^(p-1)*Sum_{k=0..p-1} T(p-1,k)*(-n-2)^k = ((n+1)^p - (-1)^p)/(n+2) = (-1)^(p-1)*Sum_{k=0..p-1} (-n-1)^k. This is equal to (-1)^(p-1)*Phi(p,-n-1), where Phi is the cyclotomic polynomial when p is an odd prime. For K_3, see A001045; for K_4, A015518; for K_5, A015521; for K_6, A015531; for K_7, A015540. - Tom Copeland, Apr 14 2014
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
See A074909 for associations among this array, the Bernoulli polynomials and their umbral compositional inverses, and the face polynomials of permutahedra and their duals (cf. A019538). - Tom Copeland, Nov 14 2014
From Wolfdieter Lang, Dec 10 2015: (Start)
A(r, n) = T(n+r-2, r-1) = risefac(n,r)/r! = binomial(n+r-1, r), for n >= 1 and r >= 1, gives the array with the number of independent components of a symmetric tensors of rank r (number of indices) and dimension n (indices run from 1 to n). Here risefac(n, k) is the rising factorial.
As(r, n) = T(n+1, r+1) = fallfac(n, r)/r! = binomial(n, r), r >= 1 and n >= 1 (with the triangle entries T(n, k) = 0 for n < k) gives the array with the number of independent components of an antisymmetric tensor of rank r and dimension n. Here fallfac is the falling factorial. (End)
The h-vectors associated to these f-vectors are given by A000012 regarded as a lower triangular matrix. Read as bivariate polynomials, the h-polynomials are the complete homogeneous symmetric polynomials in two variables, found in the compositional inverse of an e.g.f. for A008292, the h-vectors of the permutahedra. - Tom Copeland, Jan 10 2017
For a correlation between the states of a quantum system and the combinatorics of the n-simplex, see Boya and Dixit. - Tom Copeland, Jul 24 2017

Examples

			The triangle T(n, k) begins:
   n\k  0  1   2   3   4   5   6   7   8  9 10 11 ...
   0:   1
   1:   2  1
   2:   3  3   1
   3:   4  6   4   1
   4:   5 10  10   5   1
   5:   6 15  20  15   6   1
   6:   7 21  35  35  21   7   1
   7:   8 28  56  70  56  28   8   1
   8:   9 36  84 126 126  84  36   9   1
   9:  10 45 120 210 252 210 120  45  10  1
  10:  11 55 165 330 462 462 330 165  55 11  1
  11:  12 66 220 495 792 924 792 495 220 66 12  1
  ... reformatted by _Wolfdieter Lang_, Mar 23 2015
Production matrix begins
   2   1
  -1   1   1
   1   0   1   1
  -1   0   0   1   1
   1   0   0   0   1   1
  -1   0   0   0   0   1   1
   1   0   0   0   0   0   1   1
  -1   0   0   0   0   0   0   1   1
   1   0   0   0   0   0   0   0   1   1
- _Philippe Deléham_, Jan 29 2014
From _Wolfdieter Lang_, Nov 08 2018: (Start)
Recurrence [_Philippe Deléham_]: T(7, 3) = 2*35 + 35 - 15 - 20 = 70.
Recurrence from Riordan A- and Z-sequences: [1,1,repeat(0)] and [2, repeat(-1, +1)]: From Z: T(5, 0) = 2*5 - 10 + 10 - 5 + 1 = 6. From A: T(7, 3) = 35 + 35 = 70.
Boas-Buck column k=3 recurrence: T(7, 3) = (5/4)*(1 + 5 + 15 + 35) = 70. (End)
		

Crossrefs

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*1^(i-j), j = 1 .. i) od;
  • Mathematica
    Flatten[Table[CoefficientList[D[1/x ((x + 1) Exp[(x + 1) z] - Exp[z]), {z, k}] /. z -> 0, x], {k, 0, 11}]]
    CoefficientList[CoefficientList[Series[1/((1 - x)*(1 - x - x*y)), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (prod(j=0, k-1, i-j))), ", "))) \\ Derek Orr, Oct 14 2014
    
  • Sage
    Trow = lambda n: sum((x+1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=k..n} binomial(j,k) = binomial(n+1, k+1), n >= k >= 0, else 0. (Partial sum of column k of A007318 (Pascal), or summation on the upper binomial index (Graham et al. (GKP), eq. (5.10). For the GKP reference see A007318.) - Wolfdieter Lang, Aug 22 2012
E.g.f.: 1/x*((1 + x)*exp(t*(1 + x)) - exp(t)) = 1 + (2 + x)*t + (3 + 3*x + x^2)*t^2/2! + .... The infinitesimal generator for this triangle has the sequence [2,3,4,...] on the main subdiagonal and 0's elsewhere. - Peter Bala, Jul 16 2013
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
T(n,k) = A193862(n,k)/2^k. - Philippe Deléham, Jan 29 2014
G.f.: 1/((1-x)*(1-x-x*y)). - Philippe Deléham, Mar 13 2014
From Tom Copeland, Mar 26 2014: (Start)
[From Copeland's 2007 and 2008 comments]
A) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 } (same as Deleham's).
B) The infinitesimal generator for T is given in A132681 with m=1 (same as Bala's), which makes connections to the ubiquitous associated Laguerre polynomials of integer orders, for this case the Laguerre polynomials of order one L(n,-t,1).
C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,-t,1) x^n = exp[t*x/(1-x)]/(1-x)^2 = 1 + (2+t)x + (3+3*t+t^2/2!)x^2 + (4+6*t+4*t^2/2!+t^3/3!)x^3+ ... .
D) E.g.f. of row o.g.f.s: ((1+t)*exp((1+t)*x)-exp(x))/t (same as Bala's).
E) E.g.f. for T(n,k)*a(n-k): {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. For example, for a(k)=2^k, the e.g.f. for the row o.g.f.s is {(2+t) exp[(2+t)x] - 2 exp(2x)}/t.
(End)
From Tom Copeland, Apr 28 2014: (Start)
With different indexing
A) O.g.f. by row: [(1+t)^n-1]/t.
B) O.g.f. of row o.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
These generating functions are related to row e.g.f.s of A111492. (End)
From Tom Copeland, Sep 17 2014: (Start)
A) U(x,s,t)= x^2/[(1-t*x)(1-(s+t)x)] = Sum_{n >= 0} F(n,s,t)x^(n+2) is a generating function for bivariate row polynomials of T, e.g., F(2,s,t)= s^2 + 3s*t + 3t^2 (Buchstaber, 2008).
B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1-s*x) (Buchstaber, 2008).
C) U(x,s,t) = exp(t*x^2*d/dx)U(x,s,0) = U(x/(1-t*x),s,0).
D) U(x,s,t) = Sum[n >= 0, (t*x)^n L(n,-:xD:,-1)] U(x,s,0), where (:xD:)^k=x^k*(d/dx)^k and L(n,x,-1) are the Laguerre polynomials of order -1, related to normalized Lah numbers. (End)
E.g.f. satisfies the differential equation d/dt(e.g.f.(x,t)) = (x+1)*e.g.f.(x,t) + exp(t). - Vincent J. Matsko, Jul 18 2015
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = (t/(e^t - 1))^(m+1) * e^(xt). Norlund gave the relation to the factorials (x-1)!/(x-1-n)! = (x-1) ... (x-n) = NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!. - Tom Copeland, Oct 01 2015
From Wolfdieter Lang, Nov 08 2018: (Start)
Recurrences from the A- and Z- sequences for the Riordan triangle (see the W. Lang link under A006232 with references), which are A(n) = A019590(n+1), [1, 1, repeat (0)] and Z(n) = (-1)^(n+1)*A054977(n), [2, repeat(-1, 1)]:
T(0, 0) = 1, T(n, k) = 0 for n < k, and T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and T(n, k) = T(n-1, k-1) + T(n-1, k), for n >= m >= 1.
Boas-Buck recurrence for columns (see the Aug 10 2017 remark in A036521 also for references):
T(n, k) = ((2 + k)/(n - k))*Sum_{j=k..n-1} T(j, k), for n >= 1, k = 0, 1, ..., n-1, and input T(n, n) = 1, for n >= 0, (the BB-sequences are alpha(n) = 2 and beta(n) = 1). (End)
T(n, k) = [x^k] Sum_{j=0..n} (x+1)^j. - Peter Luschny, Jul 09 2019

Extensions

Edited by Tom Copeland and N. J. A. Sloane, Dec 11 2007

A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 4, 4, 1, 1, 10, 10, 5, 5, 1, 1, 20, 15, 15, 6, 6, 1, 1, 35, 35, 21, 21, 7, 7, 1, 1, 70, 56, 56, 28, 28, 8, 8, 1, 1, 126, 126, 84, 84, 36, 36, 9, 9, 1, 1, 252, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1, 462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1
Offset: 0

Views

Author

Henry Bottomley, May 17 2001

Keywords

Comments

Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe Deléham, May 21 2005
Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry, Feb 28 2006
Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna, Nov 20 2006
Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108. - Philippe Deléham, Mar 17 2007
Triangle T(n,k), 0<=k<=n, read by rows given by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 27 2007
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by Deléham. - Gerald McGarvey, Oct 15 2008
Triangle read by rows = partial sums of A053121 terms starting from the right. - Gary W. Adamson, Oct 24 2008
As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) = (1; -1, 1; 2, -1, 1; -3, 3, -1, 1; ...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331; ... such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965, ... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011
The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*cos(2*Pi/N). By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011

Examples

			The array starts:
   1, 1,  2,  3,  6, 10,  20,  35,   70,  126, ...
   1, 1,  3,  4, 10, 15,  35,  56,  126,  210, ...
   1, 1,  4,  5, 15, 21,  56,  84,  210,  330, ...
   1, 1,  5,  6, 21, 28,  84, 120,  330,  495, ...
   1, 1,  6,  7, 28, 36, 120, 165,  495,  715, ...
   1, 1,  7,  8, 36, 45, 165, 220,  715, 1001, ...
   1, 1,  8,  9, 45, 55, 220, 286, 1001, 1365, ...
   1, 1,  9, 10, 55, 66, 286, 364, 1365, 1820, ...
   1, 1, 10, 11, 66, 78, 364, 455, 1820, 2380, ...
   1, 1, 11, 12, 78, 91, 455, 560, 2380, 3060, ...
Triangle (antidiagonal) version begins:
    1;
    1,   1;
    2,   1,   1;
    3,   3,   1,   1;
    6,   4,   4,   1,   1;
   10,  10,   5,   5,   1,   1;
   20,  15,  15,   6,   6,   1,  1;
   35,  35,  21,  21,   7,   7,  1,  1;
   70,  56,  56,  28,  28,   8,  8,  1,  1;
  126, 126,  84,  84,  36,  36,  9,  9,  1,  1;
  252, 210, 210, 120, 120,  45, 45, 10, 10,  1, 1;
  462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1; ...
Matrix inverse begins:
   1;
  -1,  1;
  -1, -1,   1;
   1, -2,  -1,   1;
   1,  2,  -3,  -1,  1;
  -1,  3,   3,  -4, -1,  1;
  -1, -3,   6,   4, -5, -1,  1;
   1, -4,  -6,  10,  5, -6, -1,  1;
   1,  4, -10, -10, 15,  6, -7, -1, 1; ...
From _Paul Barry_, May 21 2009: (Start)
Production matrix is
  1, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1 (End)
		

Crossrefs

Rows are A001405, A037952, A037955, A037951, A037956, A037953, A037957 etc. Columns are truncated pairs of A000012, A000027, A000217, A000292, A000332, A000389, A000579, etc. Main diagonal is alternate values of A051036.

Programs

  • Maple
    T := proc(n, k) option remember;
    if n = k then 1 elif k < 0 or n < 0 or k > n then 0
    elif k = 0 then T(n-1, 0) + T(n-1, 1) else T(n-1, k-1) + T(n-1, k+1) fi end:
    for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, May 25 2021
  • Mathematica
    t[n_, k_] = Binomial[n, Floor[(n+1)/2 - (-1)^(n-k)*(k+1)/2]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 31 2011 *)
  • PARI
    T(n,k)=binomial(n,(n+1)\2-(-1)^(n-k)*((k+1)\2))

Formula

As a triangle: T(n,k) = binomial(n,m) where m = floor((n+1)/2 - (-1)^(n-k)*(k+1)/2).
a(0, k) = binomial(k, floor(k/2)) = A001405(k); for n>0 T(n, k) = T(n+1, k-2) + T(n-1, k).
n-th row = M^n * V, where M = the infinite tridiagonal matrix with all 1's in the super and subdiagonals and (1,0,0,0,...) in the main diagonal. V = the infinite vector [1,0,0,0,...]. Example: (3,3,1,1,0,0,0,...) = M^3 * V. - Gary W. Adamson, Nov 04 2006
Sum_{k=0..n} T(m,k)*T(n,k) = T(m+n,0) = A001405(m+n). - Philippe Deléham, Feb 26 2007
Sum_{k=0..n} T(n,k)=2^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A127361(n), A126869(n), A001405(n), A000079(n), A127358(n), A127359(n), A127360(n) for x = -2, -1, 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Dec 04 2009

Extensions

Entry revised by N. J. A. Sloane, Nov 22 2006

A087127 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of triangular numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 2*p-1, where a(i,p) satisfies Sum_{i=1..n} C(i+1,2)^p = 3 * C(n+2,3) * Sum_{i=1..2*p-1} a(i,p) * C(n-1,i-1)/(i+2).

Original entry on oeis.org

1, 1, 2, 1, 1, 8, 19, 18, 6, 1, 26, 163, 432, 564, 360, 90, 1, 80, 1135, 6354, 18078, 28800, 26100, 12600, 2520, 1, 242, 7291, 77400, 405060, 1210680, 2211570, 2520000, 1751400, 680400, 113400, 1, 728, 45199, 862218, 7667646, 38350080, 118848420
Offset: 1

Views

Author

André F. Labossière, Aug 11 2003

Keywords

Comments

From Peter Bala, Mar 08 2018: (Start)
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+2,2)^p of degree 2*p in terms of falling factorials: C(x+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the triangular numbers is also given by Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 2..2*p} A122193(p,k)*C(n+2,k+1) for p >= 1. (End)

Examples

			Row 3 contains 1,8,19,18,6, so Sum_{i=1..n} C(i+1,2)^3 = (n+2) * C(n+1,2) * [ a(1,3)/3 + a(2,3)*C(n-1,1)/4 + a(3,3)*C(n-1,2)/5 + a(4,3)*C(n-1,3)/6 + a(5,3)*C(n-1,4)/7 ] = [ (n+2)*(n+1)*n/2 ] * [ 1/3 + (8/4)*C(n-1,1) + (19/5)*C(n-1,2) + (18/6)*C(n-1,3) + (6/7)*C(n-1,4). Cf. A085438 for more details.
From _Peter Bala_, Mar 08 2018: (Start)
Table begins
n=0 |1
n=1 |1   2     1
n=2 |1   8    19    18      6
n=3 |1  26   163   432    564    360     90
n=4 |1  80  1135  6354  18078  28800  26100  12600  2520
...
Row 2: C(i+2,2)^2 = C(i,0) + 8*C(i,1) + 19*C(i,2) + 18*C(i,3) + 6*C(i,4). Hence, Sum_{i = 0..n-1} C(i+2,2)^2 =  C(n,1) + 8*C(n,2) + 19*C(n,3) + 18*C(n,4) + 6*C(n,5). (End)
		

Crossrefs

Programs

  • GAP
    Flat(List([0..6],n->List([0..2*n],k->Sum([0..k],i->(-1)^(k-i)*Binomial(k,i)*Binomial(i+2,2)^n)))); # Muniru A Asiru, Mar 22 2018
  • Maple
    seq(seq(add( (-1)^(k-i)*binomial(k,i)*binomial(i+2,2)^n, i = 0..k), k = 0..2*n), n = 0..8); # Peter Bala, Mar 08 2018
  • Mathematica
    a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 3, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 2, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 2*p - 1}]//Flatten (* G. C. Greubel, Nov 23 2017 *)
    a[i_,p_]:=(-1)^i HypergeometricPFQ[ConstantArray[3,p]~Join~{2-i},ConstantArray[1,p],1];Table[a[i,p],{p,0,10},{i,2,2 p+2}]//Flatten (* Jonathan Burns, Mar 20 2018 *)
  • PARI
    {a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 3, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 2, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 2*p-1, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017
    

Formula

a(1, p) = 1, a(2, p) = 3^(p-1)-1, a(3, p) = 3^(p-1)*[2^(p-1)-2]+1, ..., a(2*p-3, p) = [ (6*p^4-20*p^3+21*p^2-7*p)*(2*p-4)! ]/[3*2^(p-1)], a(2*p-2, p) = [ (p^2-p)*(2*p-3)! ]/2^(p-2), a(2*p-1, p) = [ (p-1)*(2*p-3)! ]/2^(p-2).
a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+3, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+2, i-2*k)^(p-1) ]
From Peter Bala, Mar 08 2018: (Start)
The following remarks assume row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*C(i+2,2)^n. Equivalently, let v_n denote the sequence (1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318. Cf. A122193.
T(n+1,k) = C(k+2,2)*T(n,k) + 2*C(k+1,2)*T(n,k-1) + C(k,2)*T(n,k-2), with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 2*n.
Let R(n,x) denote the n-th row polynomial.
R(n+1,x) = 1/2!*(1 + x)^2*(d/dx)^2 (x^2*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+2,2)^n*x^i/(1 + x)^(i+1).
R(n,x) = (1 + x)^2 o (1 + x)^2 o ... o (1 + x)^2 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^2 o ... o x^2 (n factors) is the n-th row polynomial of A122193.
x^2*R(n,x) = (1 + x)^2 * the n-th row polynomial of A122193 (End)

Extensions

Edited by Dean Hickerson, Aug 16 2003

A115067 a(n) = (3*n^2 - n - 2)/2.

Original entry on oeis.org

0, 4, 11, 21, 34, 50, 69, 91, 116, 144, 175, 209, 246, 286, 329, 375, 424, 476, 531, 589, 650, 714, 781, 851, 924, 1000, 1079, 1161, 1246, 1334, 1425, 1519, 1616, 1716, 1819, 1925, 2034, 2146, 2261, 2379, 2500, 2624, 2751, 2881, 3014, 3150, 3289, 3431, 3576
Offset: 1

Views

Author

Roger L. Bagula, Mar 01 2006

Keywords

Comments

Number of orbits of Aut(Z^7) as function of the infinity norm n of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 6720. - Philippe A.J.G. Chevalier, Dec 28 2015
a(n) is the sum of the numerator and denominator of the reduced fraction resulting from the sum A000217(n-2)/A000217(n-1) + A000217(n-1)/A000217(n), n>1. - J. M. Bergot, Jun 10 2017
For n > 1, a(n) is also the number of (not necessarily maximal) cliques in the (n-1)-Andrásfai graph. - Eric W. Weisstein, Nov 29 2017
a(n+1) is the sum of the lengths of all the segments used to draw a square of side n representing the most classic pattern for walls made of 2 X 1 bricks, known as a 1-over-2 pattern, where each joint between neighboring bricks falls over the center of the brick below. - Stefano Spezia, Jun 05 2021

Examples

			Illustrations for n = 2..7 from _Stefano Spezia_, Jun 05 2021:
       _           _ _          _ _ _
      |_|         |_|_|        |_|_ _|
                  |_ _|        |_ _|_|
                               |_|_ _|
   a(2) = 4     a(3) = 11     a(4) = 21
    _ _ _ _     _ _ _ _ _    _ _ _ _ _ _
   |_ _|_ _|   |_ _|_ _|_|  |_ _|_ _|_ _|
   |_|_ _|_|   |_|_ _|_ _|  |_|_ _|_ _|_|
   |_ _|_ _|   |_ _|_ _|_|  |_ _|_ _|_ _|
   |_|_ _|_|   |_|_ _|_ _|  |_|_ _|_ _|_|
               |_ _|_ _|_|  |_ _|_ _|_ _|
                            |_|_ _|_ _|_|
   a(5) = 34    a(6) = 50     a(7) = 69
		

Crossrefs

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.
Orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A008585, A005843, A001477, A000217.

Programs

Formula

a(n) = (3*n+2)*(n-1)/2.
a(n+1) = n*(3*n + 5)/2. - Omar E. Pol, May 21 2008
a(n) = 3*n + a(n-1) - 2 for n>1, a(1)=0. - Vincenzo Librandi, Nov 13 2010
a(n) = A095794(-n). - Bruno Berselli, Sep 02 2011
G.f.: x^2*(4-x) / (1-x)^3. - R. J. Mathar, Sep 02 2011
a(n) = A055998(2*n-2) - A055998(n-1). - Bruno Berselli, Sep 23 2016
E.g.f.: exp(x)*x*(8 + 3*x)/2. - Stefano Spezia, May 19 2021
From Amiram Eldar, Feb 22 2022: (Start)
Sum_{n>=2} 1/a(n) = Pi/(5*sqrt(3)) - 3*log(3)/5 + 21/25.
Sum_{n>=2} (-1)^n/a(n) = 4*log(2)/5 - 2*Pi/(5*sqrt(3)) + 9/25. (End)
a(n) = Sum_{j=0..n-2} (2*n-j) = Sum_{j=0..n-2} (n+2+j), for n>=1. See the trapezoid link. - Leo Tavares, May 20 2022

Extensions

Edited by N. J. A. Sloane, Mar 05 2006
Previous Showing 11-20 of 123 results. Next