cp's OEIS Frontend

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

%I A000914 M1998 N0789 #143 Mar 06 2025 14:51:19
%S A000914 0,2,11,35,85,175,322,546,870,1320,1925,2717,3731,5005,6580,8500,
%T A000914 10812,13566,16815,20615,25025,30107,35926,42550,50050,58500,67977,
%U A000914 78561,90335,103385,117800,133672,151096,170170,190995,213675,238317,265031
%N A000914 Stirling numbers of the first kind: s(n+2, n).
%C A000914 Sum of product of unordered pairs of numbers from {1..n+1}.
%C A000914 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
%C A000914 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
%C A000914 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
%C A000914 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
%C A000914 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
%D A000914 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.
%D A000914 George E. Andrews, Number Theory, Dover Publications, New York, 1971, p. 4.
%D A000914 Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
%D A000914 F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
%D A000914 H. S. Hall and S. R. Knight, Higher Algebra, Fourth Edition, Macmillan, 1891, p. 518.
%D A000914 Zhu Shijie, Jade Mirror of the Four Unknowns (Siyuan yujian), Book III Guo Duo Die Gang (Piles of Fruit), Problem number 1, 1303.
%D A000914 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A000914 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A000914 T. D. Noe, <a href="/A000914/b000914.txt">Table of n, a(n) for n = 0..1000</a>
%H A000914 M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H A000914 Karl Dienger, <a href="/A000217/a000217.pdf">Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung</a>, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]
%H A000914 Robert E. Moritz, <a href="/A001701/a001701.pdf">On the sum of products of n consecutive integers</a>, Univ. Washington Publications in Math., Vol. 1, No. 3 (1926), pp. 44-49. [Annotated scanned copy]
%H A000914 Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
%H A000914 Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992.
%H A000914 Zhu Shijie, <a href="https://archive.org/details/jademirrorofthefourunknowns2/">Jade Mirror of the Four Unknowns 2</a>, Translation by Library of Chinese classics, original from 1303.
%H A000914 Wikipedia, <a href="https://en.wikipedia.org/wiki/Jade_Mirror_of_the_Four_Unknowns">Jade Mirror of the Four Unknowns</a>.
%H A000914 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A000914 a(n) = binomial(n+2, 3)*(3*n+5)/4 = (n+1)*n*(n+2)*(3*n+5)/24.
%F A000914 E.g.f.: exp(x)*x*(48 + 84*x + 32*x^2 + 3*x^3)/24.
%F A000914 G.f.: (2*x+x^2)/(1-x)^5. - _Simon Plouffe_ in his 1992 dissertation.
%F A000914 a(n) = Sum_{i=1..n} i*(i+1)^2/2. - _Jon Perry_, Jul 31 2003
%F A000914 a(n) = A052149(n+1)/2. - _J. M. Bergot_, Jun 02 2012
%F A000914 -(3*n+2)*(n-1)*a(n) + (n+2)*(3*n+5)*a(n-1) = 0. - _R. J. Mathar_, Apr 30 2015
%F A000914 a(n) = a(n-1) + (n+1)*binomial(n+1,2) for n >= 1. - _Dennis P. Walsh_, Sep 21 2015
%F A000914 a(n) = A001296(-2-n) for all n in Z. - _Michael Somos_, Sep 04 2017
%F A000914 From _Amiram Eldar_, Jan 10 2022: (Start)
%F A000914 Sum_{n>=1} 1/a(n) = 162*log(3)/5 - 18*sqrt(3)*Pi/5 - 384/25.
%F A000914 Sum_{n>=1} (-1)^(n+1)/a(n) = 36*sqrt(3)*Pi/5 - 96*log(2)/5 - 636/25. (End)
%F A000914 a(n) = 3*A000332(n+3) - A000292(n). - _Yasser Arath Chavez Reyes_, Apr 03 2024
%e A000914 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.
%e A000914 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
%p A000914 A000914 := n -> 1/24*(n+1)*n*(n+2)*(3*n+5);
%p A000914 A000914 := proc(n)
%p A000914     combinat[stirling1](n+2,n) ;
%p A000914 end proc: # _R. J. Mathar_, May 19 2016
%t A000914 Table[StirlingS1[n+2,n],{n,0,40}] (* _Harvey P. Dale_, Aug 24 2011 *)
%t A000914 a[ n_] := n (n + 1) (n + 2) (3 n + 5) / 24; (* _Michael Somos_, Sep 04 2017 *)
%o A000914 (PARI) a(n)=sum(i=1,n+1,sum(j=1,n+1,i*j*(i<j)))
%o A000914 (PARI) a(n)=sum(i=1,n+1,sum(j=1,i-1,i*j)) \\ _Charles R Greathouse IV_, Apr 07 2015
%o A000914 (PARI) a(n) = binomial(n+2, 3)*(3*n+5)/4 \\ _Charles R Greathouse IV_, Apr 07 2015
%o A000914 (Sage) [stirling_number1(n+2, n) for n in range(41)] # _Zerinvary Lajos_, Mar 14 2009
%o A000914 (Haskell)
%o A000914 a000914 n = a000914_list !! n
%o A000914 a000914_list = scanl1 (+) a006002_list
%o A000914 -- _Reinhard Zumkeller_, Mar 25 2014
%o A000914 (Magma) [StirlingFirst(n+2, n): n in [0..40]]; // _Vincenzo Librandi_, May 28 2019
%Y A000914 Cf. A000217, A000290, A033428, A033581, A033583, A008275, A052149.
%Y A000914 Cf. similar sequences listed in A241765.
%Y A000914 Cf. A001296.
%Y A000914 Cf. A006325(n+1) (Zhu Shijie's problem number 2 uses a pyramid with square base).
%K A000914 nonn,easy,nice
%O A000914 0,2
%A A000914 _N. J. A. Sloane_
%E A000914 More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 17 2000
%E A000914 Comments from _Michael Somos_, Jan 29 2000
%E A000914 Erroneous duplicate of the polynomial formula removed by _R. J. Mathar_, Sep 15 2009