cp's OEIS Frontend

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

%I A276472 #58 Sep 17 2016 12:16:30
%S A276472 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,
%T A276472 91,144,233,521,322,203,149,160,235,377,610,1364,843,525,352,309,395,
%U A276472 612,987,1597,3571,2207,1368,877,661,704,1007,1599,2584,4181
%N A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 1<k<n. T(n,1) = T(n-1,1) + T(n,2), n>2. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.
%C A276472 The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
%C A276472 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.
%C A276472 The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
%C A276472 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).
%H A276472 Yuriy Sibirmovsky, <a href="/A276472/b276472.txt">T(n,k), read by rows as a linear sequence a(j) for j = 1..5050</a>
%H A276472 Yuriy Sibirmovsky, <a href="/A276472/a276472.jpg">Symmetrical hexagonal arrangement for initial terms of T(n,k)</a>
%H A276472 Yuriy Sibirmovsky, <a href="/A276472/a276472.png">T(n,k) compared with Pascal's triangle</a>
%H A276472 Yuriy Sibirmovsky, <a href="/A276472/a276472_1.png">Illustration for T(n,k) mod 3</a>
%F A276472 Conjectures:
%F A276472 Relations with other sequences:
%F A276472 T(n+1,1) = A002878(n-1), n>=1.
%F A276472 T(n,n) = A001519(n) = A122367(n-1), n>=1.
%F A276472 T(n+1,2) = A005248(n-1), n>=1.
%F A276472 T(n+1,n) = A001906(n) = A088305(n), n>=1.
%F A276472 T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
%F A276472 Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
%F A276472 Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
%F A276472 T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
%F A276472 T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
%F A276472 T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
%F A276472 T(2n+1,2n) = 3*A004187(n), n>=1.
%F A276472 T(2n+1,2) = 3*A049685(n-1), n>=1.
%F A276472 T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
%F A276472 T(2n+1,3) = 5*A206351(n), n>=1.
%F A276472 T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
%F A276472 T(2n,1) = 1 + 5*A081018(n-1), n>=1.
%F A276472 T(2n,2) = 2 + 5*A049684(n-1), n>=1.
%F A276472 T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
%F A276472 T(2n,3) = 3 + 5*A081016(n-2), n>=2.
%F A276472 T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
%F A276472 T(3n,1) = 4*A049629(n-1), n>=1.
%F A276472 T(3n,1) = 4 + 8*A119032(n), n>=1.
%F A276472 T(3n+1,3) = 8*A133273(n), n>=1.
%F A276472 T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
%F A276472 T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
%F A276472 T(3n+2,2) = 2 + 16*A049683(n), n>=1.
%F A276472 T(3n+2,2) = 2*A023039(n), n>=1.
%F A276472 T(2n-1,2n-1) = A033889(n-1), n>=1.
%F A276472 T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
%F A276472 T(5n-1,1) = 11*A097842(n-1), n>=1.
%F A276472 T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
%F A276472 T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
%F A276472 Relations between left and right sides:
%F A276472 T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
%F A276472 T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
%F A276472 T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.
%e A276472 Triangle T(n,k) begins:
%e A276472 n\k 1    2    3    4   5    6    7    8    9
%e A276472 1   1
%e A276472 2   1    2
%e A276472 3   4    3    5
%e A276472 4   11   7    8    13
%e A276472 5   29   18   15   21   34
%e A276472 6   76   47   33   36   55   89
%e A276472 7   199  123  80   69   91   144 233
%e A276472 8   521  322  203  149  160  235 377  610
%e A276472 9   1364 843  525  352  309  395 612  987  1597
%e A276472 ...
%e A276472 In another format:
%e A276472 __________________1__________________
%e A276472 _______________1_____2_______________
%e A276472 ____________4_____3_____5____________
%e A276472 ________11_____7_____8_____13________
%e A276472 ____29_____18_____15____21_____34____
%e A276472 _76_____47____33_____36____55_____89_
%t A276472 Nm=12;
%t A276472 T=Table[0,{n,1,Nm},{k,1,n}];
%t A276472 T[[1,1]]=1;
%t A276472 T[[2,1]]=1;
%t A276472 T[[2,2]]=2;
%t A276472 Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
%t A276472 T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
%t A276472 If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
%t A276472 {Row[#,"\t"]}&/@T//Grid
%o A276472 (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)));
%o A276472 tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ");); print()); \\ _Michel Marcus_, Sep 14 2016
%Y A276472 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.
%K A276472 nonn,tabl
%O A276472 1,3
%A A276472 _Yuriy Sibirmovsky_, Sep 12 2016