A056040 -id:A056040 - cp's OEIS frontend

A214282 Largest Euler characteristic of a downset on an n-dimensional cube.

1, 1, 1, 3, 6, 10, 15, 35, 70, 126, 210, 462, 924, 1716, 3003, 6435, 12870, 24310, 43758, 92378, 184756, 352716, 646646, 1352078, 2704156, 5200300, 9657700, 20058300, 40116600, 77558760, 145422675, 300540195, 601080390, 1166803110, 2203961430, 4537567650, 9075135300, 17672631900

Offset: 1

Terence Tao, Jul 09 2012

nonn

An m-downset is a set of subsets of 1..m such that if S is in the set, so are all subsets of S. The Euler characteristic of a downset is the number of sets in the downset with an even cardinality, minus the number with an odd cardinality.

			G.f. = x + x^2 + x^3 + 3*x^4 + 6*x^5 + 10*x^6 + 15*x^7 + 35*x^8 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Terry Tao, Optimal bounds for an alternating sum on a downset, 2012.

Cf. A214283.

a214282 n = a007318 (n - 1) (a004524 (n - 1))
-- Reinhard Zumkeller, Jul 14 2012

Table[{Binomial[n - 1, n/2], Binomial[n, n/2], Binomial[n + 1, n/2 + 1], Binomial[n + 2, n/2 + 2]}, {n, 0, 28, 4}] (* Alonso del Arte, Jul 09 2012 *)

a(n)=binomial(n-1,if(n%2,(n+1)\4*2,n/2)) \\ Charles R Greathouse IV, Jul 09 2012

{a(n) = if( n<1, 0, vecmax( Vec((1 - x)^(n-1))))}; /* Michael Somos, Apr 21 2014 */

from math import comb
def A214282(n): return comb(n-1, (n+1>>1)&(-1^(n&1))) # Chai Wah Wu, Jan 31 2024

a(n) = binomial(n - 1, n/2) when n is even, a(n) = binomial(n - 1, (n + 1)/2) when n is 3 mod 4, and a(n) = binomial(n - 1, (n - 1)/2) when n is 1 mod 4.
a(2n) = A001700(n-1). a(4n+1) = A001448(n). a(4n+3) = A186231(n).
a(n) = A214283(n) + A001405(n). - Reinhard Zumkeller, Jul 14 2012
a(n) = A007318(n-1, A004524(n-1)). - Reinhard Zumkeller, Jul 14 2012
a(n+1) = A000108([n/2])*A215495(n). - M. F. Hasler, Aug 25 2012
A214282(n) - A214283(n) is A056040(n) if n is even and A056040(n)/((n+1)/2) otherwise. - Peter Luschny, Jul 08 2016

A274706 Irregular triangle read by rows. T(n,k) (n >= 0) is a statistic on orbital systems over n sectors: the number of orbitals which have an integral whose absolute value is k.

Original entry on oeis.org

1, 1, 0, 2, 0, 4, 2, 2, 0, 2, 0, 2, 6, 4, 6, 4, 4, 4, 2, 0, 6, 0, 6, 0, 4, 0, 2, 0, 2, 6, 24, 16, 20, 14, 16, 12, 8, 6, 8, 4, 4, 2, 8, 0, 14, 0, 14, 0, 10, 0, 10, 0, 6, 0, 4, 0, 2, 0, 2, 36, 52, 68, 48, 64, 48, 48, 40, 44, 32, 36, 24, 22, 16, 16, 8, 10, 8, 4, 4, 2

Offset: 0

Peter Luschny, Jul 10 2016

For the combinatorial definitions see A232500. The absolute integral of an orbital w over n sectors is abs(Sum_{k=1..n} Sum_{i=1..k} w(i)) where w(i) are the jumps of the orbital represented by -1, 0, 1.

An orbital is balanced if its integral is 0 (A241810).

			The length of row n is 1+floor(n^2//4).
The triangle begins:
  [n] [k=0,1,2,...] [row sum]
  [0] [1] 1
  [1] [1] 1
  [2] [0, 2] 2
  [3] [0, 4, 2] 6
  [4] [2, 0, 2, 0, 2] 6
  [5] [6, 4, 6, 4, 4, 4, 2] 30
  [6] [0, 6, 0, 6, 0, 4, 0, 2, 0, 2] 20
  [7] [6, 24, 16, 20, 14, 16, 12, 8, 6, 8, 4, 4, 2] 140
  [8] [8, 0, 14, 0, 14, 0, 10, 0, 10, 0, 6, 0, 4, 0, 2, 0, 2] 70
T(5, 4) = 4 because the integral of four orbitals have the absolute value 4:
  Integral([-1, -1, 1, 1, 0]) = -4, Integral([0, -1, -1, 1, 1]) = -4,
  Integral([0, 1, 1, -1, -1]) = 4, Integral([1, 1, -1, -1, 0]) = 4.

Peter Luschny, Orbitals

Cf. A056040 (row sum), A232500, A241810 (col. 0), A242087.

Other orbital statistics: A241477 (first zero crossing), A274708 (number of peaks), A274709 (max. height), A274710 (number of turns), A274878 (span), A274879 (returns), A274880 (restarts), A274881 (ascent).

from itertools import accumulate
# Brute force counting
def unit_orbitals(n):
    sym_range = [i for i in range(-n+1, n, 2)]
    for c in Combinations(sym_range, n):
        P = Permutations([sgn(v) for v in c])
        for p in P: yield p
def orbital_integral(n):
    if n == 0: return [1]
    S = [0]*(1+floor(n^2//4))
    for u in unit_orbitals(n):
        L = list(accumulate(accumulate(u)))
        S[abs(L[-1])] += 1
    return S
for n in (0..8): print(orbital_integral(n))

A274710 A statistic on orbital systems over n sectors: the number of orbitals which make k turns.

Original entry on oeis.org

1, 1, 0, 2, 0, 0, 6, 0, 2, 2, 2, 0, 0, 6, 12, 12, 0, 2, 4, 8, 4, 2, 0, 0, 6, 24, 52, 40, 18, 0, 2, 6, 18, 18, 18, 6, 2, 0, 0, 6, 36, 120, 180, 180, 84, 24, 0, 2, 8, 32, 48, 72, 48, 32, 8, 2, 0, 0, 6, 48, 216, 480, 744, 672, 432, 144, 30, 0, 2, 10, 50, 100, 200, 200, 200, 100, 50, 10, 2

Offset: 0

Peter Luschny, Jul 10 2016

The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.
A 'turn' of an orbital w takes place where signum(w[i]) is not equal to signum(w[i+1]).
A152659 is a subtriangle.

			Triangle read by rows, n>=0. The length of row n is n for n>=1.
[n] [k=0,1,2,...]                      [row sum]
[0] [1]                                    1
[1] [1]                                    1
[2] [0, 2]                                 2
[3] [0, 0, 6]                              6
[4] [0, 2, 2,  2]                          6
[5] [0, 0, 6, 12,  12]                    30
[6] [0, 2, 4,  8,   4,   2]               20
[7] [0, 0, 6, 24,  52,  40,  18]         140
[8] [0, 2, 6, 18,  18,  18,   6,  2]      70
[9] [0, 0, 6, 36, 120, 180, 180, 84, 24] 630
T(5,2) = 6 because the six orbitals [-1, -1, 0, 1, 1], [-1, -1, 1, 1, 0], [0, -1, -1, 1, 1], [0, 1, 1, -1, -1], [1, 1, -1, -1, 0], [1, 1, 0, -1, -1] make 2 turns.

Peter Luschny, Orbitals

Cf. A056040 (row sum), A152659, A232500.

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (number of peaks), A274709 (max. height), A274878 (span), A274879 (returns), A274880 (restarts), A274881 (ascent).

# uses[unit_orbitals from A274709]
# Brute force counting
def orbital_turns(n):
    if n == 0: return [1]
    S = [0]*(n)
    for u in unit_orbitals(n):
        L = sum(0 if sgn(u[i]) == sgn(u[i+1]) else 1 for i in (0..n-2))
        S[L] += 1
    return S
for n in (0..12): print(orbital_turns(n))

For even n>0: T(n,k) = 2*C(n/2-1,(k-1+mod(k-1,2))/2)*C(n/2-1,(k-1-mod(k-1,2))/2) for k=0..n-1 (from A152659).

A274878 A statistic on orbital systems over n sectors: the number of orbitals with span k.

Original entry on oeis.org

1, 1, 0, 2, 0, 6, 0, 2, 4, 0, 10, 20, 0, 2, 12, 6, 0, 14, 84, 42, 0, 2, 28, 32, 8, 0, 18, 252, 288, 72, 0, 2, 60, 120, 60, 10, 0, 22, 660, 1320, 660, 110, 0, 2, 124, 390, 300, 96, 12, 0, 26, 1612, 5070, 3900, 1248, 156, 0, 2, 252, 1176, 1260, 588, 140, 14

Offset: 0

Peter Luschny, Jul 10 2016

The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.

The 'span' of an orbital w is the difference between the highest and the lowest level of the orbital system touched by w.

			Triangle read by rows, n>=0. The length of row n is floor((n+2)/2).
[ n] [k=0,1,2,...]          [row sum]
[ 0] [1]                        1
[ 1] [1]                        1
[ 2] [0, 2]                     2
[ 3] [0, 6]                     6
[ 4] [0, 2, 4]                  6
[ 5] [0, 10, 20]               30
[ 6] [0, 2, 12, 6]             20
[ 7] [0, 14, 84, 42]          140
[ 8] [0, 2, 28, 32, 8]         70
[ 9] [0, 18, 252, 288, 72]    630
[10] [0, 2, 60, 120, 60, 10]  252
T(6, 3) = 6 because the span of the following six orbitals is 3:
[-1, -1, -1, 1, 1, 1], [-1, -1, 1, 1, 1, -1], [-1, 1, 1, 1, -1, -1],
[1, -1, -1, -1, 1, 1], [1, 1, -1, -1, -1, 1], [1, 1, 1, -1, -1, -1].

Peter Luschny, Orbitals

Cf. A056040 (row sum), A232500.

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (number of peaks), A274709 (max. height), A274710 (number of turns), A274879 (returns), A274880 (restarts), A274881 (ascent).

# uses[unit_orbitals from A274709]
from itertools import accumulate
# Brute force counting.
def orbital_span(n):
    if n == 0: return [1]
    S = [0]*((n+2)//2)
    for u in unit_orbitals(n):
        L = list(accumulate(u))
        S[max(L) - min(L)] += 1
    return S
for n in (0..10): print(orbital_span(n))

A274879 A statistic on orbital systems over n sectors: the number of orbitals with k returns.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 4, 6, 12, 12, 4, 8, 8, 20, 40, 48, 32, 10, 20, 24, 16, 70, 140, 180, 160, 80, 28, 56, 72, 64, 32, 252, 504, 672, 672, 480, 192, 84, 168, 224, 224, 160, 64, 924, 1848, 2520, 2688, 2240, 1344, 448, 264, 528, 720, 768, 640, 384, 128

Offset: 0

Peter Luschny, Jul 11 2016

The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.
When a segment of an orbital starts at a point on the central circle this point is called a 'return' of the orbital if it is not the origin.
If an orbital touches the central circle only in the origin it is called a prime orbital. Column 0 counts the prime orbitals over n sectors.
A108747 is a subtriangle.

			Triangle read by rows, n>=0. The length of row n is floor((n+1)/2) for n>=1.
[ n] [k=0,1,2,...] [row sum]
[ 0] [1] 1
[ 1] [1] 1
[ 2] [2] 2
[ 3] [2, 4] 6
[ 4] [2, 4] 6
[ 5] [6, 12, 12] 30
[ 6] [4, 8, 8] 20
[ 7] [20, 40, 48, 32] 140
[ 8] [10, 20, 24, 16] 70
[ 9] [70, 140, 180, 160, 80] 630
[10] [28, 56, 72, 64, 32] 252
[11] [252, 504, 672, 672, 480, 192] 2772
T(6,0) = 4 because the following 4 orbitals stay above or below the central
circle: [-1, -1, -1, 1, 1, 1], [-1, -1, 1, -1, 1, 1], [1, 1, -1, 1, -1, -1],
[1, 1, 1, -1, -1, -1].

Peter Luschny, Orbitals

Cf. A056040 (row sum), A108747, A232500, A241543 (col. 0).

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (peaks), A274709 (max. height), A274710 (number of turns), A274878 (span), A274880 (restarts), A274881 (ascent).

# uses[unit_orbitals from A274709]
from itertools import accumulate
# Brute force counting
def orbital_returns(n):
    if n == 0: return [1]
    S = [0]*((n+1)//2)
    for u in unit_orbitals(n):
        L = list(accumulate(u))
        Z = len(list(filter(lambda z: z == 0, L)))
        S[Z-1] += 1  # exclude origin
    return S
for n in (0..10): print(orbital_returns(n))

For even n>0: T(n,k) = 2^(k+1)*(k+1)*binomial(n-k-1,n/2)/(n-k-1) for k=0..n/2-1 (from A108747).

A274880 A statistic on orbital systems over n sectors: the number of orbitals with k restarts.

Original entry on oeis.org

1, 1, 2, 5, 1, 4, 2, 18, 11, 1, 10, 8, 2, 65, 57, 17, 1, 28, 28, 12, 2, 238, 252, 116, 23, 1, 84, 96, 54, 16, 2, 882, 1050, 615, 195, 29, 1, 264, 330, 220, 88, 20, 2, 3300, 4257, 2915, 1210, 294, 35, 1, 858, 1144, 858, 416, 130, 24, 2, 12441, 17017, 13013, 6461, 2093, 413, 41, 1

Offset: 0

Peter Luschny, Jul 11 2016

The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.
A 'restart' of an orbital is a raise which starts from the central circle.
A118920 is a subtriangle.

			Triangle read by rows, n>=0. The length of row n is floor((n+1)/2) for n>=1.
[n] [k=0,1,2,...]                 [row sum]
[ 0] [1]                              1
[ 1] [1]                              1
[ 2] [2]                              2
[ 3] [5, 1]                           6
[ 4] [4, 2]                           6
[ 5] [18, 11, 1]                     30
[ 6] [10, 8, 2]                      20
[ 7] [65, 57, 17, 1]                140
[ 8] [28, 28, 12, 2]                 70
[ 9] [238, 252, 116, 23, 1]         630
[10] [84, 96, 54, 16, 2]            252
[11] [882, 1050, 615, 195, 29, 1]  2772
T(6, 2) = 2 because there are two orbitals over 6 segments which have 2 ascents:
[-1, 1, 1, -1, 1, -1] and [1, -1, 1, -1, 1, -1].

Peter Luschny, Orbitals

Cf. A056040 (row sum), A118920, A232500.

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (peaks), A274709 (max. height), A274710 (number of turns), A274878 (span), A274879 (returns), A274881 (ascent).

# uses[unit_orbitals from A274709]
from itertools import accumulate
# Brute force counting
def orbital_restart(n):
    if n == 0: return [1]
    S = [0]*((n+1)//2)
    for u in unit_orbitals(n):
        A = list(accumulate(u))
        L = [1 if A[i] == 0 and A[i+1] == 1  else 0 for i in (0..n-2)]
        S[sum(L)] += 1
    return S
for n in (0..12): print(orbital_restart(n))

For even n>0: T(n,k) = 4*(k+1)*binomial(n,n/2-k-1)/n for k=0..n/2-1 (from A118920).

A274881 A statistic on orbital systems over n sectors: the number of orbitals which have an ascent of length k.

Original entry on oeis.org

1, 1, 0, 2, 0, 6, 0, 3, 3, 0, 18, 12, 0, 4, 12, 4, 0, 40, 80, 20, 0, 5, 40, 20, 5, 0, 75, 375, 150, 30, 0, 6, 120, 90, 30, 6, 0, 126, 1470, 882, 252, 42, 0, 7, 350, 371, 147, 42, 7, 0, 196, 5292, 4508, 1568, 392, 56, 0, 8, 1008, 1456, 672, 224, 56, 8

Offset: 0

Peter Luschny, Jul 12 2016

The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.

The ascent of an orbital is its longest up-run.

			Triangle read by rows, n>=0. The length of row n is floor((n+2)/2).
[ n] [k=0,1,2,...]                [row sum]
[ 0] [1]                              1
[ 1] [1]                              1
[ 2] [0, 2]                           2
[ 3] [0, 6]                           6
[ 4] [0, 3, 3]                        6
[ 5] [0, 18, 12]                     30
[ 6] [0, 4, 12, 4]                   20
[ 7] [0, 40, 80, 20]                140
[ 8] [0, 5, 40, 20, 5]               70
[ 9] [0, 75, 375, 150, 30]          630
[10] [0, 6, 120, 90, 30, 6]         252
[11] [0, 126, 1470, 882, 252, 42]  2772
[12] [0, 7, 350, 371, 147, 42, 7]   924
T(6,3) = 4 because four orbitals over six sectors have a maximal up-run of length 3.
[-1,-1,-1,1,1,1], [-1,-1,1,1,1,-1], [-1,1,1,1,-1,-1], [1,1,1,-1,-1,-1].

Peter Luschny, Orbitals

Cf. A056040 (row sum), A232500.

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274708 (peaks), A274709 (max. height), A274710 (number of turns), A274878 (span), A274879 (returns), A274880 (restarts).

# uses[unit_orbitals from A274709]
# Brute force counting
def orbital_ascent(n):
    if n < 2: return [1]
    S = [0]*((n+2)//2)
    for u in unit_orbitals(n):
        B = [0]*n
        for i in (0..n-1):
            B[i] = 0 if u[i] <= 0 else B[i-1] + u[i]
        S[max(B)] += 1
    return S
for n in (0..12): print(orbital_ascent(n))

A248682 Decimal expansion of Sum_{n >= 0} (floor(n/2)!)^2/n!.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Oct 11 2014

Limit_{x -> inf} Sum {n=0..inf} (Floor[n/x])!^x/n! = e (A001113).
For A248682: x = 2; A248683: x = 3; A248684: x = 4; A248685: x = 5. - Robert G. Wilson v, Feb 22 2016
Let n} denote the swinging factorial A056040(n), then the constant equals Sum_{n>=0} 1/n} and is sometimes called the swinging constant e}. ("e}" is written in TeX $e\wr$). For a proof that it equals 3^(1/2)*(2/3)^3*Pi + 4/3 see the link to Mathematics Stack Exchange. - Peter Luschny, Jul 22 2022

			2.94559943487486031163918067345969398425250...

Georg Fischer, Table of n, a(n) for n = 1..1000 (first 149 terms from Clark Kimberling)
Jan Eerland, Answer to question 3689772, Mathematics Stack Exchange, 2021. See also question 3692793.
Index entries for transcendental numbers

Cf. A001113, A248683, A248684, A248785, A248664, A056040 (swinging factorial).

RealDigits[Sum[(Floor[n/2])!^2/n!, {n, 0, 400}], 10, 111][[1]]
RealDigits[4/3+8Pi/Sqrt[243],10,111][[1]] (* Robert G. Wilson v, Feb 10 2016 *)

suminf(n=0, ((n\2)!)^2/n!) \\ Michel Marcus, Feb 11 2016

Equals Sum_{n >= 0} (n!^2)*p(2,n)/(2*n + 1)!, where p(k,n) is defined at A248664.
Equals Sum_{n >= 0} (floor(n/2)!)^2/n! = Sum_(n >= 1) (3n^2 - 7n + 6)/C(2n, n) = 4/3 + 8*Pi/sqrt(243). - Robert G. Wilson v, Feb 11 2016
Equals 1 + Integral_{x>=0} 1/(x^2 - x + 1)^2 dx. - Amiram Eldar, Nov 16 2021

A274708 A statistic on orbital systems over n sectors: the number of orbitals with k peaks.

Original entry on oeis.org

1, 1, 2, 4, 2, 4, 2, 12, 15, 3, 10, 8, 2, 38, 68, 30, 4, 26, 30, 12, 2, 121, 272, 183, 49, 5, 70, 104, 60, 16, 2, 384, 1026, 912, 372, 72, 6, 192, 350, 260, 100, 20, 2, 1214, 3727, 4095, 2220, 650, 99, 7, 534, 1152, 1050, 520, 150, 24, 2, 3822, 13200, 17178, 11600, 4510, 1032, 130, 8

Offset: 0

Peter Luschny, Jul 10 2016

The definition of an orbital system is given in A232500 (see also the illustration there). The number of orbitals over n sectors is counted by the swinging factorial A056040.
An orbital w has a 'peak' at i+1 when signum(w[i]) < signum(w[i+1]) and signum(w[i+1]) > signum(w[i+2]).
A097692 is a subtriangle.

			Triangle read by rows, n>=0. The length of row n is floor((n+1)/2) for n>=1.
[ n] [k=0,1,2,...]               [row sum]
[ 0] [  1]                           1
[ 1] [  1]                           1
[ 2] [  2]                           2
[ 3] [  4,    2]                     6
[ 4] [  4,    2]                     6
[ 5] [ 12,   15,   3]               30
[ 6] [ 10,    8,   2]               20
[ 7] [ 38,   68,  30,   4]         140
[ 8] [ 26,   30,  12,   2]          70
[ 9] [121,  272, 183,  49,  5]     630
[10] [ 70,  104,  60,  16,  2]     252
[11] [384, 1026, 912, 372, 72, 6] 2772
[12] [192,  350, 260, 100, 20, 2]  924
T(6, 2) = 2 because the two orbitals [-1, 1, -1, 1, -1, 1] and [1, -1, 1, -1, 1, -1] have 2 peaks.

Peter Luschny, Orbitals

Cf. A025565 (even col. 0), A056040 (row sum), A097692, A232500.

Other orbital statistics: A241477 (first zero crossing), A274706 (absolute integral), A274709 (max. height), A274710 (number of turns), A274878 (span), A274879 (returns), A274880 (restarts), A274881 (ascent).

# uses[unit_orbitals from A274709]
# Brute force counting
def orbital_peaks(n):
    if n == 0: return [1]
    S = [0]*((n+1)//2)
    for u in unit_orbitals(n):
        L = [1 if sgn(u[i]) < sgn(u[i+1]) and sgn(u[i+1]) > sgn(u[i+2]) else 0 for i in (0..n-3)]
        S[sum(L)] += 1
    return S
for n in (0..12): print(orbital_peaks(n))

A211226 Triangular array: T(n,k) = f(n)/(f(k)*f(n-k)), where f(n) = (floor(n/2))!.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 6, 3, 3, 1, 1, 1, 3, 3, 3, 3, 1, 1, 1, 4, 4, 12, 6, 12, 4, 4, 1, 1, 1, 4, 4, 6, 6, 4, 4, 1, 1, 1, 5, 5, 20, 10, 30, 10, 20, 5, 5, 1, 1, 1, 5, 5, 10, 10, 10, 10, 5, 5, 1, 1, 1, 6, 6, 30, 15

Offset: 0

Peter Bala, Apr 05 2012

			Triangle begins
.n\k.|....0....1....2....3....4....5....6....7....8....9...10...11
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....1....1
..2..|....1....1....1
..3..|....1....1....1....1
..4..|....1....2....2....2....1
..5..|....1....1....2....2....1....1
..6..|....1....3....3....6....3....3....1
..7..|....1....1....3....3....3....3....1....1
..8..|....1....4....4...12....6...12....4....4....1
..9..|....1....1....4....4....6....6....4....4....1....1
.10..|....1....5....5...20...10...30...10...20....5....5....1
.11..|....1....1....5....5...10...10...10...10....5....5....1....1
...

Peter Bala, Notes on A211226

Cf. A007318, A056040, A211227 (row sums), A211228 (shallow diagonal sums), A211229 (inverse), A211230 (array squared).

T(n,k) := f(n)/(f(k)*f(n-k)), where f(n) := (floor(n/2))!.
T(2*n+1,2*k) = T(2*n+1,2*k+1) = T(2*n,2*k) = binomial(n,k);
T(2*n,2*k+1) = n*binomial(n-1,k).
Recurrence equations:
T(2*n,2*k) = T(2*n-1,2*k) + T(2*n-1,2*k-1);
T(2*n,2*k+1) = T(2*n-1,2*k+1) + (n-1)*T(2*n-1,2*k);
T(2*n+1,2*k) = T(2*n,2*k); T(2*n+1,2*k+1) = T(2*n,2*k).
The Star of David property holds:
T(n,k+1)*T(n+1,k)*T(n+2,k+2) = T(n,k)*T(n+2,k+1)*T(n+1,k+2).
O.g.f.: (1 + t*(1+x) - t^2*(1-x+x^2) - t^3*(1+x+x^2+x^3))/(1 - t^2*(1+x^2))^2 = sum {n>=0} R(n,x)*t^n = 1 + (1+x)*t + (1+x+x^2)*t^2 + (1+x+x^2+x^3)*t^3 + ....
E.g.f.: cosh(t*sqrt(1+x^2)) + (1+x+x*t/2)/sqrt(1+x^2)*sinh(t*sqrt(1+x^2)) = sum {n>=0} R(n,x)*t^n/n! = 1 + (1+x)*t + (1+x+x^2)*t^2/2! + (1+x+x^2+x^3)*t^3/3! + ....
Row generating polynomials: R(2*n+1,x) = (1+x)*(1+x^2)^n; R(2*n,x) = (1+n*x+x^2)*(1+x^2)^(n-1).
Row sums: A211227. Shallow diagonal sums A211228. Central terms T(2*n,n) equal A056040(n).
The inverse array A211229 involves the derangement numbers A000166. The squared array is A211230.

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