OFFSET
1,3
COMMENTS
The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).
LINKS
Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3
FORMULA
Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.
EXAMPLE
Triangle T(n,k) begins:
n\k 1 2 3 4 5 6 7 8 9
1 1
2 1 2
3 4 3 5
4 11 7 8 13
5 29 18 15 21 34
6 76 47 33 36 55 89
7 199 123 80 69 91 144 233
8 521 322 203 149 160 235 377 610
9 1364 843 525 352 309 395 612 987 1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_
MATHEMATICA
Nm=12;
T=Table[0, {n, 1, Nm}, {k, 1, n}];
T[[1, 1]]=1;
T[[2, 1]]=1;
T[[2, 2]]=2;
Do[T[[n, 1]]=T[[n-1, 1]]+T[[n, 2]];
T[[n, n]]=T[[n-1, n-1]]+T[[n, n-1]];
If[k!=1&&k!=n, T[[n, k]]=T[[n-1, k]]+T[[n-1, k-1]]], {n, 3, Nm}, {k, 1, n}];
{Row[#, "\t"]}&/@T//Grid
PROG
(PARI) T(n, k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1, 1) + T(n, 2))), if (k<n, T(n-1, k) + T(n-1, k-1), if (k==n, T(n, n-1) + T(n-1, n-1), 0)));
tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", "); ); print()); \\ Michel Marcus, Sep 14 2016
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Yuriy Sibirmovsky, Sep 12 2016
STATUS
approved