Displaying 1-10 of 18 results found.
1, 4, 8, 16, 32, 33, 63, 124, 136, 244, 276, 480, 560, 561, 944, 1135, 1140, 1856, 2298, 2316, 3649, 4705, 7174, 9398, 9558, 9559, 14104, 18984, 19415, 27728, 38320, 39432, 39457, 54512, 77298, 80075, 80163, 107168, 155823, 162583, 162863, 162864, 210687, 313927, 330878, 414200, 632080, 669872, 814296, 1271960, 1600864
COMMENTS
This is a dynamically defined sequence. Since the nonprimes from each row are mixed with the nonprimes of previous rows and then sorted, the value of a(n) may change each time we add a new row.
(End)
EXAMPLE
Scanning rows of A141020 or A141021 and sorting new nonprimes into the list we get: 1 yields a(1) = 1.
1 1 yields no new member.
1 2 1 yields no new member.
1 4 2 1 yields a(2) = 4.
1 8 4 2 1 yields a(3) = 8.
1 16 8 4 2 1 yields a(4) = 16.
1 32 16 8 4 2 1 yields a(5) = 32.
1 63 33 16 8 4 2 1 yields a(6) = 33 and a(7) = 63.
1 124 67 33 16 8 4 2 1 yields a(8) = 124.
1 244 136 67 33 16 8 4 2 1 yields a(9) = 136 and a(10) = 244.
1 480 276 136 67 33 16 8 4 2 1 yields a(11) = 276 and a(12) = 480.
1 944 560 276 136 67 33 16 8 4 2 1 yields a(13) = 560 and a(14) = 944.
...
In the above example, we only sort the nonprimes up to row 11; we get the same output from R. J. Mathar's program below if we say A141031(11). If, however, we include more rows in the program, the indexing of the nonprimes changes. For example, the nonprimes in the data above come from the nonprimes of 22 rows. If we include more rows, then the indexing again changes and the value of each a(n) may not stay the same.
(End)
MAPLE
A141020 := proc(n, k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; elif k=n-4 then 16 ; else procname(n-1, k)+procname(n-2, k)+procname(n-3, k)+procname(n-4, k) +procname(n-5, k)+procname(n-5, k-1) ; fi; end:
A141031 := proc(nmax) local a, n, k ; a := [] ; for n from 0 to nmax do for k from 0 to n do a141020 := A141020(n, k) ; if not isprime( a141020) and not a141020 in a then a := [op(a), a141020] ; fi; od: od: RETURN(sort(a)) ; end: A141031(30) ; # R. J. Mathar, Sep 19 2008
CROSSREFS
Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141064, A141065, A141066, A141067, A141068, A141069.
EXTENSIONS
Simplified definition, corrected values by R. J. Mathar, Sep 19 2008
Triangle G(n, k) read by rows, for 1 <= k <= n, where G(n, n) = G(n+1, 1) = 1, G(n+2, 2) = 2, G(n+3, m) = G(n+1, m-1) + G(n+1, m-2) + G(n+2, m-1) for n >= 1 and m = 3..(n+2).
+10
24
1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 5, 7, 1, 1, 2, 5, 11, 12, 1, 1, 2, 5, 12, 23, 20, 1, 1, 2, 5, 12, 28, 46, 33, 1, 1, 2, 5, 12, 29, 63, 89, 54, 1, 1, 2, 5, 12, 29, 69, 137, 168, 88, 1, 1, 2, 5, 12, 29, 70, 161, 289, 311, 143, 1, 1, 2, 5, 12, 29, 70, 168, 367, 594, 567, 232, 1, 1, 2, 5, 12, 29, 70, 169, 399, 817, 1194, 1021, 376, 1
COMMENTS
Let b(m) = lim_{n -> infinity} G(n, m) for each m >= 1. Then b(1) = 1, b(2) = 2, and b(m) = 2*b(m-1) + b(m-2) for m >= 3. (The existence of the limit can be proved by induction on m.) This means b(m) = A000129(m) for m >= 1 (known as the Pell numbers). If we want to get the second main diagonal, we let c(n) = G(n+1, n) for n >= 1. Then c(n+2) = G(n+3, n+2) = G(n+1, n+1) + G(n+1, n) + G(n+2, n+1) = 1 + c(n) + c(n+1) with c(1) = G(2, 1) = 1 and c(2) = G(3, 2) = 2, which implies that c(n) = A000071(n+2) = Fibonacci(n+2) - 1 for n >= 1. This array is the mirror image of A140998 (except for a shifting of the indices by 1). Thus, G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n. This array has index of obliqueness e = 1, while array A140998 has index of obliqueness e = 0. Both arrays have the same index of asymmetry (s = 1). (End) One of the two rectangular versions, say (RA(n,k): n,k >= 0), of this triangular array (G(n,k): 1 <= k <= n) is given by RA(n,k) = G(n+k-1,k) for n,k >= 1. Conversely, G(n,k) = RA(n-k+1, k) for 1 <= k <= n. (This assumes that the triangle G(n,k) is read from the array RA(n,k) by ascending antidiagonals.)
Note that [o.g.f of RA](x,y) = x*[o.g.f. of G](x, y/x) and [o.g.f of G](x,y) = x^(-1)*[o.g.f of RA](x,x*y).
The other rectangular version, say (RD(n,k): n,k >= 0), of this triangular array (G(n,k): 1 <= k <= n) is given by RD(n,k) = RA(k,n) = G(n+k-1,n) for n,k >= 1. Conversely, G(n,k) = RD(k,n-k+1) for 1 <= k <= n. (This assumes that the triangle G(n,k) is read from the array RD(n,k) by descending antidiagonals.)
Note that [o.g.f of RD](x,y) = y*[o.g.f. of G](y,x/y) and [o.g.f of G](x,y) = x^(-1)*[o.g.f of RD](x*y, x). (End)
FORMULA
G(n, k) = A140998(n - 1, n - k) for 1 <= k <= n. Bivariate o.g.f.: Sum_{n >= 1, k >= 1} G(n, k)*x^n*y^k = x*y*(1 - x*y -x^2*y^2 + x^3*y^2)/((1 - x) * (1 - x*y) * (1 - x*y - x^2*y - x^2*y^2)). (Here, we let G(n, k) = 0 when 1 <= n < k.)
To get the row sums, we let y = 1 in the above bivariate g.f. and simplify. We get x/(1 - 2*x), which is the g.f. of sequence ( A000079(n-1): n >= 1) = (2^(n-1): n >= 1). (End) We give formulas about the rectangular array RA(n,k).
Initial conditions: RA(1,n) = RA(n+1,1) = 1 and RA(n+1,2) = 2 for n >= 1.
Recurrence: RA(n,k) = RA(n-1,k-1) + RA(n,k-2) + RA(n,k-1) for n >= 2 and k >= 3.
The main diagonal of the array is RA(n,n) = A000129(n) (Pell numbers). Bivariate o.g.f: Sum_{n,k >= 1} RA(n,m)*x^n*y^k = x*y*(x*y^2 - y^2 - y + 1)/((1 - x)*(1 - y)*(-x*y - y^2 - y + 1)).
To obtain formulas about the other rectangular array, RD(n,k), we use the equations RD(n,k) = RA(k,n) for n,k >= 1 and [o.g.f. of RD](x,y) = [o.g.f. of RA](y,x). (End)
EXAMPLE
Triangle G(n,k) (with rows for n >= 1 and columns for 1 <= k <= n) begins:
1
1 1
1 2 1
1 2 4 1
1 2 5 7 1
1 2 5 11 12 1
1 2 5 12 23 20 1
1 2 5 12 28 46 33 1
1 2 5 12 29 63 89 54 1
1 2 5 12 29 69 137 168 88 1
1 2 5 12 29 70 161 289 311 143 1
1 2 5 12 29 70 168 367 594 567 232 1
1 2 5 12 29 70 169 399 817 1194 1021 376 1
1 2 5 12 29 70 169 407 934 1778 2355 1820 609 1
...
Rectangular array RA(n,k) (with rows for n >= 1 and columns for k >= 1) begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 2, 4, 7, 12, 20, 33, 54, 88, 143, ...
1, 2, 5, 11, 23, 46, 89, 168, 311, 567, ...
1, 2, 5, 12, 28, 63, 137, 289, 594, 1194, ...
1, 2, 5, 12, 29, 69, 161, 367, 817, 1778, ...
1, 2, 5, 12, 29, 70, 168, 399, 934, 2150, ...
1, 2, 5, 12, 29, 70, 169, 407, 975, 2316, ...
1, 2, 5, 12, 29, 70, 169, 408, 984, 2367, ...
1, 2, 5, 12, 29, 70, 169, 408, 985, 2377, ...
1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, ...
...
Reading the array RA(n,k) by ascending antidiagonals, we get triangle G(n,k) above. (End)
MAPLE
A140993 := proc(n, k) if k = n then 1; elif k = 1 then 1; elif k = 2 then 2; else procname(n-2, k-1)+procname(n-2, k-2)+procname(n-1, k-1) ; end if; end proc: seq(seq( A140993(n, k), k=1..n), n=1..15) ; # R. J. Mathar, Apr 28 2010
MATHEMATICA
t[n_, k_] := If[k == n, 1, If[k == 1, 1, If[k == 2, 2, t[n - 2, k - 1] + t[n - 2, k - 2] + t[n - 1, k - 1]]]]; Flatten[Table[ t[n, k], {n, 13}, {k, n}]] (* Robert G. Wilson v, Dec 22 2011 *)
CROSSREFS
Cf. A000071, A000079, A000129, A007318, A140994, A140995, A140996, A140997, A140998, A141020, A141021.
Triangle G(n, k), for 0 <= k <= n, read by rows, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, m) = G(n+1, m-2) + G(n+1, m-3) + G(n+2, m-2) + G(n+3, m-1) for n >= 0 and m = 3..(n+3).
+10
24
1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 4, 9, 15, 1, 1, 2, 4, 9, 19, 28, 1, 1, 2, 4, 9, 19, 40, 52, 1, 1, 2, 4, 9, 19, 41, 83, 96, 1, 1, 2, 4, 9, 19, 41, 88, 170, 177, 1, 1, 2, 4, 9, 19, 41, 88, 188, 345, 326, 1, 1, 2, 4, 9, 19, 41, 88, 189, 400, 694, 600, 1, 1, 2, 4, 9, 19, 41, 88, 189, 406, 846, 1386, 1104, 1
COMMENTS
This is a mirror image of the triangular array A140997. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 1. Array A140997 has the same index of asymmetry, but has index of obliqueness e = 0. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.) In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0). Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively). If A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k is the bivariate g.f. for this array (with G(n, k) = 0 for 0 <= n < k) and B(x, y) = Sum_{n, k} A140997(n, k)*x^n*y^k, then A(x, y) = B(x*y, y^(-1)). This can be proved using formal manipulation of double series expansions and the fact G(n, k) = A140997(n, n-k) for 0 <= k <= n. If we let b(k) = lim_{n -> infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, and b(k) = b(k-1) + 2*b(k-2) + b(k-3) for k >= 3. (The existence of the limit can be proved by induction on k.) It follows that b(k) = A141015(k) for k >= 0. (End)
FORMULA
G(n, k) = A140997(n, n-k) for 0 <= k <= n. Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (x^4*y^3 - x^3*y^3 - x^2*y^2 + x^2*y - x*y + 1)/((1- x*y)*(1 - x)*(1- x*y - x^2*y^2 - x^3*y^3 - x^3*y^2)).
(End)
EXAMPLE
Triangle begins:
1
1 1
1 2 1
1 2 4 1
1 2 4 8 1
1 2 4 9 15 1
1 2 4 9 19 28 1
1 2 4 9 19 40 52 1
1 2 4 9 19 41 83 96 1
1 2 4 9 19 41 88 170 177 1
1 2 4 9 19 41 88 188 345 326 1
1 2 4 9 19 41 88 189 400 694 600 1
1 2 4 9 19 41 88 189 406 846 1386 1104 1
E.g., G(12, 9) = G(9, 7) + G(9, 6) + G(10, 7) + G(11, 8) = 170 + 88 + 188 + 400 = 846.
MAPLE
G := proc(n, k) if k=0 or n =k then 1; elif k= 1 then 2 ; elif k =2 then 4; elif k > n or k < 0 then 0 ; else procname(n-3, k-2)+procname(n-3, k-3)+procname(n-2, k-2)+procname(n-1, k-1) ; end if; end proc: seq(seq(G(n, k), k=0..n), n=0..15) ; # R. J. Mathar, Apr 14 2010
MATHEMATICA
nlim = 50;
Do[G[n, 0] = 1, {n, 0, nlim}];
Do[G[n, n] = 1, {n, 1, nlim}];
Do[G[n + 2, 1] = 2, {n, 0, nlim}];
Do[G[n + 3, 2] = 4, {n, 0, nlim}];
Do[G[n + 4, m] =
G[n + 1, m - 2] + G[n + 1, m - 3] + G[n + 2, m - 2] +
G[n + 3, m - 1], {n, 0, nlim}, {m, 3, n + 3}];
A140994 = {}; For[n = 0, n <= nlim, n++,
For[k = 0, k <= n, k++, AppendTo[ A140994, G[n, k]]]];
CROSSREFS
Cf. A007318, A008937, A140993, A140995, A140996, A140997, A140998, A141015, A141018, A141020, A141021, A141031, A141065, A141066, A141067.
Triangle G(n,k) read by rows, for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, and G(n+4, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) for n >= 0 and m = 1..n+1.
+10
24
1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 15, 9, 4, 2, 1, 1, 28, 19, 9, 4, 2, 1, 1, 52, 40, 19, 9, 4, 2, 1, 1, 96, 83, 41, 19, 9, 4, 2, 1, 1, 177, 170, 88, 41, 19, 9, 4, 2, 1, 1, 326, 345, 188, 88, 41, 19, 9, 4, 2, 1, 1, 600, 694, 400, 189, 88, 41, 19, 9, 4, 2, 1, 1, 1104, 1386, 846, 406, 189, 88, 41, 19, 9, 4, 2, 1, 1, 2031, 2751, 1779, 871, 406, 189, 88, 41, 19, 9, 4, 2, 1, 1, 3736, 5431, 3719, 1866, 872, 406, 189, 88, 41, 19, 9, 4, 2, 1
COMMENTS
This is a mirror image of the triangular array A140994. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 0. Array A140994 has the same index of asymmetry, but has index of obliqueness e = 1. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.) In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0). Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively). (End)
FORMULA
G(n, k) = A140994(n, n-k) for 0 <= k <= n. Bivariate g.f.: Sum_{n,k >= 0} G(n,k)*x^n*y^k = (1 - x - x^2 - x^3 + x^2*y + x^4*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^3*y)).
Differentiating once w.r.t. y and setting y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3)). This is the g.f. of sequence A008937. (End)
EXAMPLE
Triangle begins:
1
1 1
1 2 1
1 4 2 1
1 8 4 2 1
1 15 9 4 2 1
1 28 19 9 4 2 1
1 52 40 19 9 4 2 1
1 96 83 41 19 9 4 2 1
1 177 170 88 41 19 9 4 2 1
1 326 345 188 88 41 19 9 4 2 1
1 600 694 400 189 88 41 19 9 4 2 1
...
E.g., G(14, 2) = G(11, 1) + G(11, 2) + G(12, 2) + G(13, 2) = 600 + 694 + 1386 + 2751 = 5431.
MATHEMATICA
nlim = 50;
Do[G[n, 0] = 1, {n, 0, nlim}];
Do[G[n + 1, n + 1] = 1, {n, 0, nlim}];
Do[G[n + 2, n + 1] = 2, {n, 0, nlim}];
Do[G[n + 3, n + 1] = 4, {n, 0, nlim}];
Do[G[n + 4, m] =
G[n + 1, m - 1] + G[n + 1, m] + G[n + 2, m] + G[n + 3, m], {n, 0,
nlim}, {m, 1, n + 1}];
A140997 = {}; For[n = 0, n <= nlim, n++,
For[k = 0, k <= n, k++, AppendTo[ A140997, G[n, k]]]];
CROSSREFS
Cf. A007318, A008937, A140993, A140994, A140995, A140996, A140998, A141015, A141018, A141020, A141021, A141065, A141066, A141067.
EXTENSIONS
Deleted extraneous term at a(29) by Robert Price, Aug 25 2019 Added 13 missing terms at a(79) by Robert Price, Aug 25 2019
Triangle G(n, k), read by rows, for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, and G(n+3, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) for n >= 0 and m = 1..n+1.
+10
24
1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 7, 5, 2, 1, 1, 12, 11, 5, 2, 1, 1, 20, 23, 12, 5, 2, 1, 1, 33, 46, 28, 12, 5, 2, 1, 1, 54, 89, 63, 29, 12, 5, 2, 1, 1, 88, 168, 137, 69, 29, 12, 5, 2, 1, 1, 143, 311, 289, 161, 70, 29, 12, 5, 2, 1, 1, 232, 567, 594, 367, 168, 70, 29, 12, 5, 2, 1
COMMENTS
According to the attached picture, the index of asymmetry here is s = 1 and the index of obliqueness (or obliquity) is e = 0.
In the picture, the equation G(n, e*n) = 1 becomes G(n, 0) = 1, while the equations G(n+x+1, n-e*n+e*x-e+1) = 2^x for 0 <= x < s = 1 become G(n+1, n+1) = 1 and G(n+2, n+1) = 2.
Also, in the picture, the recurrence G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{m=1..s+1} G(n+m, k-e*s+m*e-2*e) for k = 1..n+1 becomes G(n+3, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) for k = 1..n+1.
Except for a shifting of the indices by 1, this array is a mirror image of array A140993. We have G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n. Triangular array A140993 has the same index of asymmetry (i.e., s = 1) but index of obliqueness e = 1. (End)
FORMULA
G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n. Let A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k and B(x,y) = Sum_{n,k >= 1} A140993(n, k). Then A(x, y) = x^(-1) * B(x*y, y^(-1)). Thus, the g.f. of the current array is A(x, y) = (1 - x - x^2 + x^3*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^2*y)). To find the g.f. of the k-th column (where k >= 0), we differentiate A(x, y) k times with respect to y, divide by k!, and substitute y = 0. For example, differentiating A(x, y) once w.r.t. y and setting y = 0, we get the g.f. of the k = 1 column: x/((1 - x)*(1 - x - x^2)). This is the g.f. of sequence ( A000071(n+2): n >= 0) = (Fibonacci(n+2) - 1: n >= 0). G.f. of column k = 2 is x^2*(1 - x + x^3)/((1 - x)*(1 - x - x^2)^2). Thus, column k = 2 is a shifted version of ( A140992(n): n >= 0). (End)
EXAMPLE
Triangle begins (with rows for n >= 0 and columns for k >= 0):
1;
1, 1;
1, 2, 1;
1, 4, 2, 1;
1, 7, 5, 2, 1;
1, 12, 11, 5, 2, 1;
1, 20, 23, 12, 5, 2, 1;
1, 33, 46, 28, 12, 5, 2, 1;
1, 54, 89, 63, 29, 12, 5, 2, 1;
1, 88, 168, 137, 69, 29, 12, 5, 2, 1;
1, 143, 311, 289, 161, 70, 29, 12, 5, 2, 1;
MATHEMATICA
G[n_, k_] := G[n, k] = Which[k==0 || k==n, 1, k==n-1, 2, True, G[n-2, k-1] + G[n-2, k] + G[n-1, k]]; Table[G[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 09 2019 *)
PROG
(PARI) G(n, k) = if(k==0 || k==n, 1, if(k==n-1, 2, G(n-1, k) + G(n-2, k) + G(n-2, k-1)));
for(n=0, 12, for(k=0, n, print1(G(n, k), ", "))) \\ G. C. Greubel, Jun 09 2019 (Sage)
def G(n, k):
if (k==0 or k==n): return 1
elif (k==n-1): return 2
else: return G(n-1, k) + G(n-2, k) + G(n-2, k-1)
[[G(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 09 2019
EXTENSIONS
Indices in the definition corrected by R. J. Mathar, Aug 02 2009
Triangle G(n, k) read by rows for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) + G(n+4, m) for n >= 0 for m = 1..(n+1).
+10
23
1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 31, 17, 8, 4, 2, 1, 1, 60, 35, 17, 8, 4, 2, 1, 1, 116, 72, 35, 17, 8, 4, 2, 1, 1, 224, 148, 72, 35, 17, 8, 4, 2, 1, 1, 432, 303, 149, 72, 35, 17, 8, 4, 2, 1, 1, 833, 618, 308, 149, 72, 35, 17, 8, 4, 2, 1, 1, 1606, 1257, 636, 308, 149, 72, 35, 17, 8, 4, 2, 1
COMMENTS
This is a mirror image of the triangular array A140995. The current array has index of asymmetry s = 3 and index of obliqueness (obliquity) e = 0. Array A140995 has the same index of asymmetry, but has index of obliqueness e = 1. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.) In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0). Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), arrays A140997 and A140994 have s = 2 (with e = 0 and e = 1, respectively), and arrays A141020 and A141021 have s = 4 (with e = 0 and e = 1, respectively). (End)
FORMULA
G(n, k) = A140995(n, n - k) for 0 <= k <= n. Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (1 - x - x^2 - x^3 - x^4 + x^2*y + x^3*y + x^5*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^4 - x^4*y)).
If we take the first derivative of the bivariate g.f. w.r.t. y and set y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3 - x^4)). This is the g.f. of a shifted version of sequence A107066. Substituting y = 1 in the above bivariate function and simplifying, we get the g.f. of row sums: 1/(1 - 2*x). Hence, the row sums are powers of 2; i.e., A000079. (End)
EXAMPLE
Triangle (with rows n >= 0 and columns k >= 0) begins as follows:
1
1 1
1 2 1
1 4 2 1
1 8 4 2 1
1 16 8 4 2 1
1 31 17 8 4 2 1
1 60 35 17 8 4 2 1
1 116 72 35 17 8 4 2 1
1 224 148 72 35 17 8 4 2 1
1 432 303 149 72 35 17 8 4 2 1
1 833 618 308 149 72 35 17 8 4 2 1
...
MATHEMATICA
nlim = 100;
For[n = 0, n <= nlim, n++, G[n, 0] = 1];
For[n = 1, n <= nlim, n++, G[n, n] = 1];
For[n = 2, n <= nlim, n++, G[n, n-1] = 2];
For[n = 3, n <= nlim, n++, G[n, n-2] = 4];
For[n = 4, n <= nlim, n++, G[n, n-3] = 8];
For[n = 5, n <= nlim, n++, For[k = 1, k < n - 3, k++,
G[n, k] = G[n-4, k-1] + G[n-4, k] + G[n-3, k] + G[n-2, k] + G[n-1, k]]];
A140996 = {}; For[n = 0, n <= nlim, n++,
For[k = 0, k <= n, k++, AppendTo[ A140996, G[n, k]]]]; G[n_, k_] := G[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n - 1, 2, k == n - 2, 4, k == n - 3, 8, True, G[n - 1, k] + G[n - 2, k] + G[n - 3, k] + G[n - 4, k] + G[n - 4, k - 1]];
CROSSREFS
Cf. A007318, A107066, A140993, A140994, A140995, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A309462.
Triangle G(n, k) read by rows, for 0 <= k <= n, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, m) = G(n+1, m-3) + G(n+1, m-4) + G(n+2, m-3) + G(n+3, m-2) + G(n+4, m-1) for n >= 0 and m = 4..(n+4).
+10
22
1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 4, 8, 16, 1, 1, 2, 4, 8, 17, 31, 1, 1, 2, 4, 8, 17, 35, 60, 1, 1, 2, 4, 8, 17, 35, 72, 116, 1, 1, 2, 4, 8, 17, 35, 72, 148, 224, 1, 1, 2, 4, 8, 17, 35, 72, 149, 303, 432, 1, 1, 2, 4, 8, 17, 35, 72, 149, 308, 618, 833, 1, 1, 2, 4, 8, 17, 35, 72, 149, 308, 636, 1257, 1606, 1
COMMENTS
This is a mirror image of the triangular array A140996. The current array has index of asymmetry s = 3 and index of obliqueness (obliquity) e = 1. Array A140996 has the same index of asymmetry, but has index of obliqueness e = 0. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.) Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140997 and A140994 have s = 2 (with e = 0 and e = 1, respectively). If A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k is the bivariate g.f. for this array (with G(n, k) = 0 for 0 <= n < k) and B(x, y) = Sum_{n, k} A140996(n, k)*x^n*y^k, then A(x, y) = B(x*y, y^(-1)). This can be proved using formal manipulation of double series expansions and the fact G(n, k) = A140996(n, n-k) for 0 <= k <= n. If we let b(k) = lim_{n -> infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, b(3) = 8, and b(k) = b(k-1) + b(k-2) + 2*b(k-3) + b(k-4) for k >= 4. (The existence of the limit can be proved by induction on k.) Thus, the limiting sequence is 1, 2, 4, 8, 17, 35, 72, 149, 308, 636, 1314, 2715, 5609, 11588, 23941, 49462, 102188, 211120, 436173, ... (sequence A309462). (End)
FORMULA
G(n, k) = A140996(n, n-k) for 0 <= k <= n. Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (x^5*y^4 - x^4*y^4 - x^3*y^3 + x^3*y^2 - x^2*y^2 + x^2*y - x*y + 1)/((1- x*y) * (1- x) * (1 - x*y - x^2*y^2 -x^3*y^3 - x^4*y^4 - x^4*y^3)).
Substituting y = 1 in the above bivariate function and simplifying, we get the g.f. of row sums: 1/(1 - 2*x). Hence, the row sums are powers of 2; i.e., A000079. (End)
EXAMPLE
Triangle begins:
1
1 1
1 2 1
1 2 4 1
1 2 4 8 1
1 2 4 8 16 1
1 2 4 8 17 31 1
1 2 4 8 17 35 60 1
1 2 4 8 17 35 72 116 1
1 2 4 8 17 35 72 148 224 1
1 2 4 8 17 35 72 149 303 432 1
1 2 4 8 17 35 72 149 308 618 833 1
...
CROSSREFS
Cf. A000079, A007318, A140993, A140994, A140996, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A309462.
Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 1.
+10
20
1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 4, 8, 16, 1, 1, 2, 4, 8, 16, 32, 1, 1, 2, 4, 8, 16, 33, 63, 1, 1, 2, 4, 8, 16, 33, 67, 124, 1, 1, 2, 4, 8, 16, 33, 67, 136, 244, 1, 1, 2, 4, 8, 16, 33, 67, 136, 276, 480, 1
COMMENTS
The triangle here is A141020 with each row reversed. In the attached photograph, we see that the index of asymmetry is denoted by s (rather than y) and the index of obliqueness by e (rather than z).
The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.
For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have array A141020 (with e = 0) and the current array (with e = 1). In some of these arrays, the indices n and k are sometimes shifted. Putting k = 1 in Stepan's triangles with index of asymmetry s and index of obliqueness e = 0, we get G(n + s + 2, 1) = 1 + Sum_{1 <= m <= s+1} G(n+m, 1) for n >= 0 and k = 1..(n+1) with initial conditions G(x+1, 1) = 2^x for x = 0..s. Thus, we get a shifted version of column s+1 in array A172119. These sequences were first studied by Dunkel (1925). Thus, the second main diagonal of Stepan's triangles with index of asymmetry s and index of obliqueness e = 1 is equal to a shifted version of column s + 1 in array A172119. It follows from Eq. (20) on p. 360 in Dunkel (1925) that, for Stepan's triangles with index of asymmetry s and index of obliqueness e = 0, we have G(n, 1) = Sum_{t = 1..floor((n + s + 1)/(s + 2))} (-1)^(t + 1) * binomial(n + s - t*(s + 1), t - 1) * 2^(n + s - t*(s + 2) + 1) for n >= 0.
In a similar way, for Stepan's triangles with index of asymmetry s and index of obliqueness e = 1, we have G(n, n - 1) = Sum_{t = 1..floor((n + s + 1)/(s + 2))} (-1)^(t + 1) * binomial(n + s - t*(s + 1), t - 1) * 2^(n + s - t*(s + 2) + 1) for n >= 1.
Let A_s(x, y) be the bivariate g.f. of G(n, k) with index of asymmetry s and index of obliqueness e = 0 and let B_s(x, y) be the bivariate g.f. of the other G(n, k) with index of asymmetry s and index of obliqueness e = 1. Because the two triangular arrays are mirror images of each other, we have B_s(x, y) = A_s(x*y, y^(-1)).
(End)
FORMULA
Recurrence: G(n+6, k) = G(n+1, k-4) + G(n+1, k-5) + G(n+2, k-4) + G(n+3, k-3) + G(n+4, k-2) + G(n+5, k-1) for n >= 0 and k = 5..(n+5) with G(n+x+1, x) = 2^x for x = 0..4 and n >= 0.
Bivariate g.f.: Sum_{n,k >=0} T(n, k)*x^n*y^k = (x^6*y^5 - x^5*y^5 - x^4*y^4 + x^4*y^3 - x^3*y^3 + x^3*y^2 - x^2*y^2 + x^2*y - x*y + 1)/((1 - x*y) * (1 - x) * (1 - x*y - x^2*y^2 - x^3*y^3 - x^4*y^4 - x^5*y^4 - x^5*y^5)).
Second main diagonal: G(n, n - 1) = Sum_{t = 1..floor((n + 5)/6)} (-1)^(t + 1) * binomial(n + 4 - 5*t, t - 1) * 2^(n + 5 - 6*t) for n >= 1.
Limiting row: Let b(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then b(k) = b(k-5) + 2*b(k-4) + b(k-3) + b(k-2) + b(k-1) for k >= 5 with b(x) = 2^x for x = 0..4. This is the sequence 1, 2, 4, 8, 16, 33, 67, 136, 276, 561, 1140, 2316, 4705, 9559, 19421, 39457, 80163, 162864, 330885, 672247, ..., which is A308808. (End)
EXAMPLE
Pascal-like triangle with y = 4 and z = 1 (with rows n >= 0 and columns k >= 0) begins as follows:
1
1 1
1 2 1
1 2 4 1
1 2 4 8 1
1 2 4 8 16 1
1 2 4 8 16 32 1
1 2 4 8 16 33 63 1
1 2 4 8 16 33 67 124 1
1 2 4 8 16 33 67 136 244 1
1 2 4 8 16 33 67 136 276 480 1
1 2 4 8 16 33 67 136 276 560 944 1
...
MAPLE
A141020 := proc(n, k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; elif k=n-4 then 16 ; else procname(n-1, k) +procname(n-2, k)+procname(n-3, k)+procname(n-4, k) +procname(n-5, k)+procname(n-5, k-1) ; fi; end:
MATHEMATICA
t[n_, k_] := t[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n - 1, 2, k == n - 2, 4, k == n - 3, 8, k == n - 4, 16, True, t[n - 1, k] + t[n - 2, k] + t[n - 3, k] + t[n - 4, k] + t[n - 5, k] + t[n - 5, k - 1]];
T[n_, k_] := t[n, n - k];
CROSSREFS
Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141031, A172119, A308808.
List of different composites in Pascal-like triangles with index of asymmetry y = 2 and index of obliquity z = 0 or z = 1.
+10
17
4, 8, 9, 15, 28, 40, 52, 96, 88, 170, 177, 188, 326, 345, 189, 400, 600, 694, 406, 846, 1104, 1386, 871, 1779, 2031, 2751, 872, 1866, 3736, 6872, 7730, 10672, 4022, 8505, 12640, 15979, 20885, 4023, 8633, 18079, 23249, 32859, 40724, 42762, 67240, 18559, 39677, 78652, 80866, 153402
COMMENTS
For the Pascal-like triangle G(n, k) with index of asymmetry y = 2 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) for k = 1..(n+1). (This is array A140997.) For the Pascal-like triangle with index of asymmetry y = 1 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, k) = G(n+1, k-2) + G(n+1, k-3) + G(n+2, k-2) + G(n+3, k-1) for k = 3..(n+3). (This is array A140994.) The two triangular arrays A140997 and A140994, which are described above, are mirror images of each other. To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140997, the composites not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140994 instead. Finally, we mention that in the attached picture about the connection between Stepan's triangles and the Pascal triangle, the letter s is used to describe the index of asymmetry and the letter e is used to describe the index of obliqueness (instead of the letters y and z, respectively). The Pascal triangle A007318 has index of asymmetry s = y = 0 (and it does not matter whether we use e = 0 or e = 1 in the general formulas in the attached photograph). (End)
EXAMPLE
Pascal-like triangle with y = 2 and z = 0 (i.e., A140997) begins as follows: 1, so no composite.
1 1, so no composite.
1 2 1, so no composite.
1 4 2 1, so a(1) = 4.
1 8 4 2 1, so a(2) = 8.
1 15 9 4 2 1, so a(3) = 9 and a(4) = 15.
1 28 19 9 4 2 1, so a(5) = 28.
1 52 40 19 9 4 2 1, so a(6) = 40 and a(7) = 52.
1 96 83 41 19 9 4 2 1, so a(8) = 96.
1 177 170 88 41 19 9 4 2 1, so a(9) = 88, a(10) = 170, and a(11) = 177.
1 326 345 188 88 41 19 9 4 2 1, so a(12) = 188, a(13) = 326, and a(14) = 345.
1 600 694 400 189 88 41 19 9 4 2 1, so a(15) = 189, a(16) = 400, a(17) = 600, and a(18) = 694.
MAPLE
# Construction of array A140997 (y = 2 and z = 0): A140997 := proc(n, k) option remember; if k < 0 or n < k then 0; elif k = 0 or k = n then 1; elif k = n - 1 then 2; elif k = n - 2 then 4; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 3, k - 1); end if; end proc;
# Construction of the current sequence:
A141066 := proc(nmax) local a, b, n, k, new; a := []; for n from 0 to nmax do b := []; for k from 0 to n do new := A140997(n, k); if not (new = 1 or isprime(new) or new in a or new in b) then b := [op(b), new]; end if; end do; a := [op(a), op(sort(b))]; end do; RETURN(a); end proc;
# Generation of numbers in the current sequence:
# If one wishes to sort the numbers, then replace RETURN(a) with RETURN(sort(a)) in the above Maple code. In this case, however, the sequence is not uniquely defined because it depends on the maximum n. - Petros Hadjicostas, Jun 15 2019
CROSSREFS
Cf. A007318 (y = 0), A140993 (y = 1 and z = 1), A140994 (y = 2 and z = 1), A140995 (y = 3 and z = 1), A140996 (y = 3 and z = 0), A140997 (y = 2 and z = 0), A140998 (y = 1 and z = 0), A141020 (y = 4 and z = 0), A141021 (y = 4 and z = 1), A141064 (has primes when y = 1), A141065 (has composites when y = 1), A141067 (has primes when y = 2), A141068 (has primes when y = 3), A141069 (has composites when y = 3).
List of different primes in Pascal-like triangles with index of asymmetry y = 2 and index of obliquity z = 0 or z = 1.
+10
17
2, 19, 41, 83, 3719, 5431, 1873, 3989, 8641, 18517, 38303, 79153, 136963, 2264749, 394969, 1748039, 6633577, 14820521, 18051277, 3807953189, 126558214721, 2710968363511, 803233671719, 1723473449197, 1725438080929, 7942459030543, 145539180603829, 77442861984547
COMMENTS
For the Pascal-like triangle G(n, k) with index of asymmetry y = 2 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, and G(n+4, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) for n >= 0 and k = 1..(n+1). (This is array A140997.) For the Pascal-like triangle G(n, k) with index of asymmetry y=1 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, and G(n+4, k) = G(n+1, k-2) + G(n+1, k-3) + G(n+2, k-2) + G(n+3, k-1) for n >= 0 and k = 3..(n+3). (This is array A140994.) The two triangular arrays A140997 and A140994, which are described above, are mirror images of each other. To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140997, the primes not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140994 instead. Finally, we mention that in the attached picture about the connection between Stepan's triangles and the Pascal triangle, the letter s is used to describe the index of asymmetry and the letter e is used to describe the index of obliqueness (instead of the letters y and z, respectively). The Pascal triangle A007318 has index of asymmetry s = y = 0 (and it does not matter whether we use e = 0 or e = 1 in the general formulas in the attached photograph). (End)
EXAMPLE
Pascal-like triangle with y = 2 and z = 0 (i.e., A140997) begins as follows: 1, so no primes.
1 1, so no primes.
1 2 1, then a(1) = 2.
1 4 2 1, so no new primes.
1 8 4 2 1, so no new primes.
1 15 9 4 2 1, so no new primes.
1 28 19 9 4 2 1, so a(2) = 19.
1 52 40 19 9 4 2 1, so no new primes.
1 96 83 41 19 9 4 2 1, so a(3) = 41 and a(4) = 83.
1 177 170 88 41 19 9 4 2 1, so no new primes.
1 326 345 188 88 41 19 9 4 2 1, so no new primes.
1 600 694 400 189 88 41 19 9 4 2 1, so no new primes.
Terms a(5) = 3719 and a(6) = 5431 appear in row k = 14, while terms a(7) = 1873 and a(8) = 3989 appear in row k = 15.
MAPLE
# Construction of array A140997 (y = 2 and z = 0): A140997 := proc(n, k) option remember; if k < 0 or n < k then 0; elif k = 0 or k = n then 1; elif k = n - 1 then 2; elif k = n - 2 then 4; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 3, k - 1); end if; end proc;
# Construction of the current sequence:
A141067 := proc(nmax) local a, b, n, k, new; a := []; for n from 0 to nmax do b := []; for k from 0 to n do new := A140997(n, k); if not (new = 1 or not isprime(new) or new in a or new in b) then b := [op(b), new]; end if; end do; a := [op(a), op(sort(b))]; end do; RETURN(a); end proc;
# Generation of the current sequence:
# If one wishes to get the primes sorted, then he or she should replace RETURN(a) in the above Maple code with RETURN(sort(a)). In such a case, however, the sequence is not uniquely defined because it depends on the maximum n. - Petros Hadjicostas, Jun 15 2019
CROSSREFS
Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141031, A141066.
Search completed in 0.022 seconds
| 
