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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.
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%I #58 Sep 17 2016 12:16:30

%S 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 91,144,233,521,322,203,149,160,235,377,610,1364,843,525,352,309,395,

%U 612,987,1597,3571,2207,1368,877,661,704,1007,1599,2584,4181

%N 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 The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.

%C 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 The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).

%C 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 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 Yuriy Sibirmovsky, <a href="/A276472/a276472.jpg">Symmetrical hexagonal arrangement for initial terms of T(n,k)</a>

%H Yuriy Sibirmovsky, <a href="/A276472/a276472.png">T(n,k) compared with Pascal's triangle</a>

%H Yuriy Sibirmovsky, <a href="/A276472/a276472_1.png">Illustration for T(n,k) mod 3</a>

%F Conjectures:

%F Relations with other sequences:

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

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

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

%F T(n+1,n) = A001906(n) = A088305(n), n>=1.

%F T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].

%F Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].

%F Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.

%F T(2n,n+1) - T(2n,n) = A026671(n), n>=1.

%F T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.

%F T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.

%F T(2n+1,2n) = 3*A004187(n), n>=1.

%F T(2n+1,2) = 3*A049685(n-1), n>=1.

%F T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.

%F T(2n+1,3) = 5*A206351(n), n>=1.

%F T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.

%F T(2n,1) = 1 + 5*A081018(n-1), n>=1.

%F T(2n,2) = 2 + 5*A049684(n-1), n>=1.

%F T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.

%F T(2n,3) = 3 + 5*A081016(n-2), n>=2.

%F T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.

%F T(3n,1) = 4*A049629(n-1), n>=1.

%F T(3n,1) = 4 + 8*A119032(n), n>=1.

%F T(3n+1,3) = 8*A133273(n), n>=1.

%F T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.

%F T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.

%F T(3n+2,2) = 2 + 16*A049683(n), n>=1.

%F T(3n+2,2) = 2*A023039(n), n>=1.

%F T(2n-1,2n-1) = A033889(n-1), n>=1.

%F T(3n-1,3n-1) = 2*A007805(n-1), n>=1.

%F T(5n-1,1) = 11*A097842(n-1), n>=1.

%F T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.

%F T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.

%F Relations between left and right sides:

%F T(n,1) = T(n,n) - T(n-2,n-2), n>=3.

%F T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.

%F T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

%e Triangle T(n,k) begins:

%e n\k 1 2 3 4 5 6 7 8 9

%e 1 1

%e 2 1 2

%e 3 4 3 5

%e 4 11 7 8 13

%e 5 29 18 15 21 34

%e 6 76 47 33 36 55 89

%e 7 199 123 80 69 91 144 233

%e 8 521 322 203 149 160 235 377 610

%e 9 1364 843 525 352 309 395 612 987 1597

%e ...

%e In another format:

%e __________________1__________________

%e _______________1_____2_______________

%e ____________4_____3_____5____________

%e ________11_____7_____8_____13________

%e ____29_____18_____15____21_____34____

%e _76_____47____33_____36____55_____89_

%t Nm=12;

%t T=Table[0,{n,1,Nm},{k,1,n}];

%t T[[1,1]]=1;

%t T[[2,1]]=1;

%t T[[2,2]]=2;

%t Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];

%t T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];

%t If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];

%t {Row[#,"\t"]}&/@T//Grid

%o (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 tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ");); print()); \\ _Michel Marcus_, Sep 14 2016

%Y 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 nonn,tabl

%O 1,3

%A _Yuriy Sibirmovsky_, Sep 12 2016