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.

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A016754 Odd squares: a(n) = (2n+1)^2. Also centered octagonal numbers.

Original entry on oeis.org

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025
Offset: 0

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Keywords

Comments

The brown rat (rattus norwegicus) breeds very quickly. It can give birth to other rats 7 times a year, starting at the age of three months. The average number of pups is 8. The present sequence gives the total number of rats, when the intervals are 12/7 of a year and a young rat starts having offspring at 24/7 of a year. - Hans Isdahl, Jan 26 2008
Numbers n such that tau(n) is odd where tau(x) denotes the Ramanujan tau function (A000594). - Benoit Cloitre, May 01 2003
If Y is a fixed 2-subset of a (2n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Binomial transform of [1, 8, 8, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
All terms of this sequence are of the form 8k+1. For numbers 8k+1 which aren't squares see A138393. Numbers 8k+1 are squares iff k is a triangular number from A000217. And squares have form 4n(n+1)+1. - Artur Jasinski, Mar 27 2008
Sequence arises from reading the line from 1, in the direction 1, 25, ... and the line from 9, in the direction 9, 49, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Equals the triangular numbers convolved with [1, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
First differences: A008590(n) = a(n) - a(n-1) for n>0. - Reinhard Zumkeller, Nov 08 2009
Central terms of the triangle in A176271; cf. A000466, A053755. - Reinhard Zumkeller, Apr 13 2010
Odd numbers with odd abundance. Odd numbers with even abundance are in A088828. Even numbers with odd abundance are in A088827. Even numbers with even abundance are in A088829. - Jaroslav Krizek, May 07 2011
Appear as numerators in the non-simple continued fraction expansion of Pi-3: Pi-3 = K_{k>=1} (1-2*k)^2/6 = 1/(6+9/(6+25/(6+49/(6+...)))), see also the comment in A007509. - Alexander R. Povolotsky, Oct 12 2011
Ulam's spiral (SE spoke). - Robert G. Wilson v, Oct 31 2011
All terms end in 1, 5 or 9. Modulo 100, all terms are among { 1, 9, 21, 25, 29, 41, 49, 61, 69, 81, 89 }. - M. F. Hasler, Mar 19 2012
Right edge of both triangles A214604 and A214661: a(n) = A214604(n+1,n+1) = A214661(n+1,n+1). - Reinhard Zumkeller, Jul 25 2012
Also: Odd numbers which have an odd sum of divisors (= sigma = A000203). - M. F. Hasler, Feb 23 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective even leg b (A231100); sequence gives values c-b, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
For n>1 a(n) is twice the area of the irregular quadrilateral created by the points ((n-2)*(n-1),(n-1)*n/2), ((n-1)*n/2,n*(n+1)/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+2)*(n+3)/2,(n+1)*(n+2)/2). - J. M. Bergot, May 27 2014
Number of pairs (x, y) of Z^2, such that max(abs(x), abs(y)) <= n. - Michel Marcus, Nov 28 2014
Except for a(1)=4, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
a(n) is the sum of 2n+1 consecutive numbers, the first of which is n+1. - Ivan N. Ianakiev, Dec 21 2016
a(n) is the number of 2 X 2 matrices with all elements in {0..n} with determinant = 2*permanent. - Indranil Ghosh, Dec 25 2016
Engel expansion of Pi*StruveL_0(1)/2 where StruveL_0(1) is A197037. - Benedict W. J. Irwin, Jun 21 2018
Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; the segments on the hypotenuse {p = a(n)/A001844(n), q = A060300(n)/A001844(n) = A001844(n) - p} and their ratio p/q = a(n)/A060300(n) are irreducible fractions in Q\Z. X values are A005408, Y values are A046092, Z values are A001844. - Ralf Steiner, Feb 25 2020
a(n) is the number of large or small squares that are used to tile primitive squares of type 2 (A344332). - Bernard Schott, Jun 03 2021
Also, positive odd integers with an odd number of odd divisors (for similar sequence with 'even', see A348005). - Bernard Schott, Nov 21 2021
a(n) is the least odd number k = x + y, with 0 < x < y, such that there are n distinct pairs (x,y) for which x*y/k is an integer; for example, a(2) = 25 and the two corresponding pairs are (5,20) and (10,15). The similar sequence with 'even' is A016742 (see Comment of Jan 26 2018). - Bernard Schott, Feb 24 2023
From Peter Bala, Jan 03 2024: (Start)
The sequence terms are the exponents of q in the series expansions of the following infinite products:
1) q*Product_{n >= 1} (1 - q^(16*n))*(1 + q^(8*n)) = q + q^9 + q^25 + q^49 + q^81 + q^121 + q^169 + ....
2) q*Product_{n >= 1} (1 + q^(16*n))*(1 - q^(8*n)) = q - q^9 - q^25 + q^49 + q^81 - q^121 - q^169 + + - - ....
3) q*Product_{n >= 1} (1 - q^(8*n))^3 = q - 3*q^9 + 5*q^25 - 7*q^49 + 9*q^81 - 11*q^121 + 13*q^169 - + ....
4) q*Product_{n >= 1} ( (1 + q^(8*n))*(1 - q^(16*n))/(1 + q^(16*n)) )^3 = q + 3*q^9 - 5*q^25 - 7*q^49 + 9*q^81 + 11*q^121 - 13*q^169 - 15*q^225 + + - - .... (End)

References

  • L. Lorentzen and H. Waadeland, Continued Fractions with Applications, North-Holland 1992, p. 586.

Crossrefs

Cf. A000447 (partial sums).
Cf. A348005, A379481 [= a(A048673(n)-1)].
Partial sums of A022144.
Positions of odd terms in A341528.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = 1 + Sum_{i=1..n} 8*i = 1 + 8*A000217(n). - Xavier Acloque, Jan 21 2003; Zak Seidov, May 07 2006; Robert G. Wilson v, Dec 29 2010
O.g.f.: (1+6*x+x^2)/(1-x)^3. - R. J. Mathar, Jan 11 2008
a(n) = 4*n*(n + 1) + 1 = 4*n^2 + 4*n + 1. - Artur Jasinski, Mar 27 2008
a(n) = A061038(2+4n). - Paul Curtz, Oct 26 2008
Sum_{n>=0} 1/a(n) = Pi^2/8 = A111003. - Jaume Oliver Lafont, Mar 07 2009
a(n) = A000290(A005408(n)). - Reinhard Zumkeller, Nov 08 2009
a(n) = a(n-1) + 8*n with n>0, a(0)=1. - Vincenzo Librandi, Aug 01 2010
a(n) = A033951(n) + n. - Reinhard Zumkeller, May 17 2009
a(n) = A033996(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = (A005408(n))^2. - Zak Seidov, Nov 29 2011
From George F. Johnson, Sep 05 2012: (Start)
a(n+1) = a(n) + 4 + 4*sqrt(a(n)).
a(n-1) = a(n) + 4 - 4*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 8.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2).
(a(n+1) - a(n-1))/8 = sqrt(a(n)).
a(n+1)*a(n-1) = (a(n)-4)^2.
a(n) = 2*A046092(n) + 1 = 2*A001844(n) - 1 = A046092(n) + A001844(n).
Limit_{n -> oo} a(n)/a(n-1) = 1. (End)
a(n) = binomial(2*n+2,2) + binomial(2*n+1,2). - John Molokach, Jul 12 2013
E.g.f.: (1 + 8*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 23 2016
a(n) = A101321(8,n). - R. J. Mathar, Jul 28 2016
Product_{n>=1} A033996(n)/a(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A014105(n) + A000384(n+1). - Bruce J. Nicholson, Nov 11 2017
a(n) = A003215(n) + A002378(n). - Klaus Purath, Jun 09 2020
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 13*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/e. (End)
Sum_{n>=0} (-1)^n/a(n) = A006752. - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi/4 (A003881). (End)
From Leo Tavares, Nov 24 2021: (Start)
a(n) = A014634(n) - A002943(n). See Diamond Triangles illustration.
a(n) = A003154(n+1) - A046092(n). See Diamond Stars illustration. (End)
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 ))))).
3/2 - 2*log(2) = Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 - ... ))))).
Row 2 of A142992. (End)
From Peter Bala, Mar 26 2024: (Start)
8*a(n) = (2*n + 1)*(a(n+1) - a(n-1)).
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 1/2 - Pi/8 = 1/(9 + (1*3)/(8 + (3*5)/(8 + ... + (4*n^2 - 1)/(8 + ... )))). For the continued fraction use Lorentzen and Waadeland, p. 586, equation 4.7.9 with n = 1. Cf. A057813. (End)

Extensions

Additional description from Terrel Trotter, Jr., Apr 06 2002

A000522 Total number of ordered k-tuples (k=0..n) of distinct elements from an n-element set: a(n) = Sum_{k=0..n} n!/k!.

Original entry on oeis.org

1, 2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, 9864101, 108505112, 1302061345, 16926797486, 236975164805, 3554627472076, 56874039553217, 966858672404690, 17403456103284421, 330665665962404000, 6613313319248080001, 138879579704209680022, 3055350753492612960485, 70273067330330098091156
Offset: 0

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Author

Keywords

Comments

Total number of permutations of all subsets of an n-set.
Also the number of one-to-one sequences that can be formed from n distinct objects.
Old name "Total number of permutations of a set with n elements", or the same with the word "arrangements", both sound too much like A000142.
Related to number of operations of addition and multiplication to evaluate a determinant of order n by cofactor expansion - see A026243.
a(n) is also the number of paths (without loops) in the complete graph on n+2 vertices starting at one vertex v1 and ending at another v2. Example: when n=2 there are 5 paths in the complete graph with 4 vertices starting at the vertex 1 and ending at the vertex 2: (12),(132),(142),(1342),(1432) so a(2) = 5. - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001; comment corrected by Jonathan Coxhead, Mar 21 2003
Also row sums of Table A008279, which can be generated by taking the derivatives of x^k. For example, for y = 1*x^3, y' = 3x^2, y" = 6x, y'''= 6 so a(4) = 1 + 3 + 6 + 6 = 16. - Alford Arnold, Dec 15 1999
a(n) is the permanent of the n X n matrix with 2s on the diagonal and 1s elsewhere. - Yuval Dekel, Nov 01 2003
(A000166 + this_sequence)/2 = A009179, (A000166 - this_sequence)/2 = A009628.
Stirling transform of A006252(n-1) = [1,1,1,2,4,14,38,...] is a(n-1) = [1,2,5,16,65,...]. - Michael Somos, Mar 04 2004
Number of {12,12*,21*}- and {12,12*,2*1}-avoiding signed permutations in the hyperoctahedral group.
a(n) = b such that Integral_{x=0..1} x^n*exp(-x) dx = a-b*exp(-1). - Sébastien Dumortier, Mar 05 2005
a(n) is the number of permutations on [n+1] whose left-to-right record lows all occur at the start. Example: a(2) counts all permutations on [3] except 231 (the last entry is a record low but its predecessor is not). - David Callan, Jul 20 2005
a(n) is the number of permutations on [n+1] that avoid the (scattered) pattern 1-2-3|. The vertical bar means the "3" must occur at the end of the permutation. For example, 21354 is not counted by a(4): 234 is an offending subpermutation. - David Callan, Nov 02 2005
Number of deco polyominoes of height n+1 having no reentrant corners along the lower contour (i.e., no vertical step that is followed by a horizontal step). In other words, a(n)=A121579(n+1,0). A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: a(1)=2 because the only deco polyominoes of height 2 are the vertical and horizontal dominoes, having no reentrant corners along their lower contours. - Emeric Deutsch, Aug 16 2006
Unreduced numerators of partial sums of the Taylor series for e. - Jonathan Sondow, Aug 18 2006
a(n) is the number of permutations on [n+1] (written in one-line notation) for which the subsequence beginning at 1 is increasing. Example: a(2)=5 counts 123, 213, 231, 312, 321. - David Callan, Oct 06 2006
a(n) is the number of permutations (written in one-line notation) on the set [n + k], k >= 1, for which the subsequence beginning at 1,2,...,k is increasing. Example: n = 2, k = 2. a(2) = 5 counts 1234, 3124, 3412, 4123, 4312. - Peter Bala, Jul 29 2014
a(n) and (1,-2,3,-4,5,-6,7,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Nov 01 2007
Consider the subsets of the set {1,2,3,...,n} formed by the first n integers. E.g., for n = 3 we have {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}. Let the variable sbst denote a subset. For each subset sbst we determine its number of parts, that is nprts(sbst). The sum over all possible subsets is written as Sum_{sbst=subsets}. Then a(n) = Sum_{sbst=subsets} nprts(sbst)!. E.g., for n = 3 we have 1!+1!+1!+1!+2!+2!+2!+3!=16. - Thomas Wieder, Jun 17 2006
Equals row sums of triangle A158359(unsigned). - Gary W. Adamson, Mar 17 2009
Equals eigensequence of triangle A158821. - Gary W. Adamson, Mar 30 2009
For positive n, equals 1/BarnesG(n+1) times the determinant of the n X n matrix whose (i,j)-coefficient is the (i+j)th Bell number. - John M. Campbell, Oct 03 2011
a(n) is the number of n X n binary matrices with i) at most one 1 in each row and column and ii) the subset of rows that contain a 1 must also be the columns that contain a 1. Cf. A002720 where restriction ii is removed. - Geoffrey Critzer, Dec 20 2011
Number of restricted growth strings (RGS) [d(1),d(2),...,d(n)] such that d(k) <= k and d(k) <= 1 + (number of nonzero digits in prefix). The positions of nonzero digits determine the subset, and their values (decreased by 1) are the (left) inversion table (a rising factorial number) for the permutation, see example. - Joerg Arndt, Dec 09 2012
Number of a restricted growth strings (RGS) [d(0), d(1), d(2), ..., d(n)] where d(k) >= 0 and d(k) <= 1 + chg([d(0), d(1), d(2), ..., d(k-1)]) and chg(.) gives the number of changes of its argument. Replacing the function chg(.) by a function asc(.) that counts the ascents in the prefix gives A022493 (ascent sequences). - Joerg Arndt, May 10 2013
The sequence t(n) = number of i <= n such that floor(e*i!) is a square is mentioned in the abstract of Luca & Shparlinski. The values are t(n) = 0 for 0 <= n <= 2 and t(n) = 1 for at least 3 <= n <= 300. - R. J. Mathar, Jan 16 2014
a(n) = p(n,1) = q(n,1), where p and q are polynomials defined at A248664 and A248669. - Clark Kimberling, Oct 11 2014
a(n) is the number of ways at most n people can queue up at a (slow) ticket counter when one or more of the people may choose not to queue up. Note that there are C(n,k) sets of k people who quene up and k! ways to queue up. Since k can run from 0 to n, a(n) = Sum_{k=0..n} n!/(n-k)! = Sum_{k=0..n} n!/k!. For example, if n=3 and the people are A(dam), B(eth), and C(arl), a(3)=16 since there are 16 possible lineups: ABC, ACB, BAC, BCA, CAB, CBA, AB, BA, AC, CA, BC, CB, A, B, C, and empty queue. - Dennis P. Walsh, Oct 02 2015
As the row sums of A008279, Motzkin uses the abbreviated notation $n_<^\Sigma$ for a(n).
The piecewise polynomial function f defined by f(x) = a(n)*x^n/n! on each interval [ 1-1/a(n), 1-1/a(n+1) ) is continuous on [0,1) and lim_{x->1} f(x) = e. - Luc Rousseau, Oct 15 2019
a(n) is composite for 3 <= n <= 2015, but a(2016) is prime (or at least a strong pseudoprime): see Johansson link. - Robert Israel, Jul 27 2020 [a(2016) is prime, ECPP certificate generated with CM 0.4.3 and checked by factordb. - Jason H Parker, Jun 15 2025]
In general, sequences of the form a(0)=a, a(n) = n*a(n-1) + k, n>0, will have a closed form of n!*a + floor(n!*(e-1))*k. - Gary Detlefs, Oct 26 2020
From Peter Bala, Apr 03 2022: (Start)
a(2*n) is odd and a(2*n+1) is even. More generally, a(n+k) == a(n) (mod k) for all n and k. It follows that for each positive integer k, the sequence obtained by reducing a(n) modulo k is periodic, with the exact period dividing k. Various divisibility properties of the sequence follow from this; for example, a(5*n+2) == a(5*n+4) == 0 (mod 5), a(25*n+7) == a(25*n+19) == 0 (mod 25) and a(13*n+4) == a(13*n+10)== a(13*n+12) == 0 (mod 13). (End)
Number of possible ranking options on a typical ranked choice voting ballot with n candidates (allowing undervotes). - P. Christopher Staecker, May 05 2024
From Thomas Scheuerle, Dec 28 2024: (Start)
Number of decorated permutations of size n.
Number of Le-diagrams with bounding box semiperimeter n, for n > 0.
By counting over all k = 1..n and n > 0, the number of positroid cells for the totally nonnegative real Grassmannian Gr(k, n), equivalently the number of Grassmann necklaces of type (k, n). (End)

Examples

			G.f. = 1 + 2*x + 5*x^2 + 16*x^3 + 65*x^4 + 326*x^5 + 1957*x^6 + 13700*x^7 + ...
With two objects we can form 5 sequences: (), (a), (b), (a,b), (b,a), so a(2) = 5.
From _Joerg Arndt_, Dec 09 2012: (Start)
The 16 arrangements of the 3-set and their RGS (dots denote zeros) are
  [ #]       RGS        perm. of subset
  [ 1]    [ . . . ]      [  ]
  [ 2]    [ . . 1 ]      [ 3 ]
  [ 3]    [ . 1 . ]      [ 2 ]
  [ 4]    [ . 1 1 ]      [ 2 3 ]
  [ 5]    [ . 1 2 ]      [ 3 2 ]
  [ 6]    [ 1 . . ]      [ 1 ]
  [ 7]    [ 1 . 1 ]      [ 1 3 ]
  [ 8]    [ 1 . 2 ]      [ 3 1 ]
  [ 9]    [ 1 1 . ]      [ 1 2 ]
  [10]    [ 1 1 1 ]      [ 1 2 3 ]
  [11]    [ 1 1 2 ]      [ 1 3 2 ]
  [12]    [ 1 1 3 ]      [ 2 3 1 ]
  [13]    [ 1 2 . ]      [ 2 1 ]
  [14]    [ 1 2 1 ]      [ 2 1 3 ]
  [15]    [ 1 2 2 ]      [ 3 1 2 ]
  [16]    [ 1 2 3 ]      [ 3 2 1 ]
(End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 75, Problem 9.
  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 65, p. 23, Ellipses, Paris 2008.
  • J. M. Gandhi, On logarithmic numbers, Math. Student, 31 (1963), 73-83.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E11.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 16.
  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.
  • 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

Average of n-th row of triangle in A068424 [Corrected by N. J. A. Sloane, Feb 29 2024].
Row sums of A008279 and A094816.
First differences give A001339. Second differences give A001340.
Partial sums are in A001338, A002104.
A row of the array in A144502.
See also A370973, Nearest integer to e*n!.

Programs

  • Haskell
    import Data.List (subsequences, permutations)
    a000522 = length . choices . enumFromTo 1 where
    choices = concat . map permutations . subsequences
    -- Reinhard Zumkeller, Feb 21 2012, Oct 25 2010
    
  • Magma
    [1] cat [n eq 1 select (n+1) else n*Self(n-1)+1: n in [1..25]]; // Vincenzo Librandi, Feb 15 2015
    
  • Maple
    a(n):= exp(1)*int(x^n*exp(-x)*Heaviside(x-1), x=0..infinity); # Karol A. Penson, Oct 01 2001
    A000522 := n->add(n!/k!,k=0..n);
    G(x):=exp(x)/(1-x): f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..20);
    # Zerinvary Lajos, Apr 03 2009
    G:=exp(z)/(1-z): Gser:=series(G,z=0,21):
    for n from 0 to 20 do a(n):=n!*coeff(Gser,z,n): end do
    # Paul Weisenhorn, May 30 2010
    k := 1; series(hypergeom([1,k],[],x/(1-x))/(1-x), x=0, 20); # Mark van Hoeij, Nov 07 2011
    # one more Maple program:
    a:= proc(n) option remember;
          `if`(n<0, 0, 1+n*a(n-1))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 13 2019
    seq(simplify(KummerU(-n, -n, 1)), n = 0..23); # Peter Luschny, May 10 2022
  • Mathematica
    Table[FunctionExpand[Gamma[n + 1, 1]*E], {n, 0, 24}]
    nn = 20; Accumulate[Table[1/k!, {k, 0, nn}]] Range[0, nn]! (* Jan Mangaldan, Apr 21 2013 *)
    FoldList[#1*#2 + #2 &, 0, Range@ 23] + 1 (* or *)
    f[n_] := Floor[E*n!]; f[0] = 1; Array[f, 20, 0] (* Robert G. Wilson v, Feb 13 2015 *)
    RecurrenceTable[{a[n + 1] == (n + 1) a[n] + 1, a[0] == 1}, a, {n, 0, 12}] (* Emanuele Munarini, Apr 27 2017 *)
    nxt[{n_,a_}]:={n+1,a(n+1)+1}; NestList[nxt,{0,1},30][[All,2]] (* Harvey P. Dale, Jan 29 2023 *)
  • Maxima
    a(n) := if n=0 then 1 else n*a(n-1)+1; makelist(a(n),n,0,12); /* Emanuele Munarini, Apr 27 2017 */
  • PARI
    {a(n) = local(A); if( n<0, 0, A = vector(n+1); A[1]=1; for(k=1, n, A[k+1] = k*A[k] + 1); A[n+1])}; /* Michael Somos, Jul 01 2004 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( x +x * O(x^n)) / (1 - x), n))}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1/(1-x)^2+x^2*deriv(A)/(1-x)); polcoeff(A, n) \\ Paul D. Hanna, Sep 03 2008
    
  • PARI
    {a(n)=local(X=x+x*O(x^n));polcoeff(sum(m=0,n,(m+2)^m*x^m/(1+(m+1)*X)^(m+1)),n)} /* Paul D. Hanna */
    
  • PARI
    a(n)=sum(k=0,n,binomial(n,k)*k!); \\ Joerg Arndt, Dec 14 2014
    
  • Sage
    # program adapted from Alois P. Heinz's Maple code in A022493
    @CachedFunction
    def b(n, i, t):
        if n <= 1:
            return 1
        return sum(b(n - 1, j, t + (j == i)) for j in range(t + 2))
    def a(n):
        return b(n, 0, 0)
    v000522 = [a(n) for n in range(33)]
    print(v000522)
    # Joerg Arndt, May 11 2013
    

Formula

a(n) = n*a(n-1) + 1, a(0) = 1.
a(n) = A007526(n-1) + 1.
a(n) = A061354(n)*A093101(n).
a(n) = n! * Sum_{k=0..n} 1/k! = n! * (e - Sum_{k>=n+1} 1/k!). - Michael Somos, Mar 26 1999
a(0) = 1; for n > 0, a(n) = floor(e*n!).
E.g.f.: exp(x)/(1-x).
a(n) = 1 + Sum_{n>=k>=j>=0} (k-j+1)*k!/j! = a(n-1) + A001339(n-1) = A007526(n) + 1. Binomial transformation of n!, i.e., A000142. - Henry Bottomley, Jun 04 2001
a(n) = floor(2/(n+1))*A009578(n+1)-1. - Emeric Deutsch, Oct 24 2001
Integral representation as n-th moment of a nonnegative function on a positive half-axis: a(n) = e*Integral_{x>=0} x^n*e^(-x)*Heaviside(x-1) dx. - Karol A. Penson, Oct 01 2001
Formula, in Mathematica notation: Special values of Laguerre polynomials, a(n)=(-1)^n*n!*LaguerreL[n, -1-n, 1], n=1, 2, ... . This relation cannot be checked by Maple, as it appears that Maple does not incorporate Laguerre polynomials with second index equal to negative integers. It does check with Mathematica. - Karol A. Penson and Pawel Blasiak ( blasiak(AT)lptl.jussieu.fr), Feb 13 2004
G.f.: Sum_{k>=0} k!*x^k/(1-x)^(k+1). a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*k^(n-k)*(k+1)^k. - Vladeta Jovovic, Aug 18 2002
a(n) = Sum_{k=0..n} A008290(n, k)*2^k. - Philippe Deléham, Dec 12 2003
a(n) = Sum_{k=0..n} A046716(n, k). - Philippe Deléham, Jun 12 2004
a(n) = e*Gamma(n+1,1) where Gamma(z,t) = Integral_{x>=t} e^(-x)*x^(z-1) dx is incomplete gamma function. - Michael Somos, Jul 01 2004
a(n) = Sum_{k=0..n} P(n, k). - Ross La Haye, Aug 28 2005
Given g.f. A(x), then g.f. A059115 = A(x/(1-x)). - Michael Somos, Aug 03 2006
a(n) = 1 + n + n*(n-1) + n*(n-1)*(n-2) + ... + n!. - Jonathan Sondow, Aug 18 2006
a(n) = Sum_{k=0..n} binomial(n,k) * k!; interpretation: for all k-subsets (sum), choose a subset (binomial(n,k)), and permutation of subset (k!). - Joerg Arndt, Dec 09 2012
a(n) = Integral_{x>=0} (x+1)^n*e^(-x) dx. - Gerald McGarvey, Oct 19 2006
a(n) = Sum_{k=0..n} A094816(n, k), n>=0 (row sums of Poisson-Charlier coefficient matrix). - N. J. A. Sloane, Nov 10 2007
From Tom Copeland, Nov 01 2007, Dec 10 2007: (Start)
Act on 1/(1-x) with 1/(1-xDx) = Sum_{j>=0} (xDx)^j = Sum_{j>=0} x^j*D^j*x^j = Sum_{j>=0} j!*x^j*L(j,-:xD:,0) where Lag(n,x,0) are the Laguerre polynomials of order 0, D the derivative w.r.t. x and (:xD:)^j = x^j*D^j. Truncating the operator series at the j = n term gives an o.g.f. for a(0) through a(n) consistent with Jovovic's.
These results and those of Penson and Blasiak, Arnold, Bottomley and Deleham are related by the operator A094587 (the reverse of A008279), which is the umbral equivalent of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j)*j!*x^(n-j) = Sum_{j=0..n} (n!/j!)*x^j. Umbral substitution of b(.) for x and then letting b(n)=1 for all n connects the results. See A132013 (the inverse of A094587) for a connection between these operations and 1/(1-xDx).
(End)
From Peter Bala, Jul 15 2008: (Start)
a(n) = n!*e - 1/(n + 1/(n+1 + 2/(n+2 + 3/(n+3 + ...)))).
Asymptotic result (Ramanujan): n!*e - a(n) ~ 1/n - 1/n^3 + 1/n^4 + 2/n^5 - 9/n^6 + ..., where the sequence [1,0,-1,1,2,-9,...] = [(-1)^k*A000587(k)], for k>=1.
a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). For fixed k, define the derived sequence a_k(n) = (a(n+k)-a(k))/n, n = 1,2,3,... . Then a_k(n) is also a difference divisibility sequence.
For example, the derived sequence a_0(n) is just a(n-1). The set of integer sequences satisfying the difference divisibility property forms a ring with term-wise operations of addition and multiplication.
Recurrence relations: a(0) = 1, a(n) = (n-1)*(a(n-1) + a(n-2)) + 2, for n >= 1. a(0) = 1, a(1) = 2, D-finite with recurrence: a(n) = (n+1)*a(n-1) - (n-1)*a(n-2) for n >= 2. The sequence b(n) := n! satisfies the latter recurrence with the initial conditions b(0) = 1, b(1) = 1. This leads to the finite continued fraction expansion a(n)/n! = 1/(1-1/(2-1/(3-2/(4-...-(n-1)/(n+1))))), n >= 2.
Limit_{n->oo} a(n)/n! = e = 1/(1-1/(2-1/(3-2/(4-...-n/((n+2)-...))))). This is the particular case m = 0 of the general result m!/e - d_m = (-1)^(m+1) *(1/(m+2 -1/(m+3 -2/(m+4 -3/(m+5 -...))))), where d_m denotes the m-th derangement number A000166(m).
For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to A001339 (r=1), A082030 (r=2), A095000 (r=3) and A095177 (r=4).
For the corresponding results for the constants log(2), zeta(2) and zeta(3) refer to A142992, A108625 and A143007 respectively.
(End)
G.f. satisfies: A(x) = 1/(1-x)^2 + x^2*A'(x)/(1-x). - Paul D. Hanna, Sep 03 2008
From Paul Barry, Nov 27 2009: (Start)
G.f.: 1/(1-2*x-x^2/(1-4*x-4*x^2/(1-6*x-9*x^2/(1-8*x-16*x^2/(1-10*x-25*x^2/(1-... (continued fraction);
G.f.: 1/(1-x-x/(1-x/(1-x-2*x/(1-2*x/(1-x-3*x/(1-3*x/(1-x-4*x/(1-4*x/(1-x-5*x/(1-5*x/(1-... (continued fraction).
(End)
O.g.f.: Sum_{n>=0} (n+2)^n*x^n/(1 + (n+1)*x)^(n+1). - Paul D. Hanna, Sep 19 2011
G.f. hypergeom([1,k],[],x/(1-x))/(1-x), for k=1,2,...,9 is the generating function for A000522, A001339, A082030, A095000, A095177, A096307, A096341, A095722, and A095740. - Mark van Hoeij, Nov 07 2011
G.f.: 1/U(0) where U(k) = 1 - x - x*(k+1)/(1 - x*(k+1)/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 14 2012
E.g.f.: 1/U(0) where U(k) = 1 - x/(1 - 1/(1 + (k+1)/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 16 2012
G.f.: 1/(1-x)/Q(0), where Q(k) = 1 - x/(1-x)*(k+1)/(1 - x/(1-x)*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
G.f.: 2/(1-x)/G(0), where G(k) = 1 + 1/(1 - x*(2*k+2)/(x*(2*k+3) - 1 + x*(2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: (B(x)+ 1)/(2-2*x) = Q(0)/(2-2*x), where B(x) be g.f. A006183, Q(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + (1-x)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 08 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 30 2013
E.g.f.: e^x/(1-x) = (1 - 12*x/(Q(0) + 6*x - 3*x^2))/(1-x), where Q(k) = 2*(4*k+1)*(32*k^2 + 16*k + x^2 - 6) - x^4*(4*k-1)*(4*k+7)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
G.f.: conjecture: T(0)/(1-2*x), where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
0 = a(n)*(+a(n+1) - 3*a(n+2) + a(n+3)) + a(n+1)*(+a(n+1) - a(n+3)) + a(n+2)*(+a(n+2)) for all n>=0. - Michael Somos, Jul 04 2014
From Peter Bala, Jul 29 2014: (Start)
a(n) = F(n), where the function F(x) := Integral_{0..infinity} e^(-u)*(1 + u)^x du smoothly interpolates this sequence to all real values of x. Note that F(-1) = G and for n = 2,3,... we have F(-n) = (-1)^n/(n-1)! *( A058006(n-2) - G ), where G = 0.5963473623... denotes Gompertz's constant - see A073003.
a(n) = n!*e - e*( Sum_{k >= 0} (-1)^k/((n + k + 1)*k!) ).
(End)
a(n) = hypergeometric_U(1, n+2, 1). - Peter Luschny, Nov 26 2014
a(n) ~ exp(1-n)*n^(n-1/2)*sqrt(2*Pi). - Vladimir Reshetnikov, Oct 27 2015
a(n) = A038155(n+2)/A000217(n+1). - Anton Zakharov, Sep 08 2016
a(n) = round(exp(1)*n!), n > 1 - Simon Plouffe, Jul 28 2020
a(n) = KummerU(-n, -n, 1). - Peter Luschny, May 10 2022
a(n) = (e/(2*Pi))*Integral_{x=-oo..oo} (n+1+i*x)!/(1+i*x) dx. - David Ulgenes, Apr 18 2023
Sum_{i=0..n} (-1)^(n-i) * binomial(n, i) * a(i) = n!. - Werner Schulte, Apr 03 2024

Extensions

Additional comments from Michael Somos

A002162 Decimal expansion of the natural logarithm of 2.

Original entry on oeis.org

6, 9, 3, 1, 4, 7, 1, 8, 0, 5, 5, 9, 9, 4, 5, 3, 0, 9, 4, 1, 7, 2, 3, 2, 1, 2, 1, 4, 5, 8, 1, 7, 6, 5, 6, 8, 0, 7, 5, 5, 0, 0, 1, 3, 4, 3, 6, 0, 2, 5, 5, 2, 5, 4, 1, 2, 0, 6, 8, 0, 0, 0, 9, 4, 9, 3, 3, 9, 3, 6, 2, 1, 9, 6, 9, 6, 9, 4, 7, 1, 5, 6, 0, 5, 8, 6, 3, 3, 2, 6, 9, 9, 6, 4, 1, 8, 6, 8, 7
Offset: 0

Views

Author

Keywords

Comments

Newton calculated the first 16 terms of this sequence.
Area bounded by y = tan x, y = cot x, y = 0. - Clark Kimberling, Jun 26 2020
Choose four values independently and uniformly at random from the unit interval [0,1]. Sort them, and label them a,b,c,d from least to greatest (so that a b^2+c^2. - Akiva Weinberger, Dec 02 2024
Define the trihyperboloid to be the intersection of the three solid hyperboloids x^2+y^2-z^2<1, x^2-y^2+z^2<1, and -x^2+y^2+z^2<1. This fits perfectly within the cube [-1,1]^3. Then this is the ratio of the volume of the trihyperboloid to its bounding cube. - Akiva Weinberger, Dec 02 2024

Examples

			0.693147180559945309417232121458176568075500134360255254120680009493393...
		

References

  • G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
  • Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 250.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3, 2.21, 6.2, and 7.2.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25 and appendix A, equations 25:14:3 and A:7:3 at pages 232, 670.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

Crossrefs

Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.

Programs

  • Mathematica
    RealDigits[N[Log[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
    RealDigits[Log[2],10,120][[1]] (* Harvey P. Dale, Jan 25 2024 *)
  • PARI
    { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009

Formula

log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = ((Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2)+396)/576. - Alexander R. Povolotsky, Jan 14 2011
From Alexander R. Povolotsky, Dec 16 2008: (Start)
log(2) = 105*(319/44100 - Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7))).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63). (End)
log(2) = Sum_{k>=1} A191907(2,k)/k. - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03 2015: (Start)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
From Peter Bala, Oct 30 2016: (Start)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
log(2) = Sum_{n>=1} A006519(n) / ((1 + 2^A006519(n)) * A000265(n) * (1 + A000265(n))). - Nicolas Nagel, Mar 19 2018
From Amiram Eldar, Jul 02 2020: (Start)
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Sum_{k>=1} 1/A002939(k).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
From Peter Bala, Nov 14 2020: (Start)
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
From Peter Bala, Oct 22 2023: (Start)
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2 = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
From Peter Bala, Nov 13 2023: (Start)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
= 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
= 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
log(2) = lim_{n->oo} Sum_{k=1..n} 1/(n + k) = lim_{x->0} (2^x - 1)/x = lim_{x->0} (2^x - 2^(-x))/(2*x) (see Finch). - Stefano Spezia, Oct 19 2024
From Colin Linzer, Nov 08 2024: (Start)
log(2) = Integral_{t=0...oo} (1 - tanh(t)) dt.
log(2) = Integral_{t=0...1} arctanh(t) dt.
log(2) = (1/2) * Integral_{t=-1...1} |arctanh(t)| dt. (End)
log(2) = 1 + Sum_{n >= 1} (-1)^n/(n*(4*n^2 - 1)) = 1/2 + (1/2)*Sum_{n >= 1} 1/(n*(4*n^2 - 1)). - Peter Bala, Jan 07 2025
log(2) = Integral_{x=0..1} Integral_{y=0..1} 1/((1 - x*y)*(1 + x)*(1 + y)) dy dx. - Kritsada Moomuang, May 22 2025

A008288 Square array of Delannoy numbers D(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653
Offset: 0

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Keywords

Comments

In the Formula section, some contributors use T(n,k) = D(n-k, k) (for 0 <= k <= n), which is the triangular version of the square array (D(n,k): n,k >= 0). Conversely, D(n,k) = T(n+k,k) for n,k >= 0. - Petros Hadjicostas, Aug 05 2020
Also called the tribonacci triangle [Alladi and Hoggatt (1977)]. - N. J. A. Sloane, Mar 23 2014
D(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), (1,1). - Joerg Arndt, Jul 01 2011 [Corrected by N. J. A. Sloane, May 30 2020]
Or, triangle read by rows of coefficients of polynomials P[n](x) defined by P[0] = 1, P[1] = x+1; for n >= 2, P[n] = (x+1)*P[n-1] + x*P[n-2].
D(n, k) is the number of k-matchings of a comb-like graph with n+k teeth. Example: D(1, 3) = 7 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has seven 3-matchings: four triples of three teeth and the three triples {Aa, Bb, CD}, {Aa, Dd, BC}, {Cc, Dd, AB}. Also D(3, 1)=7, the 1-matchings of the same graph being the seven edges: {AB}, {BC}, {CD}, {Aa}, {Bb}, {Cc}, {Dd}. - Emeric Deutsch, Jul 01 2002
Sum of n-th antidiagonal of the array D is A000129(n+1). - Reinhard Zumkeller, Dec 03 2004 [Edited by Petros Hadjicostas, Aug 05 2020 so that the counting of antidiagonals of D starts at n = 0. That is, the sum of the terms in the n-th row of the triangles T is A000129(n+1).]
The A-sequence for this Riordan type triangle (see one of Paul Barry's comments under Formula) is A112478 and the Z-sequence the trivial: {1, 0, 0, 0, ...}. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. This leads to the recurrence for the triangle given below. - Wolfdieter Lang, Jan 21 2008
The triangle or chess sums, see A180662 for their definitions, link the Delannoy numbers with twelve different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 and Kn15 have been added. It is remarkable that all knight sums are related to the tribonacci numbers, that is, A000073 and A001590, but none of the others. - Johannes W. Meijer, Sep 22 2010
This sequence, A008288, is jointly generated with A035607 as an array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Row n, for n > 0, of Roger L. Bagula's triangle in the Example section shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the numerator of the n-th convergent of the continued fraction [k, k, k, ...], where k = sqrt(x) + 1/sqrt(x); see A230000. - Clark Kimberling, Nov 13 2013
In an n-dimensional hypercube lattice, D(n,k) gives the number of nodes situated at a Minkowski (Manhattan) distance of k from a given node. In cellular automata theory, the cells at Manhattan distance k are called the von Neumann neighborhood of radius k. For k=1, see A005843. - Dmitry Zaitsev, Dec 10 2015
These numbers appear as the coefficients of series relating spherical and bispherical harmonics, in the solutions of Laplace's equation in 3D. [Majic 2019, Eq. 22] - Matt Majic, Nov 24 2019
From Peter Bala, Feb 19 2020: (Start)
The following remarks assume an offset of 1 in the row and column indices of the triangle.
The sequence of row polynomials T(n,x), beginning with T(1,x) = x, T(2,x) = x + x^2, T(3,x) = x + 3*x^2 + x^3, ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd(T(n,x), T(m,x)) = T(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the sequence (T(n,x): n >= 1) is a divisibility sequence in the polynomial ring Z[x]; that is, if n divides m then T(n,x) divides T(m,x) in Z[x].
Let S(x) = 1 + 2*x + 6*x^2 + 22*x^3 + ... denote the o.g.f. for the large Schröder numbers A006318. The power series (x*S(x))^n, n = 2, 3, 4, ..., can be expressed as a linear combination with polynomial coefficients of S(x) and 1: (x*S(x))^n = T(n-1,-x) - T(n,-x)*S(x). The result can be extended to negative integer n if we define T(0,x) = 0 and T(-n,x) = (-1)^(n+1) * T(n,x)/x^n. Cf. A115139.
[In the previous two paragraphs, D(n,x) was replaced with T(n,x) because the contributor is referring to the rows of the triangle T(n,k), not the rows of the array D(n,k). - Petros Hadjicostas, Aug 05 2020] (End)
Named after the French amateur mathematician Henri-Auguste Delannoy (1833-1915). - Amiram Eldar, Apr 15 2021
D(i,j) = D(j,i). With this and Dmitry Zaitsev's Dec 10 2015 comment, D(i,j) can be considered the number of points at L1 distance <= i in Z^j or the number of points at L1 distance <= j in Z^i from any given point. The rows and columns of D(i,j) are the crystal ball sequences on cubic lattices. See the first example below. The n-th term in the k-th crystal ball sequence can be considered the number of points at distance <= n from any point in a k-dimensional cubic lattice, or the number of points at distance <= k from any point in an n-dimensional cubic lattice. - Shel Kaphan, Jan 01 2023 and Jan 07 2023
Dimensions of hom spaces Hom(R^{(i)}, R^{(j)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023

Examples

			The square array D(i,j) (i >= 0, j >= 0) begins:
  1, 1,  1,   1,   1,   1,    1,    1,    1,    1, ... = A000012
  1, 3,  5,   7,   9,  11,   13,   15,   17,   19, ... = A005408
  1, 5, 13,  25,  41,  61,   85,  113,  145,  181, ... = A001844
  1, 7, 25,  63, 129, 231,  377,  575,  833, 1159, ... = A001845
  1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
  ...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section:  D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
   0,   0,   0,   0,   1,   0,  11,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   1,   0,   3,   0,  13,   0,  63,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   0,   0,   1,   0,  11,   0, ...
   [Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
   1;
   1,  1;
   1,  3,   1;
   1,  5,   5,   1;
   1,  7,  13,   7,    1;
   1,  9,  25,  25,    9,    1;
   1, 11,  41,  63,   41,   11,    1;
   1, 13,  61, 129,  129,   61,   13,   1;
   1, 15,  85, 231,  321,  231,   85,  15,   1;
   1, 17, 113, 377,  681,  681,  377, 113,  17,  1;
   1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
   ... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   1,  0;
   1,  1,  0;
   1,  3,  1,  0;
   1,  5,  5,  1,  0;
   1,  7, 13,  7,  1,  0;
   1,  9, 25, 25,  9,  1, 0;
   1, 11, 41, 63, 41, 11, 1, 0;
   ...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   0, 1;
   0, 1,  1;
   0, 1,  3,  1;
   0, 1,  5,  5,  1;
   0, 1,  7, 13,  7,  1;
   0, 1,  9, 25, 25,  9,  1;
   0, 1, 11, 41, 63, 41, 11, 1;
   ... (End)
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Mathematica, 26 (1963), 223-229.
  • G. Picou, Note #2235, L'Intermédiaire des Mathématiciens, 8 (1901), page 281. - N. J. A. Sloane, Mar 02 2022
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.

Crossrefs

Sums of antidiagonals: A000129 (Pell numbers).
Main diagonal: A001850 (central Delannoy numbers), which has further information and references.
A002002, A026002, and A190666 are +-k-diagonals for k=1, 2, 3 resp. - Shel Kaphan, Jan 01 2023
See also A027618.
Cf. A059446.
Has same main diagonal as A064861. Different from A100936.
Read mod small primes: A211312, A211313, A211314, A211315.
Triangle sums (see the comments): A000129 (Row1); A056594 (Row2); A000073 (Kn11 & Kn21); A089068 (Kn12 & Kn22); A180668 (Kn13 & Kn23); A180669 (Kn14 & Kn24); A180670 (Kn15 & Kn25); A099463 (Kn3 & Kn4); A116404 (Fi1 & Fi2); A006498 (Ca1 & Ca2); A006498(3*n) (Ca3 & Ca4); A079972 (Gi1 & Gi2); A079972(4*n) (Gi3 & Gi4); A079973(3*n) (Ze1 & Ze2); A079973(2*n) (Ze3 & Ze4).
Cf. A102413, A128966. (D(n,1)) = A005843. Cf. A115139.

Programs

  • Haskell
    a008288 n k = a008288_tabl !! n !! k
    a008288_row n = a008288_tabl !! n
    a008288_tabl = map fst $ iterate
        (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                           zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
    -- Reinhard Zumkeller, Jul 21 2013
    
  • Maple
    A008288 := proc(n, k) option remember; if k = 0 then 1 elif n=k then 1 else procname(n-1, k-1) + procname(n-2, k-1) + procname(n-1, k) end if; end proc: seq(seq(A008288(n,k),k=0..n), n=0..10); # triangular indices n and k
    P[0]:=1; P[1]:=x+1; for n from 2 to 12 do P[n]:=expand((x+1)*P[n-1]+x*P[n-2]); lprint(P[n]); lprint(seriestolist(series(P[n],x,200))); end do:
  • Mathematica
    (* Next, A008288 jointly generated with A035607 *)
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
    v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A008288 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A035607 *)
    (* Clark Kimberling, Mar 09 2012 *)
    d[n_, k_] := Binomial[n+k, k]*Hypergeometric2F1[-k, -n, -n-k, -1]; A008288 = Flatten[Table[d[n-k, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Apr 05 2012, after 3rd formula *)
  • Python
    from functools import cache
    @cache
    def delannoy_row(n: int) -> list[int]:
        if n == 0: return [1]
        if n == 1: return [1, 1]
        rov = delannoy_row(n - 2)
        row = delannoy_row(n - 1) + [1]
        for k in range(n - 1, 0, -1):
            row[k] += row[k - 1] + rov[k - 1]
        return row
    for n in range(10): print(delannoy_row(n))  # Peter Luschny, Jul 30 2023
  • Sage
    for k in range(8):  # seen as an array, read row by row
        a = lambda n: hypergeometric([-n, -k], [1], 2)
        print([simplify(a(n)) for n in range(11)]) # Peter Luschny, Nov 19 2014
    

Formula

D(n, 0) = 1 = D(0, n) for n >= 0; D(n, k) = D(n, k-1) + D(n-1, k-1) + D(n-1, k).
Bivariate o.g.f.: Sum_{n >= 0, k >= 0} D(n, k)*x^n*y^k = 1/(1 - x - y - x*y).
D(n, k) = Sum_{d = 0..min(n,k)} binomial(k, d)*binomial(n+k-d, k) = Sum_{d=0..min(n,k)} 2^d*binomial(n, d)*binomial(k, d). [Edited by Petros Hadjicostas, Aug 05 2020]
Seen as a triangle read by rows: T(n, 0) = T(n, n) = 1 for n >= 0 and T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-1, k), 0 < k < n and n > 1. - Reinhard Zumkeller, Dec 03 2004
Read as a number triangle, this is the Riordan array (1/(1-x), x(1+x)/(1-x)) with T(n, k) = Sum_{j=0..n-k} C(n-k, j) * C(k, j) * 2^j. - Paul Barry, Jul 18 2005
T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k). - Paul Barry, May 21 2006
Let y^k(n) be the number of Khalimsky-continuous functions f from [0,n-1] to Z such that f(0) = 0 and f(n-1) = k. Then y^k(n) = D(i,j) for i = (1/2)*(n-1-k) and j = (1/2)*(n-1+k) where n-1+k belongs to 2Z. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007
Recurrence for triangle from A-sequence (see the Wolfdieter Lang comment above): T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j), n >= 1, k >= 1. [For k > n, the sum is empty, in which case T(n,k) = 0.]
From Peter Bala, Jul 17 2008: (Start)
The n-th row of the square array is the crystal ball sequence for the product lattice A_1 x ... x A_1 (n copies). A035607 is the table of the associated coordination sequences for these lattices.
The polynomial p_n(x) := Sum {k = 0..n} 2^k * C(n,k) * C(x,k) = Sum_{k = 0..n} C(n,k) * C(x+k,n), whose values [p_n(0), p_n(1), p_n(2), ... ] give the n-th row of the square array, is the Ehrhart polynomial of the n-dimensional cross polytope (the hyperoctahedron) [Bump et al. (2000), Theorem 6].
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The reciprocity law p_n(m) = p_m(n) reflects the symmetry of the table.
The polynomial p_n(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1) - x*f(x-1) = (2*n+1)*f(x), normalized so that f(0) = 1.
These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis [Bump et al. (2000), Theorem 4]. The o.g.f. for the p_n(x) is (1 + t)^x/(1 - t)^(x + 1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ... .
The square array of Delannoy numbers has a close connection with the constant log(2). The entries in the n-th row of the array occur in the series acceleration formula log(2) = (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n) + (-1)^n * Sum_{k>=1} (-1)^(k+1)/(k*D(n,k-1)*D(n,k)). [T(n,k) was replaced with D(n,k) in the formula to agree with the beginning of the paragraph. - Petros Hadjicostas, Aug 05 2020]
For example, the fourth row of the table (n = 3) gives the series log(2) = 1 - 1/2 + 1/3 - 1/(1*1*7) + 1/(2*7*25) - 1/(3*25*63) + 1/(4*63*129) - ... . See A142979 for further details.
Also the main diagonal entries (the central Delannoy numbers) give the series acceleration formula Sum_{n>=1} 1/(n*D(n-1,n-1)*D(n,n)) = (1/2)*log(2), a result due to Burnside. [T(n,n) was replaced here with D(n,n) to agree with the previous paragraphs. - Petros Hadjicostas, Aug 05 2020]
Similar relations hold between log(2) and the crystal ball sequences of the C_n lattices A142992. For corresponding results for the constants zeta(2) and zeta(3), involving the crystal ball sequences for root lattices of type A_n and A_n x A_n, see A108625 and A143007 respectively. (End)
From Peter Bala, Oct 28 2008: (Start)
Hilbert transform of Pascal's triangle A007318 (see A145905 for the definition of this term).
D(n+a,n) = P_n(a,0;3) for all integer a such that a >= -n, where P_n(a,0;x) is the Jacobi polynomial with parameters (a,0) [Hetyei]. The related formula A(n,k) = P_k(0,n-k;3) defines the table of asymmetric Delannoy numbers, essentially A049600. (End)
Seen as a triangle read by rows: T(n, k) = Hyper2F1([k-n, -k], [1], 2). - Peter Luschny, Aug 02 2014, Oct 13 2024.
From Peter Bala, Jun 25 2015: (Start)
O.g.f. for triangle T(n,k): A(z,t) = 1/(1 - (1 + t)*z - t*z^2) = 1 + (1 + t)*z + (1 + 3*t + t^2)*z^2 + (1 + 5*t + 5*t^2 + t^3)*z^3 + ....
1 + z*d/dz(A(z,t))/A(z,t) is the o.g.f. for A102413. (End)
E.g.f. for the n-th subdiagonal of T(n,k), n >= 0, equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 4*x + 4*x^2/2) = 1 + 5*x + 13*x^2/2! + 25*x^3/3! + 41*x^4/4! + 61*x^5/5! + .... - Peter Bala, Mar 05 2017 [The n-th subdiagonal of triangle T(n,k) is the n-th row of array D(n,k).]
Let a_i(n) be multiplicative with a_i(p^e) = D(i, e), p prime and e >= 0, then Sum_{n > 0} a_i(n)/n^s = (zeta(s))^(2*i+1)/(zeta(2*s))^i for i >= 0. - Werner Schulte, Feb 14 2018
Seen as a triangle read by rows: T(n,k) = Sum_{i=0..k} binomial(n-i, i) * binomial(n-2*i, k-i) for 0 <= k <= n. - Werner Schulte, Jan 09 2019
Univariate generating function: Sum_{k >= 0} D(n,k)*z^k = (1 + z)^n/(1 - z)^(n+1). [Dziemianczuk (2013), Eq. 5.3] - Matt Majic, Nov 24 2019
(n+1)*D(n+1,k) = (2*k+1)*D(n,k) + n*D(n-1,k). [Majic (2019), Eq. 22] - Matt Majic, Nov 24 2019
For i, j >= 1, D(i,j) = D(i,j-1) + 2*Sum_{k=0..i-1} D(k,j-1), or, because D(i,j) = D(j,i), D(i,j) = D(i-1,j) + 2*Sum_{k=0..j-1} D(i-1,k). - Shel Kaphan, Jan 01 2023
Sum_{k=0..n} T(n,k)^2 = A026933(n). - R. J. Mathar, Nov 07 2023
Let S(x) = (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x) denote the g.f. of the sequence of large Schröder numbers A006318. Read as a lower triangular array, the signed n-th row polynomial R(n, -x) = 1/sqrt(1 - 6*x + x^2) *( 1/S(x)^(n+1) + (x*S(x))^(n+1) ). For example, R(4, -x) = 1 - 7*x + 13*x^2 - 7*x^3 + x^4 = 1/sqrt(1 - 6*x + x^2) * ( 1/S(x)^5 + (x*S(x))^5 ). Cf. A102413. - Peter Bala, Aug 01 2024

Extensions

Expanded description from Clark Kimberling, Jun 15 1997
Additional references from Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 28 2001
Changed the notation to make the formulas more precise. - N. J. A. Sloane, Jul 01 2002

A143007 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the 2*n-dimensional lattice A_n x A_n.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 13, 13, 1, 1, 25, 73, 25, 1, 1, 41, 253, 253, 41, 1, 1, 61, 661, 1445, 661, 61, 1, 1, 85, 1441, 5741, 5741, 1441, 85, 1, 1, 113, 2773, 17861, 33001, 17861, 2773, 113, 1, 1, 145, 4873, 46705, 142001, 142001, 46705, 4873, 145, 1
Offset: 0

Views

Author

Peter Bala, Jul 22 2008

Keywords

Comments

The A_n lattice consists of all vectors v = (x_1,...,x_(n+1)) in Z^(n+1) such that x_1 + ... + x_(n+1) = 0. The lattice is equipped with the norm ||v|| = 1/2*(|x_1| + ... + |x_(n+1)|). Pairs of lattice points (v,w) in the product lattice A_n x A_n have norm ||(v,w)|| = ||v|| + ||w||. Then the k-th term in the crystal ball sequence for the A_n x A_n lattice gives the number of such pairs (v,w) for which ||(v,w)|| is less than or equal to k.
This array has a remarkable relationship with Apery's constant zeta(3). The row (or column) and main diagonal entries of the array occur in series acceleration formulas for zeta(3). For row n entries there holds zeta(3) = (1+1/2^3+...+1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)). Also, as consequence of Apery's proof of the irrationality of zeta(3), we have a series acceleration formula along the main diagonal of the table: zeta(3) = 6 * sum {n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Apery's result appears to generalize to the other diagonals of the table. Calculation suggests the following result may hold: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2 +3*n*k +k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).
For the corresponding results for the constant zeta(2), related to the crystal ball sequences of the lattices A_n, see A108625. For corresponding results for log(2), coming from either the crystal ball sequences of the hypercubic lattices A_1 x ... x A_1 or the lattices of type C_n, see A008288 and A142992 respectively.

Examples

			The table begins
n\k|0...1.....2......3.......4.......5
======================================
0..|1...1.....1......1.......1.......1
1..|1...5....13.....25......41......61 A001844
2..|1..13....73....253.....661....1441 A143008
3..|1..25...253...1445....5741...17861 A143009
4..|1..41...661...5741...33001..142001 A143010
5..|1..61..1441..17861..142001..819005 A143011
........
Example row 1 [1,5,13,...]:
The lattice A_1 x A_1 is equivalent to the square lattice of all integer lattice points v = (x,y) in Z x Z equipped with the taxicab norm ||v|| = (|x| + |y|). There are 4 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 8 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the A_1 x A_1 lattice begins 1, 1+4 = 5, 1+4+8 = 13, ... .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 1 2 . . . .
. . . 2 1 0 1 2 . . .
. . . . 2 1 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
Row 1 = [1,5,13,...] is the sequence of partial sums of A008574; row 2 = [1,13,73,...] is the sequence of partial sums of A008530, so row 2 is the crystal ball sequence for the lattice A_2 x A_2 (the 4-dimensional di-isohexagonal orthogonal lattice).
Read as a triangle the array begins
n\k|0...1....2....3...4...5
===========================
0..|1
1..|1...1
2..|1...5....1
3..|1..13...13....1
4..|1..25...73...25...1
5..|1..41..253..253..41...1
		

Crossrefs

Cf. A001844 (row 1), A005259 (main diagonal), A008288, A008530 (first differences of row 2), A008574 (first differences of row 1), A085478, A108625, A142992, A143003, A143004, A143005, A143006, A143008 (row 2), A143009 (row 3), A142010 (row 4), A143011 (row 5).
Cf. A227845 (antidiagonal sums), A246464.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Magma
    A:= func< n,k | (&+[(Binomial(n,j)*Binomial(n+k-j,k-j))^2: j in [0..n]]) >; // Array
    A143007:= func< n,k | A(n-k,k) >; // Antidiagonal triangle
    [A143007(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    with(combinat): T:= (n,k) -> add(binomial(n+j,2*j)*binomial(2*j,j)^2*binomial(k+j,2*j), j = 0..n): for n from 0 to 9 do seq(T(n,k),k = 0..9) end do;
  • Mathematica
    T[n_, k_]:= HypergeometricPFQ[{-k, k+1, -n, n+1}, {1, 1, 1}, 1]; Table[T[n-k, k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    /* Print as a square array: */
    {T(n, k)=sum(j=0, n, binomial(n+j, 2*j)*binomial(2*j, j)^2*binomial(k+j, 2*j))}
    for(n=0, 10, for(k=0,10, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (1) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m * y^m / (1-x +x*O(x^n))^(2*m+1) * sum(k=0, m, binomial(m, k)^2*x^k)^2 ); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (2) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m/(1-x*y +x*O(x^n))^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^k * y^k)^2 ); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (3) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m , k)^2 * y^k * sum(j=0, k, binomial(k, j)^2 * x^j)+x*O(x^n))); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (4) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * y^(m-k) * sum(j=0, k, binomial(k, j)^2 * x^j * y^j)+x*O(x^n))); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    /* End */
    
  • SageMath
    def A(n,k): return sum((binomial(n,j)*binomial(n+k-j,k-j))^2 for j in range(n+1)) # array
    def A143007(n,k): return A(n-k,k) # antidiagonal triangle
    flatten([[A143007(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n,k) = Sum_{j = 0..n} C(n+j,2*j)*C(2*j,j)^2*C(k+j,2*j).
The array is symmetric T(n,k) = T(k,n).
The main diagonal [1,5,73,1445,...] is the sequence of Apery numbers A005259.
The entries in the k-th column satisfy the Apery-like recursion n^3*T(n,k) + (n-1)^3*T(n-2,k) = (2*n-1)*(n^2-n+1+2*k^2+2*k)*T(n-1,k).
The LDU factorization of the square array is L * D * transpose(L), where L is the lower triangular array A085478 and D is the diagonal matrix diag(C(2n,n)^2). O.g.f. for row n: The generating function for the coordination sequence of the lattice A_n is [Sum_{k = 0..n} C(n,k)^2*x^k ]/(1-x)^n. Thus the generating function for the coordination sequence of the product lattice A_n x A_n is {[Sum_{k = 0..n} C(n,k)^2*x^k]/(1-x)^n}^2 and hence the generating function for row n of this array, the crystal ball sequence of the lattice A_n x A_n, equals [Sum_{k = 0..n} C(n,k)^2*x^k]^2/(1-x)^(2n+1) = 1/(1-x)*[Legendre_P(n,(1+x)/(1-x))]^2. See [Conway & Sloane].
Series acceleration formulas for zeta(3): Row n: zeta(3) = (1 + 1/2^3 + ... + 1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)), n = 0,1,2,... . For example, the fourth row of the table (n = 3) gives zeta(3) = (1 + 1/2^3 + 1/3^3) + 1/(1^3*1*25) + 1/(2^3*25*253) + 1/(3^3*253*1445) + ... . See A143003 for further details.
Main diagonal: zeta(3) = 6 * Sum_{n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Conjectural result for other diagonals: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2+3*n*k+k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).
Sum_{k=0..n} T(n-k,k) = A227845(n) (antidiagonal sums). - Paul D. Hanna, Aug 27 2014
The main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) == S(m*p^(r-1) - 1) (mod p^(3*r)) for prime p >= 5 and m, r in N (this is true: see A352653. - Peter Bala, Apr 16 2022).
From Paul D. Hanna, Aug 27 2014: (Start)
G.f. A(x,y) = Sum_{n>=0, k=0..n} T(n,k)*x^n*y^k can be expressed by:
(1) Sum_{n>=0} x^n * y^n / (1-x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k]^2,
(2) Sum_{n>=0} x^n / (1 - x*y)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k * y^k]^2,
(3) Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * y^k * Sum_{j=0..k} C(k,j)^2 * x^j,
(4) Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * y^(n-k) * Sum_{j=0..k} C(k,j)^2 * x^j * y^j. (End)
From Peter Bala, Jun 23 2023: (Start)
T(n,k) = Sum_{j = 0..n} C(n,j)^2 * C(n+k-j,k-j)^2.
T(n,k) = binomial(n+k,k)^2 * hypergeom([-n, -n, -k, -k],[-n - k, -n - k, 1], 1).
T(n,k) = hypergeom([n+1, -n, k+1, -k], [1, 1, 1], 1). (End)
From Peter Bala, Jun 28 2023: (Start)
T(n,k) = the coefficient of (x*z)^n*(y*t)^k in the expansion of 1/( (1 - x - y)*(1 - z - t) - x*y*z*t ).
T(n,k) = A(n, k, n, k) in the notation of Straub, equation 7.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, k+1, -k], [1, 1, 1], 1) allows the table indexing to be extended to negative values of n and k; we have T(-n,k) = T(n-1,k) and T(n,-k) = T(n,k-1) leading to T(-n,-k) = T(n-1, k-1). (End)
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n, k) = T(n-k, k) be the antidiagonal triangle, then:
t(n, k) = t(n, n-k).
Sum_{k=0..floor(n/2)} t(n-k,k) = A246563(n).
t(2*n+1, n+1) = A352653(n+1). (End)
From Peter Bala, Sep 27 2024: (Start)
The square array = A063007 * transpose(A063007) (LU factorization).
Let L denote the lower triangular array (l(n,k))n,k >= 0, where l(n, k) = (-1)^(n+k) * binomial(n, k)*binomial(n+k, k). (L is a signed version of A063007 and L = A063007 * A007318 ^(-1).)
Then the square array = L * transpose(A108625).
L^2 * transpose(A108625) = the Hadamard product of A108625 with itself (both identities can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package to find recurrences for the double sums involved). (End)

Extensions

Spelling/notation corrections by Charles R Greathouse IV, Mar 18 2010

A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 12 2005

Keywords

Comments

Compare to the corresponding array A108553 of crystal ball sequences for D_n lattice.
From Peter Bala, Jul 18 2008: (Start)
Row reverse of A099608.
This array has a remarkable relationship with the constant zeta(2). The row, column and diagonal entries of the array occur in series acceleration formulas for zeta(2).
For the entries in row n we have zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k)). For example, n = 4 gives zeta(2) = 2*(1 - 1/4 + 1/9 - 1/16) + 1/(1*21) + 1/(4*21*131) + 1/(9*131*471) + ... . See A142995 for further details.
For the entries in column k we have zeta(2) = (1 + 1/4 + 1/9 + ... + 1/k^2) + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k)). For example, k = 4 gives zeta(2) = (1 + 1/4 + 1/9 + 1/16) + 2*(1/(1*9) - 1/(4*9*61) + 1/(9*61*309) - ... ). See A142999 for further details.
Also, as consequence of Apery's proof of the irrationality of zeta(2), we have a series acceleration formula along the main diagonal of the table: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n,n)*T(n-1,n-1)) = 5*(1/3 - 1/(2^2*3*19) + 1/(3^2*19*147) - ...).
There also appear to be series acceleration results along other diagonals. For example, for the main subdiagonal, calculation supports the result zeta(2) = 2 - Sum_{n >= 1} (-1)^(n+1)*(n^2+(2*n+1)^2)/(n^2*(n+1)^2*T(n,n-1)*T(n+1,n)) = 2 - 10/(2^2*7) + 29/(6^2*7*55) - 58/(12^2*55*471) + ..., while for the main superdiagonal we appear to have zeta(2) = 1 + Sum_{n >= 1} (-1)^(n+1)*((n+1)^2 + (2*n+1)^2)/(n^2*(n+1)^2*T(n-1,n)*T(n,n+1)) = 1 + 13/(2^2*5) - 34/(6^2*5*37) + 65/(12^2*37*309) - ... .
Similar series acceleration results hold for Apery's constant zeta(3) involving the crystal ball sequences for the product lattices A_n x A_n; see A143007 for further details. Similar results also hold between the constant log(2) and the crystal ball sequences of the hypercubic lattices A_1 x...x A_1 and between log(2) and the crystal ball sequences for lattices of type C_n ; see A008288 and A142992 respectively for further details. (End)
This array is the Hilbert transform of triangle A008459 (see A145905 for the definition of the Hilbert transform). - Peter Bala, Oct 28 2008

Examples

			Square array begins:
  1,   1,    1,     1,      1,       1,       1, ... A000012;
  1,   3,    5,     7,      9,      11,      13, ... A005408;
  1,   7,   19,    37,     61,      91,     127, ... A003215;
  1,  13,   55,   147,    309,     561,     923, ... A005902;
  1,  21,  131,   471,   1251,    2751,    5321, ... A008384;
  1,  31,  271,  1281,   4251,   11253,   25493, ... A008386;
  1,  43,  505,  3067,  12559,   39733,  104959, ... A008388;
  1,  57,  869,  6637,  33111,  124223,  380731, ... A008390;
  1,  73, 1405, 13237,  79459,  350683, 1240399, ... A008392;
  1,  91, 2161, 24691, 176251,  907753, 3685123, ... A008394;
  1, 111, 3191, 43561, 365751, 2181257, ...      ... A008396;
  ...
As a triangle:
  [0]  1
  [1]  1,  1
  [2]  1,  3,   1
  [3]  1,  7,   5,    1
  [4]  1, 13,  19,    7,    1
  [5]  1, 21,  55,   37,    9,    1
  [6]  1, 31, 131,  147,   61,   11,   1
  [7]  1, 43, 271,  471,  309,   91,  13,   1
  [8]  1, 57, 505, 1281, 1251,  561, 127,  15,  1
  [9]  1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
       ...
Inverse binomial transform of rows yield rows of triangle A063007:
  1;
  1,  2;
  1,  6,   6;
  1, 12,  30,  20;
  1, 20,  90, 140,  70;
  1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
  1;
  1,  1;
  1,  4,   1;
  1,  9,   9,   1;
  1, 16,  36,  16,  1;
  1, 25, 100, 100, 25, 1;
  ...
		

Crossrefs

Rows include: A003215 (row 2), A005902 (row 3), A008384 (row 4), A008386 (row 5), A008388 (row 6), A008390 (row 7), A008392 (row 8), A008394 (row 9), A008396 (row 10).
Cf. A063007, A099601 (n-th term of A_{2n} lattice), A108553.
Cf. A008459 (h-vectors type B associahedra), A145904, A145905.
Cf. A005258 (main diagonal), A108626 (antidiagonal sums).

Programs

  • Magma
    T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array
    A108625:= func< n,k | T(n-k,k) >; // antidiagonals
    [A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1):
    seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018
  • Mathematica
    T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *)
  • PARI
    T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))
    
  • SageMath
    def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array
    def A108625(n,k): return T(n-k, k) # antidiagonals
    flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
The square array = A063007 * transpose(A007318).
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A063496 a(n) = (2*n - 1)*(8*n^2 - 8*n + 3)/3.

Original entry on oeis.org

1, 19, 85, 231, 489, 891, 1469, 2255, 3281, 4579, 6181, 8119, 10425, 13131, 16269, 19871, 23969, 28595, 33781, 39559, 45961, 53019, 60765, 69231, 78449, 88451, 99269, 110935, 123481, 136939, 151341, 166719, 183105, 200531, 219029, 238631, 259369, 281275, 304381
Offset: 1

Views

Author

N. J. A. Sloane, Aug 01 2001

Keywords

Comments

Number of potential flows in a 2 X 2 matrix with integer velocities in -n..n, i.e., number of 2 X 2 matrices with adjacent elements differing by no more than n, counting matrices differing by a constant only once. - R. H. Hardin, Feb 27 2002
Number of ordered quadruples (a,b,c,d), -(n-1) <= a,b,c,d <= n-1, such that a+b+c+d = 0. - Benoit Cloitre, Jun 14 2003
If Y and Z are 2-blocks of a (2n+1)-set X then a(n-1) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
Equals binomial transform of [1, 18, 48, 32, 0, 0, 0, ...]. - Gary W. Adamson, Jul 19 2008

Crossrefs

(1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Programs

  • Magma
    [(2*n-1)*(8*n^2-8*n+3)/3: n in [1..40]]; // Wesley Ivan Hurt, May 09 2014
  • Maple
    A063496:=n->(2*n-1)*(8*n^2-8*n+3)/3; seq(A063496(n), n=1..40); # Wesley Ivan Hurt, May 09 2014
  • Mathematica
    Table[(2*n - 1)*(8*n^2 - 8*n + 3)/3, {n, 40}] (* Wesley Ivan Hurt, May 09 2014 *)
    LinearRecurrence[{4,-6,4,-1}, {1,19,85,231}, 30] (* G. C. Greubel, Dec 01 2017 *)
  • PARI
    a(n) = { (2*n - 1)*(8*n^2 - 8*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009
    
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+24*x^2+16*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017
    

Formula

From Peter Bala, Jul 18 2008: (Start)
The following remarks about the C_3 lattice assume the sequence offset is 0.
Partial sums of A010006. So this sequence is the crystal ball sequence for the C_3 lattice - row 3 of A142992. The lattice C_3 consists of all integer lattice points v = (a,b,c) in Z^3 such that a + b + c is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c|).
The crystal ball sequence of C_3 gives the number of lattice points v in C_3 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].
For example, a(1) = 19 because the origin has norm 0 and the 18 lattice points in Z^3 of norm 1 (as defined above) are +-(2,0,0), +-(0,2,0), +-(0,0,2), +-(1,1,0), +-(1,0,1), +-(0,1,1), +-(1,-1,0), +-(1,0,-1) and +-(0,1,-1). These 18 vectors form a root system of type C_3.
O.g.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4 = x/(1 - x) * T(3, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
2*log(2) = 4/3 + Sum_{n >= 1} 1/(n*a(n)*a(n+1)). (End)
a(n+1) = (1/Pi) * Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^4. - Yalcin Aktar, Nov 02 2011, corrected by R. J. Mathar, Dec 01 2011
From G. C. Greubel, Dec 01 2017: (Start)
G.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4.
E.g.f.: (-3 + 6*x + 24*x^2 + 16*x^3)*exp(x)/3 + 1. (End)
a(n) = A005900(2n-1). - Ivan N. Ianakiev, Mar 27 2022
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k+1)) = 1/(19 - 3/(27 - 60/(43 - 315/(67 - ... -n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*3^2))))).
E.g.f.: exp(x)*(1 + 18*x + 48*x^2/2! + 32*x^3/3!). Note that -T(6, i*sqrt(x)) = 1 + 18*x + 48*x^2 + 32*x^3, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See A008310. (End)

A143003 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).

Original entry on oeis.org

0, 1, 21, 1091, 114520, 21298264, 6410456640, 2923097201856, 1920450126458880, 1747596822651334656, 2133806329230225408000, 3405545462439659704320000, 6950705677729940374290432000, 17807686090745585163974737920000
Offset: 0

Views

Author

Peter Bala, Jul 19 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1)*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^3 for Apery's constant zeta(3). For other cases see A066989 (m=0), A143004 (m=2), A143005 (m=3) and A143006 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^3*p_m(n)*Sum_{k = 1..n} 1/(k^3*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..n} C(2*k,k)^2*C(n+k,2*k)*C(x+k,2*k) is a polynomial in x of degree 2*m.
The first few are p_0(x) = 1, p_1(x) = 2*x^2 + 2*x + 1, p_2(x) = (3*x^4 + 6*x^3 + 9*x^2 + 6*x + 2)/2 and p_3(x) = (10*x^6 + 30*x^5 + 85*x^4 + 120*x^3 + 121*x^2 + 66*x + 18)/18. For fixed n, the sequence [p_n(k)]k>=0 is the crystal ball sequence for the product lattice A_n x A_n. See A143007 for the table of values [p_n(k)] n,k >= 0. Observe that [p_n(n)] n >= 0 is the sequence of Apery numbers A005259.
The reciprocity law p_m(n) = p_n(m) holds for nonnegative integers m and n. In particular we have p_m(1) = 2*m^2 + 2*m + 1 and p_m(2) = (3*m^4 + 6*m^3 + 9*m^2 + 6*m + 2)/2.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^3*f(x+1) + x^3*f(x-1) = (2*x+1)*(x^2+x+2*m^2+2*m+1)*f(x), normalized so that f(0) = 1. The reciprocity law now yields the Apery-like recursion m^3*p_m(x) + (m-1)^3*p_(m-2)(x) = (2*m-1)*(m^2-m+1+2*x^2+2*x)*p_(m-1)(x).
The polynomial functions p_m(x) have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^3*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m^2+2*m+1. Hence the behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*p_m(k-1)*p_m(k)) = 1/((2*m^2+2*m+1) - 1^6/(3*(2*m^2+2*m+3) - 2^6/(5*(2*m^2+2*m+7) - 3^6/(7*(2*m^2+2*m+13) - ...)))) = Sum_{k>=1} 1/(m+k)^3. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii)].
For the corresponding results for the constant zeta(2) see A142995. For corresponding results for the constant log(2) see A142979 and A142992.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Maple
    p := n -> 2*n^2+2*n+1: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)
  • Mathematica
    RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+5)a[n]- n^6 a[n-1]}, a[n],{n,15}] (* Harvey P. Dale, Jun 20 2011 *)

Formula

a(n) = n!^3*p(n)*Sum_{k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n).
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).
The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n-1)^6/((2*n-1)*(n^2-n+5)))))), for n >= 2. The behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*(4*k^4 + 1)) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - n^6/((2*n+1)*(n^2+n+5) - ...))))) = zeta(3) - 1, where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 1].

A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005
Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005
Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005
Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008
From Peter Bala, Oct 23 2008: (Start)
Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term).
(End)
Diagonal sums: A108479. - Philippe Deléham, Sep 08 2009
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010
Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A086646(n,k). - R. J. Mathar, Mar 12 2013
Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013
Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016

Examples

			From _Peter Bala_, Oct 23 2008: (Start)
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
From _Peter Bala_, Aug 06 2013: (Start)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Crossrefs

Cf. A008459, A108558, A127674, A142992. - Peter Bala, Oct 23 2008
Cf. A103327 (binomial(2n+1, 2k+1)), A103328 (binomial(2n, 2k+1)), A091042 (binomial(2n+1, 2k)). -Wolfdieter Lang, Jan 06 2013
Cf. A086646 (unsigned matrix inverse), A103327.
Cf. A034839.

Programs

  • Magma
    /* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016
  • Maple
    A086645:=(n,k)->binomial(2*n,2*k): seq(seq(A086645(n,k),k=0..n),n=0..12);
  • Mathematica
    Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)
  • Maxima
    create_list(binomial(2*n,2*k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    {T(n, k) = binomial(2*n, 2*k)};
    
  • PARI
    {T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */
    

Formula

T(n, k) = (2*n)!/((2*(n-k))!*(2*k)!) row sums = A081294. COLUMNS: A000012, A000384
Sum_{k>=0} T(n, k)*A000364(k) = A000795(n) = (2^n)*A005647(n).
Sum_{k>=0} T(n, k)*2^k = A001541(n). Sum_{k>=0} T(n, k)*3^k = 2^n*A001075(n). Sum_{k>=0} T(n, k)*4^k = A083884(n). - Philippe Deléham, Feb 29 2004
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009
Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
From Paul Barry, Sep 28 2010: (Start)
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Sep 22 2021: (Start)
n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459.
R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)

A142978 Table of figurate numbers for the n-dimensional cross polytopes.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 19, 16, 5, 1, 10, 33, 44, 25, 6, 1, 12, 51, 96, 85, 36, 7, 1, 14, 73, 180, 225, 146, 49, 8, 1, 16, 99, 304, 501, 456, 231, 64, 9, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10
Offset: 1

Views

Author

Peter Bala, Jul 15 2008

Keywords

Comments

The n-th row entries for this array are the regular polytope numbers for the n-dimensional cross polytope as defined by [Kim]. The rows are the partial sums of the rows of the square array of Delannoy numbers A008288.
The odd numbered rows of this array form A142977. For a triangular version of this table see A104698. Cf. also A101603.
The n-th row of the array is the binomial transform of n-th row of triangle A081277, followed by zeros. Example: row 4 (1, 6, 19, 44, 85, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 17 2008
The main diagonal of the array T(n,k) is A047781 Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600. The link from A099193 to J. V. Post, Table of polytope numbers, Sorted, Through 1,000,000, includes all n-D Hyperoctahedron (n-Cross Polytope) Numbers through 10-Cross(20) = 1669752016. - Jonathan Vos Post, Jul 16 2008

Examples

			The square array A(n, k) begins:
  n\k| 1   2    3     4     5       6
  ---+-------------------------------
   1 | 1   2    3     4      5      6    A000027
   2 | 1   4    9    16     25     36    A000290
   3 | 1   6   19    44     85    146    A005900
   4 | 1   8   33    96    225    456    A014820
   5 | 1  10   51   180    501   1182    A069038
   6 | 1  12   73   304    985   2668    A069039
   7 | 1  14   99   476   1765   5418    A099193
		

Crossrefs

Cf. A008288 (Delannoy numbers), A005900 (row 3), A014820 (row 4), A069038 (row 5), A069039 (row 6), A099193 (row 7), A099195 (row 8), A099196 (row 9), A099197 (row 10), A101603, A104698 (triangle version), A142977, A142983.

Programs

  • Haskell
    a142978 n k = a142978_tabl !! (n-1) !! (k-1)
    a142978_row n = a142978_tabl !! (n-1)
    a142978_tabl = map reverse a104698_tabl
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    with(combinat): T:=(n,k) -> add(binomial(n-1,i)*binomial(k+i,n),i = 0..n-1); for n from 1 to 10 do seq(T(n,k),k = 1..10) end do; # Program restored by Peter Bala, Oct 02 2008
    A := (n, k) -> k*hypergeom([1 - n, 1 - k], [2], 2):
    seq(print(seq(simplify(A(n, k)), k = 1..9)), n=1..7); # Peter Luschny, Mar 23 2023
  • Mathematica
    t[n_, k_] := Sum[ Binomial[n-1, i]*Binomial[k+i, n], {i, 0, n-1}]; Table[t[n-k, k], {n, 1, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n-1} C(n-1,i)*C(k+i,n).
Reciprocity law: n*T(n,k) = k*T(k,n).
Recurrence relation: T(n,1) = 1, T(1,k) = k, T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k), n,k > 1.
O.g.f. row n: x*(1 + x)^(n-1)/(1 - x)^(n+1).
O.g.f. for array: Sum_{n >= 1, k >= 1} T(n, k)*x^k*y^n = x*y/((1 - x)*(1 - x - y - x*y)).
The n-th row entries are the values [p_n(k)], k >= 1, of the polynomial function p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(x,k). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_n(x) is the unique polynomial solution of the difference equation x*( f(x+1) - f(x-1) ) = 2*n*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_n(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (-1)^(n+1) * (2*n)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n)). For example, n = 3 gives log(2) = 4/6 + (1/6)*(1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ...). See A142983 for further details.
From Peter Bala, Oct 02 2008: (Start)
The odd-indexed columns of this array form the array A142992 of crystal ball sequences for lattices of type C_n.
Conjectural congruences for main diagonal entries: Put A(n) = T(n,n). Calculation suggests the following congruences: for prime p > 3 and m, r >= 1, A(m*p^r) == A(m*p^(r-1)) (mod p^(3*r));
Sum_{k = 0..p-1} A(k)^2 == 0 (mod p) if p is a prime of the form 8*n+1 or 8*n+7;
Sum_{k = 0..p-1} A(k)^2 == -1 (mod p) if p is a prime of the form 8*n+3 or 8*n+5.
(End)
From Peter Bala, Sep 27 2021: (Start)
T(n,k) = (1/2)*Sum_{i = 0..k} binomial(k,i)*binomial(n+k-1-i,k-1).
T(n,k) = (1/2)*[x^n] ((1+x)/(1-x))^k = (1/2)*(k/n)*[x^k] ((1+x)/(1-x))^n.
n*T(n,k) = 2*k*T(n-1,k) + (n - 2)*T(n-2,k). (End)
A(n,k) = k*hypergeom([1 - n, 1 - k], [2], 2). - Peter Luschny, Mar 23 2023
T(n,k) = 2*(Sum_{j=1..k-1} T(n-1,j)) + T(n-1,k) for n > 1. - Robert FERREOL, Jun 25 2024
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