Search: seq:1,1,1,2,2,1,6,7,3,1
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A135880
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Triangle P, read by rows, where column k of P^2 equals column 0 of P^(2k+2) such that column 0 of P^2 equals column 0 of P shift one place left, with P(0,0)=1.
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+30
23
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1, 1, 1, 2, 2, 1, 6, 7, 3, 1, 25, 34, 15, 4, 1, 138, 215, 99, 26, 5, 1, 970, 1698, 814, 216, 40, 6, 1, 8390, 16220, 8057, 2171, 400, 57, 7, 1, 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1, 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9, 1, 14563135
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OFFSET
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0,4
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COMMENTS
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Amazingly, column 0 (A135881) also equals column 0 of tables A135878 and A135879, both of which have unusual recurrences seemingly unrelated to this triangle.
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LINKS
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FORMULA
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Denote this triangle by P and define as follows.
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix Q = A135885 may be defined by
[Q]_k = [P^(2k+2)]_0, for k>=0, such that
(1) Q = P^2 and (2) [Q]_0 = [P]_0 shifted left.
Define the dual triangular matrix R = A135894 by
[R]_k = [P^(2k+1)]_0, for k>=0.
Then columns of P may be formed from powers of R:
[P]_k = [R^(k+1)]_0, for k>=0.
Further, columns of powers of P, Q and R satisfy:
[R^(j+1)]_k = [P^(2k+1)]_j,
[Q^(j+1)]_k = [P^(2k+2)]_j,
[Q^(j+1)]_k = [Q^(k+1)]_j,
[P^(2j+2)]_k = [P^(2k+2)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
R * [P]_k = [P]_{k+1},
Q * [Q]_k = [Q]_{k+1},
Q * [R]_k = [R]_{k+1},
P^2 * [Q]_k = [Q]_{k+1},
P^2 * [R]_k = [R]_{k+1}, for all k>=0.
Other identities include the matrix products:
P^-1*R (A135898) = P shifted right one column;
P*R^-1*P (A135899) = Q shifted down one row;
R^-1*Q (A135900) = R shifted down one row.
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EXAMPLE
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Triangle P begins:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1;
8390, 16220, 8057, 2171, 400, 57, 7, 1;
86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1;
1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9,
1;
14563135, 35219202, 19003467, 5352788, 1004176, 140908, 15855,
1504, 126, 10, 1;
where column k of P equals column 0 of R^(k+1) where R =
1;
2, 1;
6, 4, 1;
25, 20, 6, 1;
138, 126, 42, 8, 1;
970, 980, 351, 72, 10, 1;
8390, 9186, 3470, 748, 110, 12, 1;
86796, 101492, 39968, 8936, 1365, 156, 14, 1;
1049546, 1296934, 528306, 121532, 19090, 2250, 210, 16, 1; ...
where column k of Q equals column 0 of Q^(k+1) for k>=0;
thus column k of P^2 equals column 0 of P^(2k+2).
1;
1, 1;
2, 3, 1;
6, 12, 5, 1;
25, 63, 30, 7, 1;
138, 421, 220, 56, 9, 1;
970, 3472, 1945, 525, 90, 11, 1;
8390, 34380, 20340, 5733, 1026, 132, 13, 1;
86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1; ...
where column k of R equals column 0 of P^(2k+1) for k>=0.
Surprisingly, column 0 is also found in triangle A135879:
1;
1,1;
2,2,1,1;
6,6,4,4,2,2,1;
25,25,19,19,13,13,9,5,5,3,1,1;
138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ...
and is generated by a process that seems completely unrelated.
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PROG
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(PARI) {T(n, k)=local(P=Mat(1), R, PShR); if(n>0, for(n=0, n, PShR=matrix(#P, #P, r, c, if(r>=c, if(r==c, 1, if(c==1, 0, P[r-1, c-1])))); R=P*PShR; R=matrix(#P+1, #P+1, r, c, if(r>=c, if(r<#P+1, R[r, c], if(c==1, (P^2)[ #P, 1], (P^(2*c-1))[r-c+1, 1])))); P=matrix(#R, #R, r, c, if(r>=c, if(r<#R, P[r, c], (R^c)[r-c+1, 1]))))); P[n+1, k+1]}
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A343847
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T(n, k) = (n - k)! * [x^(n-k)] exp(k*x/(1 - x))/(1 - x). Triangle read by rows, T(n, k) for 0 <= k <= n.
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+30
2
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1, 1, 1, 2, 2, 1, 6, 7, 3, 1, 24, 34, 14, 4, 1, 120, 209, 86, 23, 5, 1, 720, 1546, 648, 168, 34, 6, 1, 5040, 13327, 5752, 1473, 286, 47, 7, 1, 40320, 130922, 58576, 14988, 2840, 446, 62, 8, 1, 362880, 1441729, 671568, 173007, 32344, 4929, 654, 79, 9, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,4
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LINKS
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FORMULA
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T(n, k) = (-1)^(n - k)*U(k - n, 1, -k), where U is the Kummer U function.
T(n, k) = (n - k)! * L(n - k, -k), where L is the Laguerre polynomial function.
T(n, k) = (n - k)! * Sum_{j = 0..n - k} binomial(n - k, j) k^j / j!.
T(n, k) = (2*n-k-1)*T(n-1, k) - (n-k-1)^2*T(n-2, k) for n - k >= 2.
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EXAMPLE
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Triangle starts:
0: 1;
1: 1, 1;
2: 2, 2, 1;
3: 6, 7, 3, 1;
4: 24, 34, 14, 4, 1;
5: 120, 209, 86, 23, 5, 1;
6: 720, 1546, 648, 168, 34, 6, 1;
7: 5040, 13327, 5752, 1473, 286, 47, 7, 1;
8: 40320, 130922, 58576, 14988, 2840, 446, 62, 8, 1;
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Array whose upward read antidiagonals are the rows of the triangle.
n\k 0 1 2 3 4 5 6
-----------------------------------------------------------------
0: 1, 1, 1, 1, 1, 1, 1, ...
1: 1, 2, 3, 4, 5, 6, 7, ...
2: 2, 7, 14, 23, 34, 47, 62, ...
3: 6, 34, 86, 168, 286, 446, 654, ...
4: 24, 209, 648, 1473, 2840, 4929, 7944, ...
5: 120, 1546, 5752, 14988, 32344, 61870, 108696, ...
6: 720, 13327, 58576, 173007, 414160, 866695, 1649232, ...
7: 5040, 130922, 671568, 2228544, 5876336, 13373190, 27422352, ...
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MAPLE
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T := proc(n, k) option remember;
if n = k then return 1 elif n = k+1 then return k+1 fi;
(2*n-k-1)*T(n-1, k) - (n-k-1)^2*T(n-2, k) end:
seq(print(seq(T(n , k), k = 0..n)), n = 0..7);
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MATHEMATICA
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T[n_, k_] := (-1)^(n - k) HypergeometricU[k - n, 1, -k];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
(* Alternative: *)
TL[n_, k_] := (n - k)! LaguerreL[n - k, -k];
Table[TL[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
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PROG
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(PARI)
T(n, k) = (n - k)!*sum(j=0, n - k, binomial(n - k, j) * k^j / j!)
for(n=0, 9, for(k=0, n, print(T(n, k))))
(SageMath) # Columns of the array.
def column(k, len):
R.<x> = PowerSeriesRing(QQ, default_prec=len)
f = exp(k * x / (1 - x)) / (1 - x)
return f.egf_to_ogf().list()
for col in (0..6): print(column(col, 20))
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CROSSREFS
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Columns: A000142, A002720, A087912, A277382, A289147, A289211, A289212, A289213, A289214, A289215, A289216.
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KEYWORD
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AUTHOR
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STATUS
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approved
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