A167560 The ED2 array read by ascending antidiagonals.
1, 2, 1, 6, 4, 1, 24, 16, 6, 1, 120, 80, 32, 8, 1, 720, 480, 192, 54, 10, 1, 5040, 3360, 1344, 384, 82, 12, 1, 40320, 26880, 10752, 3072, 680, 116, 14, 1, 362880, 241920, 96768, 27648, 6144, 1104, 156, 16, 1
Offset: 1
Examples
The ED2 array begins with: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 6, 16, 32, 54, 82, 116, 156, 202, 254, 312 24, 80, 192, 384, 680, 1104, 1680, 2432, 3384, 4560 120, 480, 1344, 3072, 6144, 11160, 18840, 30024, 45672, 66864 720, 3360, 10752, 27648, 61440, 122880, 226800, 392832, 646128, 1018080
Links
- Johannes W. Meijer, The four Escher-Droste arrays, Mar 08 2013.
Crossrefs
A000012, A005843 (n>=1), 2*A104249 (n>=1), A167561, A167562 and A167563 equal the first sixth rows of the array.
A000142 equals the first column of the array.
A047053 equals the a(n, n) diagonal of the array.
2*A034177 equals the a(n+1, n) diagonal of the array.
A167570 equals the a(n+2, n) diagonal of the array,
A167564 equals the row sums of the ED2 array read by antidiagonals.
A167565 is a triangle related to the a(n) formulas of the rows of the ED2 array.
A167568 is a triangle related to the GF(z) formulas of the rows of the ED2 array.
A167569 is the lower left triangle of the ED2 array.
Cf. A162005 (EG1 triangle).
Programs
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Maple
nmax:=10; mmax:=10; for n from 1 to nmax do for m from 1 to n do a(n,m) := 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! od; for m from n+1 to mmax do a(n,m):= n! + sum((-1)^(k-1)*binomial(n-1,k)*a(n,m-k),k=1..n-1) od; od: for n from 1 to nmax do for m from 1 to n do d(n,m):=a(n-m+1,m) od: od: T:=1: for n from 1 to nmax do for m from 1 to n do a(T):= d(n,m): T:=T+1: od: od: seq(a(n),n=1..T-1); # alternative A167560 := proc(n,m) option remember ; if m > n then n!+add( (-1)^(k-1)*binomial(n-1,k)*procname(n,m-k),k=1..n-1) ; else 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! ; end if; end proc: seq( seq(A167560(d-m,m),m=1..d-1),d=2..12) ; # R. J. Mathar, Jun 28 2024
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Mathematica
nmax = 10; mmax = 10; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[n, m] = 4^(m - 1)*(m - 1)!*((n + m - 1)!/(2*m - 1)!)]; For[m = n + 1, m <= mmax, m++, a[n, m] = n! + Sum[(-1)^(k - 1)*Binomial[n - 1, k]*a[n, m - k], {k, 1, n - 1}]]; ]; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, d[n, m] = a[n - m + 1, m]]; ]; t = 1; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[t] = d[n, m]; t = t + 1]]; Table[a[n], {n, 1, t - 1}] (* Jean-François Alcover, Dec 20 2011, translated from Maple *)
Formula
a(n,m) = ((m-1)!/((m-n-1)!))*int(sinh(y*(2*n))/(cosh(y))^(2*m),y=0..infinity) for m>n.
The (n-1)-differences of the n-th array row lead to the recurrence relation
sum((-1)^k*binomial(n-1,k)*a(n-1,m-k),k=0..n-1) = n!
which in its turn leads to, see A167569,
a(n,m) = 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! if m<=n.
Comments