Displaying 1-10 of 39 results found.
Alkane (or paraffin) numbers l(7,n).
(Formerly M2774)
+0
9
1, 3, 9, 19, 38, 66, 110, 170, 255, 365, 511, 693, 924, 1204, 1548, 1956, 2445, 3015, 3685, 4455, 5346, 6358, 7514, 8814, 10283, 11921, 13755, 15785, 18040, 20520, 23256, 26248, 29529, 33099, 36993, 41211, 45790, 50730, 56070, 61810, 67991
COMMENTS
Equals A000217 (1, 3, 6, 10, 15, ...) convolved with A193356 (1, 0, 3, 0, 5, ...). - Gary W. Adamson, Feb 16 2009 F(1,4,n) is the number of bracelets with 1 blue, 4 red and n black beads. If F(1,4,1)=3 and F(1,4,2)=9 taken as a base;
F(1,4,n) = n(n+1)(n+2)/6+F(1,2,n) + F(1,4,n-2). [F(1,2,n) is the number of bracelets with 1 blue, 2 red and n black beads. If F(1,2,1)=2 and F(1,2,2)=4 taken as a base F(1,2,n)=n+1+F(1,2,n-2)]. - Ata Aydin Uslu and Hamdi G. Ozmenekse, Jan 11 2012
REFERENCES
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
G.f.: (1+x^2)/((1-x)^3*(1-x^2)^2) = (1+x^2)/((1-x)^5*(1+x)^2).
l(c, r) = 1/2 C(c+r-3, r) + 1/2 d(c, r), where d(c, r) is C((c + r - 3)/2, r/2) if c is odd and r is even, 0 if c is even and r is odd, C((c + r - 4)/2, r/2) if c is even and r is even, C((c + r - 4)/2, (r - 1)/2) if c is odd and r is odd.
Euler transform of length 4 sequence [3, 3, 0, -1]. - Michael Somos, Mar 08 2007 a(n) = 3a(n-1) - a(n-2) - 5a(n-3) + 5a(n-4) + a(n-5) - 3a(n-6) + a(n-7), with a(0)=1, a(1)=3, a(2)=9, a(4)=19, a(5)=38, a(6)=66, a(7)=110. - Harvey P. Dale, May 02 2011 a(n) = (1/48)*(n+1)*(n+3)*((n+2)*(n+4)+3)+1/32*(2*n+5)*(1+(-1)^n). - Yosu Yurramendi, Jun 20 2013
MAPLE
a:= n -> (Matrix([[1, 0$4, 1, 3]]). Matrix(7, (i, j)-> if (i=j-1) then 1 elif j=1 then [3, -1, -5, 5, 1, -3, 1][i] else 0 fi)^n)[1, 1]: seq (a(n), n=0..40); # Alois P. Heinz, Jul 31 2008
MATHEMATICA
LinearRecurrence[{3, -1, -5, 5, 1, -3, 1}, {1, 3, 9, 19, 38, 66, 110}, 50] (* or *) CoefficientList[Series[(1+x^2)/((1-x)^3(1-x^2)^2), {x, 0, 50}], x] (* Harvey P. Dale, May 02 2011 *) nn=45; With[{a=Accumulate[Range[nn]], b=Riffle[Range[1, nn, 2], 0]}, Flatten[ Table[ListConvolve[Take[a, n], Take[b, n]], {n, nn}]]] (* Harvey P. Dale, Nov 11 2011 *)
PROG
(PARI) {a(n)=if(n<-4, n=-5-n); polcoeff( (1+x^2)/((1-x)^3*(1-x^2)^2)+x*O(x^n), n)} /* Michael Somos, Mar 08 2007 */ import Data.List (inits, intersperse)
a005994 n = a005994_list !! n
a005994_list = map (sum . zipWith (*) (intersperse 0 [1, 3 ..]) . reverse) $
tail $ inits $ tail a000217_list
The number of binary pattern classes in the (2,n)-rectangular grid with 3 '1's and (2n-3) '0's: two patterns are in same class if one of them can be obtained by a reflection or 180-degree rotation of the other.
+0
4
0, 0, 1, 6, 14, 32, 55, 94, 140, 208, 285, 390, 506, 656, 819, 1022, 1240, 1504, 1785, 2118, 2470, 2880, 3311, 3806, 4324, 4912, 5525, 6214, 6930, 7728, 8555, 9470, 10416, 11456, 12529, 13702, 14910, 16224, 17575, 19038, 20540, 22160, 23821, 25606, 27434, 29392
COMMENTS
Also the edge count of the n X n black bishop graph. - Eric W. Weisstein, Jun 26 2017
FORMULA
a(n) = A000330(n) + A142150(n) = (n-1)*(4*n^2 - 2*n - 3*(-1)^n + 3)/12. a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) with n > 5, a(0)=0, a(1)=0, a(2)=1, a(3)=6, a(4)=14, a(5)=32.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 4*(n-4)*(-1)^n with n > 3, a(0)=0, a(1)=0, a(2)=1, a(3)=6.
G.f.: x^2*(1 + 4*x + x^2 + 2*x^3)/((1+x)^2*(1-x)^4). - Bruno Berselli, May 29 2013
MATHEMATICA
Table[(n - 1)*(4*n^2 - 2*n - 3*(-1)^n + 3)/12, {n, 0, 40}] (* Bruno Berselli, May 29 2013 *) CoefficientList[Series[x^2 (1 + 4 x + x^2 + 2 x^3) / ((1 + x)^2 (1 - x)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 04 2013 *) LinearRecurrence[{2, 1, -4, 1, 2, -1}, {0, 1, 6, 14, 32, 55}, 20] (* Eric W. Weisstein, Jun 27 2017 *)
PROG
(R) a <- vector()
for(n in 0:40) a[n] <- (1/4)*(choose(2*(n-1), 3) + 2*choose(n-2, 1)*(1/2)*(1+(-1)^n))
(Magma) [(1/4)*(Binomial(2*(n-1), 3)+2*Binomial(n-2, 1)*(1/2)*(1+(-1)^n)): n in [1..50]]; // Vincenzo Librandi, Sep 04 2013
CROSSREFS
Cf. A289179 (edge count of white bishop graph).
Rows of Losanitsch's triangle T(n, k), n >= 0, 0 <= k <= n.
+0
68
1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 6, 6, 3, 1, 1, 3, 9, 10, 9, 3, 1, 1, 4, 12, 19, 19, 12, 4, 1, 1, 4, 16, 28, 38, 28, 16, 4, 1, 1, 5, 20, 44, 66, 66, 44, 20, 5, 1, 1, 5, 25, 60, 110, 126, 110, 60, 25, 5, 1, 1, 6, 30, 85, 170, 236, 236, 170, 85, 30, 6, 1, 1, 6, 36, 110, 255
COMMENTS
Sometimes erroneously called "Lossnitsch's triangle". But the author's name is Losanitsch (I have seen the original paper in Chem. Ber.). This is a German version of the Serbian name Lozanic. - N. J. A. Sloane, Jun 29 2008 For n >= 3, a(n-3,k) is the number of series-reduced (or homeomorphically irreducible) trees which become a path P(k+1) on k+1 nodes, k >= 0, when all leaves are omitted (see illustration). Proof by Pólya's enumeration theorem. - Wolfdieter Lang, Jun 08 2001 The number of ways to put beads of two colors in a line, but take symmetry into consideration, so that 011 and 110 are considered the same. - Yong Kong (ykong(AT)nus.edu.sg), Jan 04 2005
Alternating row sums are 1,0,1,0,2,0,4,0,8,0,16,0,... - Gerald McGarvey, Oct 20 2008 The triangle sums, see A180662 for their definitions, link Losanitsch's triangle A034851 with several sequences, see the crossrefs. We observe that the Ze3 and Ze4 sums link Losanitsch's triangle with A005683, i.e., R. K. Guy's Twopins game. - Johannes W. Meijer, Jul 14 2011 T(n-(L-1)k, k) is the number of ways to cover an n-length line by exactly k L-length segments excluding symmetric covers. For L=2 it is corresponds to A102541, for L=3 to A228570 and for L=4 to A228572. - Philipp O. Tsvetkov, Nov 08 2013 Also the number of equivalence classes of ways of placing k 1 X 1 tiles in an n X 1 rectangle under all symmetry operations of the rectangle. - Christopher Hunt Gribble, Feb 16 2014 T(n, k) is the number of non-isomorphic outer planar graphs of order n+3, size n+3+k, and maximum degree k+2. - Christian Barrientos, Oct 18 2018 T(n, k) is the sum of even-degree coefficients of the Gaussian polynomial [n, k]_q. The area below a NE lattice path between (0,0) and (k, n-k) is even for T(n, k) paths and odd for A034852(n, k) of them. For a (non-reversible) string of k black and n-k white beads, consider the minimum number of bead transpositions needed to place the black ones to the left and the white ones to the right (in other words, the number of inversions of the permutation obtained by labeling the black beads by integers 1,...,k and the white ones by k+1,...,n, in the same order they take on the string). It is even for T(n, k) strings and odd for A034852(n, k) cases. (End)
Named after the Serbian chemist, politician and diplomat Simeon Milivoje "Sima" Lozanić (1847-1935). - Amiram Eldar, Jun 10 2021 T(n, k) is the number of caterpillars with a perfect matching, with 2n+2 vertices and diameter 2n-1-k. - Christian Barrientos, Sep 12 2023
LINKS
Tewodros Amdeberhan, Mahir Bilen Can and Victor H. Moll, Broken bracelets, Molien series, paraffin wax and the elliptic curve of conductor 48, SIAM Journal of Discrete Math., Vol. 25, No. 4 (2011), p. 1843-1859; arXiv preprint, arXiv:1106.4693 [math.CO], 2011. See Theorem 2.8. Stephen G. Hartke and A. J. Radcliffe, Signatures of Strings, Annals of Combinatorics, Vol. 17, No. 1 (March, 2013), pp. 131-150.
FORMULA
G.f. for k-th column (if formatted as lower triangular matrix a(n, k)): x^k*Pe(floor((k+1)/2), x^2)/(((1-x)^(k+1))*(1+x)^(floor((k+1)/2))), where Pe(n, x^2) := Sum_{m=0..floor(n/2)} A034839(n, m)*x^(2*m) (row polynomials of Pascal array even numbered columns). - Wolfdieter Lang, May 08 2001 a(n, k) = a(n-1, k-1) + a(n-1, k) - C(n/2-1, (k-1)/2), where the last term is present only if n is even and k is odd (see Sloane link).
T(n, k) = T(n-2, k-2) + T(n-2, k) + C(n-2, k-1), n > 1.
Let P(n, x, y) = Sum_{m=0..n} a(n, m)*x^m*y^(n-m), then for x > 0, y > 0 we have P(n, x, y) = (x+y)*P(n-1, x, y) for n odd and P(n, x, y) = (x+y)*P(n-1, x, y) - x*y*(x^2+y^2)^((n-2)/2) for n even. - Gerald McGarvey, Feb 15 2005 It appears that:
T(n,k) = C(n,k)/2, n even, k odd;
T(n,k) = (C(n,k) + C(n/2,k/2))/2, n even, k even;
T(n,k) = (C(n,k) + C((n-1)/2,(k-1)/2))/2, n odd, k odd;
T(n,k) = (C(n,k) + C((n-1)/2,k/2))/2, n odd, k even.
(End)
EXAMPLE
Triangle begins
1;
1, 1;
1, 1, 1;
1, 2, 2, 1;
1, 2, 4, 2, 1;
1, 3, 6, 6, 3, 1;
1, 3, 9, 10, 9, 3, 1;
1, 4, 12, 19, 19, 12, 4, 1;
1, 4, 16, 28, 38, 28, 16, 4, 1;
1, 5, 20, 44, 66, 66, 44, 20, 5, 1;
MAPLE
A034851 := proc(n, k) option remember; local t; if k = 0 or k = n then return(1) fi; if n mod 2 = 0 and k mod 2 = 1 then t := binomial(n/2-1, (k-1)/2) else t := 0; fi; A034851(n-1, k-1)+ A034851(n-1, k)-t; end: seq(seq( A034851(n, k), k=0..n), n=0..11);
MATHEMATICA
t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]](* Jean-François Alcover, Feb 07 2012, after PARI *)
PROG
(PARI) {T(n, k) = (1/2) *(binomial(n, k) + binomial(n%2, k%2) * binomial(n\2, k\2))}; /* Michael Somos, Oct 20 1999 */ (Haskell)
a034851 n k = a034851_row n !! k
a034851_row 0 = [1]
a034851_row 1 = [1, 1]
a034851_row n = zipWith (-) (zipWith (+) ([0] ++ losa) (losa ++ [0]))
([0] ++ a204293_row (n-2) ++ [0])
where losa = a034851_row (n-1)
a034851_tabl = map a034851_row [0..]
CROSSREFS
Triangle sums (see the comments): A005418 (Row), A011782 (Related to Row2), A102526 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A005207 (Kn3, Kn4), A005418 (Fi1, Fi2), A102543 (Ca1, Ca2), A192928 (Gi1, Gi2), A005683 (Ze3, Ze4). Sums of squares of terms in rows equal A211208.
Triangle in which row n is generated from (1,2,3,...,n) dot (n, n-1,...,1) with subtractive carryovers.
+0
2
1, 2, 0, 3, 1, 2, 4, 2, 4, 0, 5, 3, 6, 2, 3, 6, 4, 8, 4, 6, 0, 7, 5, 10, 6, 9, 3, 4, 8, 6, 12, 8, 12, 6, 8, 0, 9, 7, 14, 10, 15, 9, 12, 4, 5, 10, 8, 16, 12, 18, 12, 16, 8, 10, 0, 11, 9, 18, 14, 21, 15, 20, 12, 15, 5, 6, 12, 10, 20, 16, 24, 18, 24, 16, 20, 10, 12, 0
COMMENTS
The subtractive carryover dot product of two vectors (a(1),a(2),...,a(n)) dot (b(1),b(2),...,b(n)) = (c(1),...,c(n)) is defined by c(1) = a(1)*b(1) and c(i) = a(i)*b(i)-c(i-1), i>1.
A177877 = analogous triangle with additive carryovers.
A160770 = the analogous triangle using the triangular series as the generating vector.
FORMULA
By rows, dot product of (1,2,3,...) and (...3,2,1) with subtractive carryovers; such that current row product subtracts previous product.
EXAMPLE
Row 3 = (4, 2, 4, 0) = (1, 2, 3, 4) dot (4, 3, 2, 1) with subtractive carryovers = (4), then (2*3 - 4 = 2), (3*2 - 2 = 4), and (4*1 - 4 = 0).
First few rows of the triangle:
1;
2, 0;
3, 1, 2;
4, 2, 4, 0;
5, 3, 6, 2, 3;
6, 4, 8, 4, 6, 0;
7, 5, 10, 6, 9, 3, 4;
8, 6, 12, 8, 12, 6, 8, 0;
9, 7, 14, 10, 15, 9, 12, 4, 5;
10, 8, 16, 12, 18, 12, 16, 8, 10, 0;
11, 9, 18, 14, 21, 15, 20, 12, 15, 5, 6;
12, 10, 20, 16, 24, 18, 24, 16, 20, 10, 12, 0;
...
Number of different isotemporal classes of diasters with n peripheral edges.
+0
2
1, 1, 4, 7, 15, 23, 38, 53, 77, 101, 136, 171, 219, 267, 330, 393, 473, 553, 652, 751, 871, 991, 1134, 1277, 1445, 1613, 1808, 2003, 2227, 2451, 2706, 2961, 3249, 3537, 3860, 4183, 4543, 4903, 5302, 5701, 6141, 6581, 7064, 7547, 8075, 8603
COMMENTS
See A092481 for the definition of isotemporal classes.
REFERENCES
Benjamin de Bivort, Isotemporal classes of diasters, beachballs and daisies, preprint, 2005.
FORMULA
a(n=2k) = 1 + (Sum_{i=1..(n/2)-1} n*i-i^2+n+1) + (1/2)*((n/2)^2+3*(n/2)+2). a(n=2k+1) = 1 + (Sum_{i=1..(n-1)/2} n*i-i^2+n+1). [Corrected by Sean A. Irvine after private communication with Benjamin de Bivort, Feb 13 2012]
EXAMPLE
A diaster is defined to be any graph with a central edge with vertices of degree j and k and j+k peripheral edges connected to the central edge each terminating in a vertex of degree 1. a(5)=23 refers to diasters with 5 peripheral edges. These can be uniquely arranged with 0, 1 or 2 peripheral edges on a particular side, yielding 1, 10 and 12 isotemporal classes respectively each.
AUTHOR
Benjamin de Bivort (bivort(AT)fas.harvard.edu), Aug 02 2005
Triangle read by rows: T(n,m) = number of planar partitions of n with trace m.
+0
7
1, 2, 1, 3, 2, 1, 4, 6, 2, 1, 5, 10, 6, 2, 1, 6, 19, 14, 6, 2, 1, 7, 28, 28, 14, 6, 2, 1, 8, 44, 52, 33, 14, 6, 2, 1, 9, 60, 93, 64, 33, 14, 6, 2, 1, 10, 85, 152, 127, 70, 33, 14, 6, 2, 1, 11, 110, 242, 228, 142, 70, 33, 14, 6, 2, 1, 12, 146, 370, 404, 272, 149, 70, 33, 14, 6, 2, 1, 13
COMMENTS
Also number of partitions of n objects of 2 colors into k parts, each part containing at least one black object.
T(n+m, m) = A005380(n), n >= 1, for all m >= n. T(m, m) = 1 for m >= 1. See the Stanley reference Exercise 7.99. With offset n=0 a column for m=0 with the only non-vanishing entry T(0, 0) = 1 could be added. - Wolfdieter Lang, Mar 09 2015
REFERENCES
G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976 (Ch. 11, Example 5 and Ch. 12, Example 5).
R. P. Stanley, Enumerative Combinatorics, Cambridge University Press, Vol. 2, 1999; p. 365 and Exercise 7.99, p. 484 and pp. 548-549.
FORMULA
G.f.: Product_{k>=1} 1/(1-q*x^k)^k (with offset n=0 in x powers).
EXAMPLE
The triangle T(n,m) begins:
n\m 1 2 3 4 5 6 7 8 9 10 11 12 ...
1: 1
2: 2 1
3: 3 2 1
4: 4 6 2 1
5: 5 10 6 2 1
6: 6 19 14 6 2 1
7: 7 28 28 14 6 2 1
8: 8 44 52 33 14 6 2 1
9: 9 60 93 64 33 14 6 2 1
10: 10 85 152 127 70 33 14 6 2 1
11: 11 110 242 228 142 70 33 14 6 2 1
12: 12 146 370 404 272 149 70 33 14 6 2 1
MAPLE
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, add(b(n-i*j, i-1)*x^j*
binomial(i+j-1, j), j=0..n/i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n$2)):
MATHEMATICA
b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, i - 1]*x^j*Binomial[i + j - 1, j], {j, 0, n/i}]]]];
T[n_] := Table[Coefficient[#, x, i], {i, 1, Exponent[#, x]}]]& @ b[n, n];
Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...
+0
26
1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
COMMENTS
Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002 The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002 The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2* A014301. - Claude MorinRiordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004 Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014 Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc. The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0).
End
FORMULA
G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014 O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014 T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002 T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002 T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004 T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005 T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017 Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024 The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below.
(End)
EXAMPLE
Triangle begins
1;
0, 1;
1, 1, 1;
0, 2, 2, 1;
1, 2, 4, 3, 1;
0, 3, 6, 7, 4, 1;
1, 3, 9, 13, 11, 5, 1;
0, 4, 12, 22, 24, 16, 6, 1;
1, 4, 16, 34, 46, 40, 22, 7, 1;
0, 5, 20, 50, 80, 86, 62, 29, 8, 1;
Sequences obtained with Miquel A. Fiol's Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6): ------- ------- ------- ------- ------- ------- A062136*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
22--12---24---14---44
| \ / |
23 11 34
\ | /
13
|
33
MAPLE
read transforms; 1/(1-y-x*y-x^2); SERIES2(%, x, y, 12); SERIES2TOLIST(%, x, y, 12);
MATHEMATICA
t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)
PROG
(Sage)
@cached_function
def prec(n, k):
if k==n: return 1
if k==0: return 0
return -prec(n-1, k-1)-sum(prec(n, k+i-1) for i in (2..n-k+1))
return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
(PARI) T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
for(n=0, 12, for(k=0, n, print1(T(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 11 2017 (Python)
from sympy import binomial
def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
CROSSREFS
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001
Triangle read by rows T(n,k) in which column k lists the partial sums of the k-th column of triangle A236104.
+0
8
1, 5, 14, 1, 30, 2, 55, 6, 91, 10, 1, 140, 19, 2, 204, 28, 3, 285, 44, 7, 385, 60, 11, 1, 506, 85, 15, 2, 650, 110, 24, 3, 819, 146, 33, 4, 1015, 182, 42, 8, 1240, 231, 58, 12, 1, 1496, 280, 74, 16, 2, 1785, 344, 90, 20, 3, 2109, 408, 115, 29, 4, 2470, 489, 140, 38, 5, 2870, 570, 165, 47, 9, 3311, 670, 201, 56, 13, 1
COMMENTS
Alternating sum of row n equals A175254(n), i.e., Sum_{k=1.. A003056(n)} (-1)^(k-1)*T(n,k) = A175254(n), which is also the volume (or the total number of units cubes) in the first n levels of the stepped pyramid described in A245092. Row n has length A003056(n) hence the first element of column k is in row A000217(k).
EXAMPLE
Triangle begins:
1;
5;
14, 1;
30, 2;
55, 6;
91, 10, 1;
140, 19, 2;
204, 28, 3;
285, 44, 7;
385, 60, 11, 1;
506, 85, 15, 2;
650, 110, 24, 3;
819, 146, 33, 4;
1015, 182, 42, 8;
1240, 231, 58, 12, 1;
1496, 280, 74, 16, 2;
1785, 344, 90, 20, 3;
2109, 408, 115, 29, 4;
2470, 489, 140, 38, 5;
2870, 570, 165, 47, 9;
3311, 670, 201, 56, 13, 1;
3795, 770, 237, 72, 17, 2;
4324, 891, 273, 88, 21, 3;
4900, 1012, 322, 104, 25, 4;
...
For n = 6 we have that A175254(6) = [1] + [1 + 3] + [1 + 3 + 4] + [1 + 3 + 4 + 7] + [1 + 3 + 4 + 7 + 6] + [1 + 3 + 4 + 7 + 6 + 12] = 1 + 4 + 8 + 15 + 21 + 33 = 82. On the other hand the alternating sum of the 6th row of the triangle is 91 - 10 + 1 = 82, equaling A175254(6).
CROSSREFS
Cf. A000203, A000217, A003056, A024916, A175254, A196020, A235791, A236104, A237048, A237591, A237593, A237270, A237271, A245092, A261699, A262626.
Odd-numbered columns of Losanitsch triangle A034851 formatted as triangle with an additional first column.
+0
3
1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 2, 6, 3, 1, 0, 3, 10, 12, 4, 1, 0, 3, 19, 28, 20, 5, 1, 0, 4, 28, 66, 60, 30, 6, 1, 0, 4, 44, 126, 170, 110, 42, 7, 1, 0, 5, 60, 236, 396, 365, 182, 56, 8, 1, 0, 5, 85, 396, 868, 1001, 693, 280, 72, 9, 1
COMMENTS
Because the sequence of column m=2*k, k >= 1, of A034851 is the partial sum sequence of the one of column m=2*k-1 the present triangle is essentially Losanitsch's triangle A034851.
FORMULA
T(n, m) = 0 if n<m; T(0, 0)=1, T(n, 0)=0 if n >= 1; T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1) if n+m even and T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1)-binomial((n+m-3)/2, m-1) if n+m odd, n >= m >= 1.
G.f. for column m: x^m*Pe(m, x^2)/(((1-x)^(2*m))*(1+x)^m), m >= 0, with Pe(m, x^2)= sum( A034839(m, k)*x^(2*k), k=0..floor(n/2)), the row polynomial of array A034839 (even-indexed entries of the rows of Pascal's triangle).
EXAMPLE
Triangle begins:
{1};
{0,1};
{0,1,1};
{0,2,2,1};
...
Pe(4,x^2)=1+6*x^2+x^4.
MATHEMATICA
t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2; Flatten[Table[t[n - 1 + m, n - m], {n, 0, 12}, {m, 0, n}]] (* Michael De Vlieger, Sep 28 2024, after Jean-François Alcover at A034851 *)
Triangle T(n,m) = 2*(min(n - m + 1, m))^2 read by rows.
+0
4
2, 2, 2, 2, 8, 2, 2, 8, 8, 2, 2, 8, 18, 8, 2, 2, 8, 18, 18, 8, 2, 2, 8, 18, 32, 18, 8, 2, 2, 8, 18, 32, 32, 18, 8, 2, 2, 8, 18, 32, 50, 32, 18, 8, 2, 2, 8, 18, 32, 50, 50, 32, 18, 8, 2, 2, 8, 18, 32, 50, 72, 50, 32, 18, 8, 2, 2, 8, 18, 32, 50, 72, 72, 50, 32, 18, 8, 2, 2, 8, 18, 32, 50, 72, 98, 72
COMMENTS
The flattened triangle is simply 2 followed by A137508.
EXAMPLE
The table starts in row n=1 with columns 1<=m<=n as:
2;
2,2;
2,8,2;
2,8,8,2;
2,8,18,8,2;
2,8,18,18,8,2;
...
MAPLE
A168281 := proc(n, m) 2*(min(n+1-m, m))^2 ; end proc:
seq(seq( A168281(n, m), m=1..n), n=1..20) ;
MATHEMATICA
Table[Map[2 Min[n + # - 1, #]^2 &, Drop[#, -Boole@ EvenQ@ n] ~Join~ Reverse@ # &@ Range@ Floor[n/2]], {n, 2, 14}] // Flatten (* Michael De Vlieger, Jul 19 2016 *)
EXTENSIONS
Rephrased all comments in terms of a triangle by R. J. Mathar, Nov 24 2010
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