A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Author

Yuriy Sibirmovsky, Sep 12 2016

Keywords

• nonn
• tabl

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).

Examples

			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_
		

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

Crossrefs

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Programs

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

PARI

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Formula

Conjectures:

Relations with other sequences:

T(n+1,1) = A002878(n-1), n>=1.

T(n,n) = A001519(n) = A122367(n-1), n>=1.

T(n+1,2) = A005248(n-1), n>=1.

T(n+1,n) = A001906(n) = A088305(n), 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.