A000914 Stirling numbers of the first kind: s(n+2, n).

0, 2, 11, 35, 85, 175, 322, 546, 870, 1320, 1925, 2717, 3731, 5005, 6580, 8500, 10812, 13566, 16815, 20615, 25025, 30107, 35926, 42550, 50050, 58500, 67977, 78561, 90335, 103385, 117800, 133672, 151096, 170170, 190995, 213675, 238317, 265031

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

Sum of product of unordered pairs of numbers from {1..n+1}.

Number of edges of a complete k-partite graph of order k*(k+1)/2 (A000217), K_1,2,3,...,k. - Roberto E. Martinez II, Oct 18 2001

This sequence holds the x^(n-2) coefficient of the characteristic polynomial of the N X N matrix A formed by MAX(i,j), where i is the row index and j is the column index of element A[i][j], 1 <= i,j <= N. Here N >= 2. - Paul Max Payton, Sep 06 2005

The sequence contains the partial sums of A006002, which represent the areas beneath lines created by the triangular numbers plotted (t(1),t(2)) connected to (t(2),t(3)) then (t(3),t(4))...(t(n-1),t(n)) and the x-axis. - J. M. Bergot, May 05 2012

Number of functions f from [n+2] to [n+2] with f(x)=x for exactly n elements x of [n+2] and f(x)>x for exactly two elements x of [n+2]. To prove this, let the two elements of [n+2] with a larger image be labeled i and j. Note both i and j must be less than n+2. Then there are (n+2-i) choices for f(i) and (n+2-j) choices for f(j). Summing the product of the number of choices over all sets {i,j} gives us "Sum of product of unordered pairs of numbers from {1..n+1}" in the first line of the Comments Section. See the example in the Example Section below. - Dennis P. Walsh, Sep 06 2017

Zhu Shijie gives in his Magnus Opus "Jade Mirror of the Four Unknowns" the problem: "Apples are piled in the form of a triangular pyramid. The top apple is worth 2 and the price of the whole is 1320. Each apple in one layer costs 1 less than an apple in the next layer below." We find the solution 9 to this problem in this sequence 1320 = a(9). Zhu Shijie gave the solution polynomial: "Let the element tian be the number of apples in a side of the base. From the statement we have 31680 for the negative shi, 10 for the positive fang, 21 for the positive first lian, 14 for the positive last lian, and 3 for the positive yu." This translates into the polynomial equation: 3*x^4 + 14*x^3 + 21*x^2 + 10*x - 31680 = 0. - Thomas Scheuerle, Feb 10 2025

Examples

			Examples include E(K_1,2,3) = s(2+2,2) = 11 and E(K_1,2,3,4,5) = s(4+2,4) = 85, where E is the function that counts edges of graphs.
For n=2 the a(2)=11 functions f:[4]->[4] with exactly two f(x)=x and two f(x)>x are given by the 11 image vectors of form <f(1),f(2),f(3),f(4)> that follow: <1,3,4,4>, <1,4,4,4>, <2,2,4,4>, <3,2,4,4>, <4,2,4,4>, <2,3,3,4>, <2,4,3,4>, <3,3,3,4>, <3,4,3,4>, <4,3,3,4>, and <4,4,3,4>. - _Dennis P. Walsh_, Sep 06 2017
		

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. 833.
• George E. Andrews, Number Theory, Dover Publications, New York, 1971, p. 4.
• Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
• F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
• H. S. Hall and S. R. Knight, Higher Algebra, Fourth Edition, Macmillan, 1891, p. 518.
• Zhu Shijie, Jade Mirror of the Four Unknowns (Siyuan yujian), Book III Guo Duo Die Gang (Piles of Fruit), Problem number 1, 1303.
• 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).

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
• Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]
• Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., Vol. 1, No. 3 (1926), pp. 44-49. [Annotated scanned copy]
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• Zhu Shijie, Jade Mirror of the Four Unknowns 2, Translation by Library of Chinese classics, original from 1303.
• Wikipedia, Jade Mirror of the Four Unknowns.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000217, A000290, A033428, A033581, A033583, A008275, A052149.

Cf. similar sequences listed in A241765.

Cf. A001296.

Cf. A006325(n+1) (Zhu Shijie's problem number 2 uses a pyramid with square base).

Programs

Haskell

a000914 n = a000914_list !! n
a000914_list = scanl1 (+) a006002_list
-- Reinhard Zumkeller, Mar 25 2014

Magma

[StirlingFirst(n+2, n): n in [0..40]]; // Vincenzo Librandi, May 28 2019

Maple

A000914 := n -> 1/24*(n+1)*n*(n+2)*(3*n+5);
A000914 := proc(n)
    combinat[stirling1](n+2,n) ;
end proc: # R. J. Mathar, May 19 2016

Mathematica

Table[StirlingS1[n+2,n],{n,0,40}] (* Harvey P. Dale, Aug 24 2011 *)
a[ n_] := n (n + 1) (n + 2) (3 n + 5) / 24; (* Michael Somos, Sep 04 2017 *)

PARI

a(n)=sum(i=1,n+1,sum(j=1,n+1,i*j*(i
				

PARI

a(n)=sum(i=1,n+1,sum(j=1,i-1,i*j)) \\ Charles R Greathouse IV, Apr 07 2015

PARI

a(n) = binomial(n+2, 3)*(3*n+5)/4 \\ Charles R Greathouse IV, Apr 07 2015

Sage

[stirling_number1(n+2, n) for n in range(41)] # Zerinvary Lajos, Mar 14 2009

Formula

a(n) = binomial(n+2, 3)*(3*n+5)/4 = (n+1)*n*(n+2)*(3*n+5)/24.

E.g.f.: exp(x)*x*(48 + 84*x + 32*x^2 + 3*x^3)/24.

G.f.: (2*x+x^2)/(1-x)^5. - Simon Plouffe in his 1992 dissertation.

a(n) = Sum_{i=1..n} i*(i+1)^2/2. - Jon Perry, Jul 31 2003

a(n) = A052149(n+1)/2. - J. M. Bergot, Jun 02 2012

-(3*n+2)*(n-1)*a(n) + (n+2)*(3*n+5)*a(n-1) = 0. - R. J. Mathar, Apr 30 2015

a(n) = a(n-1) + (n+1)*binomial(n+1,2) for n >= 1. - Dennis P. Walsh, Sep 21 2015

a(n) = A001296(-2-n) for all n in Z. - Michael Somos, Sep 04 2017

From Amiram Eldar, Jan 10 2022: (Start)

Sum_{n>=1} 1/a(n) = 162*log(3)/5 - 18*sqrt(3)*Pi/5 - 384/25.

Sum_{n>=1} (-1)^(n+1)/a(n) = 36*sqrt(3)*Pi/5 - 96*log(2)/5 - 636/25. (End)

a(n) = 3*A000332(n+3) - A000292(n). - Yasser Arath Chavez Reyes, Apr 03 2024

Extensions

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 17 2000

Comments from Michael Somos, Jan 29 2000

Erroneous duplicate of the polynomial formula removed by R. J. Mathar, Sep 15 2009