A072405 -id:A072405 - cp's OEIS frontend

A122218 Pascal array A(n,p,k) for selection of k elements from two sets L and U with n elements in total whereat the nl = n - p elements in L are labeled and the nu = p elements in U are unlabeled and (in this example) with p = 2 (read by rows).

0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 1, 1, 4, 7, 7, 4, 1, 1, 5, 11, 14, 11, 5, 1, 1, 6, 16, 25, 25, 16, 6, 1, 1, 7, 22, 41, 50, 41, 22, 7, 1, 1, 8, 29, 63, 91, 91, 63, 29, 8, 1, 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1

Offset: 0

Thomas Wieder, Aug 27 2006

See Maple's command choose applied to lists (not sets): e.g., with p = 3 choose([1,2,a,a],3) gives [[1, 2, a], [1, a, a], [2, a, a]]. Furthermore nops(choose([1,2,a,a],3)) gives 3. For p = 0 one gets the usual Pascal triangle. For p=2 and k=2 we have the sequence 1,2,4,7,11,16,22,29,37,... = A000124 = Central polygonal numbers. For p=2 and k=3 we have the sequence 3,7,14,25,41,63,92,... = A004006 = C(n,1)+C(n,2)+C(n,3), or n*(n^2+5)/6. For p=2 the row sums form sequence A007283 = 3*2^n.
This is a triangular array like the Pascal triangle. [I.e., T(n+1,k) = T(n,k-1) +  T(n,k) for n <> 1, cf. formulas. - M. F. Hasler, Jan 06 2024]
It appears that for n > 2, a(n) = A072405(n) (C(n,k)-C(n-2,k-1)). - Gerald McGarvey, Sep 30 2008

			From _M. F. Hasler_, Jan 06 2024: (Start)
The triangle T(n,k) := A(n,2,k) starts:
  n |row(n) = (A(n,2,0), ..., A(n,2,n))
----+------------------------------------
  0 | 0,
  1 | 0, 0,
  2 | 1, 1,  1,
  3 | 1, 2,  2,  1,
  4 | 1, 3,  4,  3,   1,
  5 | 1, 4,  7,  7,   4,   1,
  6 | 1, 5, 11, 14,  11,   5,   1
  7 | 1, 6, 16, 25,  25,  16,   6,  1,
  8 | 1, 7, 22, 41,  50,  41,  22,  7,  1,
  9 | 1, 8, 29, 63,  91,  91,  63, 29,  8, 1,
  10| 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1
(End)
For n = 4 and p = 2 we have nl = 2, nu = 2 and we have the sets L = {1,2} and U = {a,a}, or L+U = {1,2,a,a}.
Then for k = 1 we have A(4,2,1) = 3 because we can select {1}, {2}, {a}.
Then for k = 2 we have A(4,2,2) = 4 because we can select {1,2}, {1,a}, {2,a}, {a,a}.
Then for k = 3 we have A(4,2,3) = 3 because we can select {1,2,a}, {1,a,a,}, {2,a,a}.
Then for k = 4 we have A(4,2,4) = 1 because we can select {1,2,a,a}.
For n = 4 and p = 3 we have nl = 1, nu = 3 and we have the sets L = {1} and U = {a,a,a}, or L+U = {1,a,a,a}.
Then for k = 1 we have A(4,3,1) = 2 because we can select {1}, {a}.
Then for k = 2 we have A(4,3,2) = 2 because we can select {1,a}, {a,a}.
Then for k = 3 we have A(4,3,3) = 2 because we can select {1,a,a}, {a,a,a,}.
Then for k = 4 we have A(4,3,4) = 1 because we can select {1,a,a,a}.

H. S. Wilf, Lecture notes on combinatorics and Maple.

Cf. A007318, A000124, A004006, A007283.

Cf. A072405 (essentially the same).

CallPascalLU := proc() local n,p,k,nl,nv; global result,ierr;
for n from 0 to 10 do p:=2; nl:=n-p; nv:=p; for k from 0 to n do PascalLU(n,nl,nv,k,result,ierr); if ierr <> 0 then print("An error has occured!"); fi; print("CallPascalLU: n, p, k, C(n,p,k):",n,p,k,result); end do; end do; end proc;
PascalLU := proc(n::integer,nl::integer,nv::integer,k::integer)
local i,l,u,prttn,prttnlst,swap;
global result,ierr;
ierr:=0;
if nl+nv <> n or k > n or n < 0 or k < 0 then ierr=1; return; fi;
prttnlst:=NULL;
result:=0;
if k>=2 then
prttnlst:=PartitionList(k,2);
prttnlst:=op(prttnlst);
end if;
prttnlst:=prttnlst,[k,0];
prttnlst:=[prttnlst];
#print("PascalLU: n, k, prttnlst:",n,k,prttnlst);
for i from 1 to nops(prttnlst) do
prttn:=op(i,prttnlst);
l:=op(1,prttn);
u:=op(2,prttn);
#print("PascalLU: i, prttn, l, u:",i,prttn,l,u);
if l <= nl and u <= nv then
result:=result+binomial(nl,l)*binomial(u,u);
end if;
swap:=u; u:=l; l:=swap;
if l <> u and l <= nl and u <= nv then
result:=result+binomial(nl,l)*binomial(u,u);
end if;
end do;
#print("n,k,result",n,k,summe)
end proc;
PartitionList := proc (n, k)
# Herbert S. Wilf and Joanna Nordlicht,
# Lecture Notes "East Side West Side,..."
# Available from Wilf link.
# Calculates the partitions of n into k parts.
# E.g. PartitionList(5,2) --> [[4, 1], [3, 2]].
local East, West;
if n < 1 or k < 1 or n < k then
RETURN([])
elif n = 1 then
RETURN([[1]])
else if n < 2 or k < 2 or n < k then
West := []
else
West := map(proc (x) options operator, arrow;
[op(x), 1] end proc,PartitionList(n-1,k-1)) end if;
if k <= n-k then
East := map(proc (y) options operator, arrow;
map(proc (x) options operator, arrow; x+1 end proc,y) end proc,PartitionList(n-k,k))
else East := [] end if;
RETURN([op(West), op(East)])
end if;
end proc;

A122218(n,k) = if(n>1, binomial(n,k)-binomial(n-2,k-1), 0) \\ M. F. Hasler, Jan 06 2024

Let Sum_{l+u = k, l <= nl, u <= nu} denote the sum over all integer partitions [l,u] of k into the 2 parts l and u with the following properties:
1.) l <= nl, u <= nu,
2.) [l,u] and [u,l] are considered as two different partitions,
3.) but the partition [l=k/2, u=k/2], i.e., if l=u, is taken only once,
4.) [l=k,0] and [0, u=k] are considered to be partitions of k into 2 parts also. As usual, C(nl,l) and C(u,u) are binomial coefficients ("nl choose l" and "u choose u"). The Pascal array A(nl,l,nu,u,k) = A(n,p,k) gives the number of possible sets which can be taken from L and U (with elements either from both sets L and U or just from one of the sets L or U). Then A(n,p,k) = Sum_{l+u=k, l<=nl, u<=nu} C(n-p,l,k) C(u,u).
From M. F. Hasler, Jan 06 2024: (Start)
T(n,k) = A(n,2,k) = C(n,k) - C(n-2,k-1) except for (n,k) = (0,0) and (1,0).
Pascal-type triangle: T(n+1,k) = T(n,k-1)+ T(n,k) for all n <> 1, with T(n,k) = 0 for k < 0 or k > n. (End)

A199881 Triangle T(n,k), read by rows, given by (1,-1,0,0,0,0,0,0,0,0,0,...) DELTA (1,0,-1,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 3, 1, 0, 0, 1, 4, 4, 1, 0, 0, 0, 3, 7, 5, 1, 0, 0, 0, 1, 7, 11, 6, 1, 0, 0, 0, 0, 4, 14, 16, 7, 1, 0, 0, 0, 0, 1, 11, 25, 22, 8, 1, 0, 0, 0, 0, 0, 5, 25, 41, 29, 9, 1, 0, 0, 0, 0, 0, 1, 16, 50, 63, 37, 10, 1

Offset: 0

Philippe Deléham, Nov 11 2011

The nonzero entries of column k give row k+1 in A072405.

			Triangle begins:
  1;
  1, 1;
  0, 1, 1;
  0, 1, 2, 1; (key row for starting the recurrence)
  0, 0, 2, 3, 1;
  0, 0, 1, 4, 4,  1;
  0, 0, 0, 3, 7,  5,  1;
  0, 0, 0, 1, 7, 11,  6, 1;
  0, 0, 0, 0, 4, 14, 16, 7, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000931 (diagonal sums), A042950 (column sums), A055389 (row sums).

Cf. A000045, A028310, A072405, A084938, A104402.

T[n_, k_]:= T[n, k]= If[n<2, 1, If[k==0, 0, If[k==n, 1, If[n==2 && k==1, 1, T[n-1, k-1] +T[n-2, k-1] ]]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 28 2021 *)

def T(n,k): return binomial(k, n-k) + binomial(k+1, n-k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2021

T(n,k) = T(n-1,k-1) + T(n-2,k-1) starting with T(0,0) = T(1,0) = T(1,1) = T(2,1) = T(2,2) = 1 and T(2,0) = 0.
G.f.: (1+x-y*x^2)/(1-y*x-y*x^2).
T(2n,n) = A028310(n).
From G. C. Greubel, Apr 28 2021: (Start)
T(n, k) = binomial(k, n-k) + binomial(k+1, n-k-1).
T(n, k) = (-1)^(n-k)*A104402(n, k). (End)
From G. C. Greubel, Apr 30 2021: (Start)
Sum_{k=0..n} T(n, k) = 2*Fibonacci(n) + [n=0].
Sum_{n=k..2*k+1} T(n,k) = 3*2^(n-1) + (1/2)*[n=0]. (End)

A072406 Number of values of k for which C(n,k)-C(n-2,k-1) is odd.

Original entry on oeis.org

1, 2, 3, 2, 4, 4, 6, 4, 6, 8, 6, 4, 8, 8, 12, 8, 10, 16, 6, 4, 8, 8, 12, 8, 12, 16, 12, 8, 16, 16, 24, 16, 18, 32, 6, 4, 8, 8, 12, 8, 12, 16, 12, 8, 16, 16, 24, 16, 20, 32, 12, 8, 16, 16, 24, 16, 24, 32, 24, 16, 32, 32, 48, 32, 34, 64, 6, 4, 8, 8, 12, 8, 12, 16, 12, 8, 16, 16, 24, 16

Offset: 0

Henry Bottomley, Jun 16 2002

nonn

			a(5)=4 since the values of C(5,k)-C(3,k-1) are 1-0, 5-1, 10-3, 10-3, 5-1, 1-0, i.e. 1,4,7,7,4,1 of which 4 are odd.

Cf. A001316, A072405.

Cf. A094959.

Join[{1,2},Table[Count[Table[Binomial[n,k]-Binomial[n-2,k-1],{k,0,n}],?OddQ],{n,2,80}]] (* _Harvey P. Dale, Mar 03 2024 *)

If 2*2^m+1

A096646 Triangle (read by rows) where the number of row entries increases by steps of 2 and the entries are stacked in a rectangular fashion. The end entries = 1. Rest of entries in the n-th row are the sum of the entries directly above and to the left and right in all previous rows (total of 3*(n-1) entries).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 4, 3, 1, 1, 5, 11, 14, 11, 5, 1, 1, 7, 22, 41, 50, 41, 22, 7, 1, 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1, 1, 11, 56, 175, 375, 582, 672, 582, 375, 175, 56, 11, 1

Offset: 1

Gerald McGarvey, Aug 14 2004

The row sums are 1,3, then 2^(2*(n-2)) * 3. (I.e., A002001 a(n) = 3*4^(n-1), n>0; a(0)=1.) The n-th row is the (2n-1)st row of A072405 (Triangle of C(n,k)-C(n-2,k-1)).

			......................1....................
..................1...1...1................
..............1...3...4...3...1............
..........1...5..11..14..11...5...1........
......1...7..22..41..50..41..22...7..1.....
...1..9..37..92.154.182.154..92..37..9..1..
1.11.56.175.375.582.672.582.375.175.56.11.1

Cf. A002001, A072405.

G.f.: 1/[(1-z(1+w+w^2))(1-wz)]. Partial sums of trinomial array A027907. - Ralf Stephan, Jan 09 2005

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A122218 Pascal array A(n,p,k) for selection of k elements from two sets L and U with n elements in total whereat the nl = n - p elements in L are labeled and the nu = p elements in U are unlabeled and (in this example) with p = 2 (read by rows).

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A199881 Triangle T(n,k), read by rows, given by (1,-1,0,0,0,0,0,0,0,0,0,...) DELTA (1,0,-1,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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A072406 Number of values of k for which C(n,k)-C(n-2,k-1) is odd.

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