This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A295555 #36 Aug 19 2022 23:27:24
%S A295555 1,1,1,1,1,1,1,1,1,1,1,2,2,2,1,1,2,3,3,2,1,1,2,3,4,3,2,1,1,3,5,7,7,5,
%T A295555 3,1,1,3,6,9,10,9,6,3,1,1,3,6,10,12,12,10,6,3,1,1,4,9,16,22,24,22,16,
%U A295555 9,4,1,1,4,10,19,28,34,34,28,19,10,4,1,1,4,10,20,31,40,44,40,31,20,10,4,1
%N A295555 Generalized Pascal triangle read by rows: add the four terms that are right above you, three rows back.
%C A295555 To explain the rule of formation, consider the first few rows of the triangle, which are:
%C A295555 1
%C A295555 1, 1
%C A295555 1, 1, 1
%C A295555 1, 1, 1, 1
%C A295555 A, B, C, D, 1
%C A295555 1, 2, 3, 3, 2, 1
%C A295555 1, 2, 3, 4, 3, 2, 1
%C A295555 1, 3, 5, E, 7, 5, 3, 1
%C A295555 The left and right edge are all 1's, the entries outside the triangle are all 0's, and the first 3 rows are all 1's.
%C A295555 Thereafter each term E (say) is the sum of the 4 terms A, B, C, D exactly above it three rows back.
%C A295555 Adding the two entries just above you in the previous row gives Pascal's triangle, A007318. Adding the three entries just above you two rows back gives A169623.
%C A295555 From _Peter Bala_, Aug 19 2021: (Start)
%C A295555 Let M denote the lower unit triangular array with the sequence [1,0,0,1,0,0,1,...] on all the subdiagonals. For k = 0,1,2,..., define M(k) to be the lower unit triangular block array
%C A295555 /I_k 0\
%C A295555 \ 0 M/
%C A295555 having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section below. The proof uses the hockey-stick identities from the Formula section. (End)
%H A295555 Robert Israel, <a href="/A295555/b295555.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 141, flattened)
%H A295555 Wikipedia, <a href="https://en.wikipedia.org/wiki/Hockey-stick_identity">Hockey-stick identity</a>.
%F A295555 T(n,0)=T(n,n)=1, T(n,k)=0 if k<0 or k>n, also T(2,1)=1; thereafter T(n,k) = T(n-3,k-3) + T(n-3,k-2) + T(n-3,k-1) + T(n-3,k).
%F A295555 From _Peter Bala_, Aug 19 2021: (Start)
%F A295555 T(3*n,k) = T(3*n-2,k) + T(3*n-2,k-2).
%F A295555 T(3*n+1,k) = T(3*n,k) + T(3*n,k-1).
%F A295555 T(3*n+2,k) = T(3*n+1,k-1) + T(3*n,k).
%F A295555 Hockey-stick identities (relate row k entries to entries in row k-1):
%F A295555 T(3*n,k) = T(3*n-1,k-1) + T(3*n-4,k-1) + T(3*n-7,k-1) + ....
%F A295555 T(3*n+1,k) = T(3*n,k-1) + ( T(3*n-1,k-1) + T(3*n-4,k-1) + T(3*n-7,k-1) + ... ).
%F A295555 T(3*n+2,k) = T(3*n+1,k-1) + ( T(3*n-1,k-1) + T(3*n-4,k-1) + T(3*n-7,k-1) + ... ).
%F A295555 Row polynomials:
%F A295555 R(3*n,x) = R(3,x)^n = (1 + x + x^2 + x^3)^n.
%F A295555 R(3*n+1,x) = R(1,x)*R(3,x)^n = (1 + x)*(1 + x + x^2 + x^3)^n.
%F A295555 R(3*n+2,x) = R(2,x)*R(3,x)^n = (1 + x + x^2)*(1 + x + x^2 + x^3)^n. (End)
%e A295555 Triangle begins:
%e A295555 1
%e A295555 1, 1
%e A295555 1, 1, 1
%e A295555 1, 1, 1, 1
%e A295555 1, 2, 2, 2, 1
%e A295555 1, 2, 3, 3, 2, 1
%e A295555 1, 2, 3, 4, 3, 2, 1
%e A295555 1, 3, 5, 7, 7, 5, 3, 1
%e A295555 1, 3, 6, 9, 10, 9, 6, 3, 1
%e A295555 1, 3, 6, 10, 12, 12, 10, 6, 3, 1
%e A295555 ...
%e A295555 From _Peter Bala_, Aug 19 2021: (Start)
%e A295555 With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
%e A295555 /1 \/1 \/1 \ /1 \
%e A295555 |1 1 ||0 1 ||0 1 | |1 1 |
%e A295555 |1 0 1 ||0 1 1 ||0 0 1 | |1 1 1 |
%e A295555 |1 0 0 1 ||0 1 0 1 ||0 0 1 1 |... = |1 1 1 1 |
%e A295555 |1 0 0 1 1 ||0 1 0 0 1 ||0 0 1 0 1 | |1 2 2 2 1 |
%e A295555 |1 0 0 1 0 1 ||0 1 0 0 1 1 ||0 0 1 0 0 1 | |... |
%e A295555 |1 0 0 1 0 0 1||0 1 0 0 1 0 1||0 0 1 0 0 1 1|
%e A295555 |... ||... ||... |
%e A295555 (End)
%p A295555 T:=proc(n,k) option remember;
%p A295555 if n >= 0 and k = 0 then 1
%p A295555 elif n >= 0 and k = n then 1
%p A295555 elif (k < 0 or k > n) then 0
%p A295555 elif n=2 then 1
%p A295555 else T(n-3,k-3)+T(n-3,k-2)+T(n-3,k-1)+T(n-3,k);
%p A295555 fi;
%p A295555 end;
%p A295555 for n from 0 to 14 do lprint([seq(T(n,k),k=0..n)]); od:
%t A295555 T[n_, k_] := T[n, k] = Which[
%t A295555 n >= 0 && k == 0, 1,
%t A295555 n >= 0 && k == n, 1,
%t A295555 k < 0 || k > n, 0,
%t A295555 n == 2, 1,
%t A295555 True, T[n-3, k-3] + T[n-3, k-2] + T[n-3, k-1] + T[n-3, k]];
%t A295555 Table[T[n, k], {n, 0, 14}, { k, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 19 2022, after Maple code *)
%Y A295555 Cf. A007318, A169623.
%Y A295555 Row sums are A133464.
%K A295555 nonn,tabl
%O A295555 0,12
%A A295555 _N. J. A. Sloane_, Nov 23 2017