A157407 The integer partitions of n taken as digits in base n+1 and listed in the reflected Hindenburg order.

0, 1, 4, 2, 21, 6, 3, 156, 32, 12, 8, 4, 1555, 260, 50, 45, 15, 10, 5, 19608, 2802, 408, 114, 402, 66, 24, 60, 18, 12, 6, 299593, 37450, 4690, 658, 4683, 595, 147, 91, 588, 84, 28, 77, 21, 14, 7

Offset: 0

Author

Peter Luschny, Mar 11 2009

Keywords

• easy
• nonn
• tabf
• base

Comments

The rows are enumerated 0,1,2,... Converting the numbers in the n-th row (n>0) to base n+1 gives all partitions of n in the 'reflected Hindenburg order'. The term 'reflected Hindenburg order' is not standard and refers to the partition generating algorithm of C. F. Hindenburg (1779).

The offset of row n (n>0) is A000070[n+1], the length of row n is A000041[n]. The left hand side of the triangle 0,1,4,21,156,... is A060072.

Examples

			[0] <-> [[ ]]
[1] <-> [[1]]
[4,2] <-> [[1,1],[2]]
[21,6,3] <-> [[1,1,1],[2,1],[3]]
[156,32,12,8,4] <-> [[1,1,1,1],[2,1,1],[2,2],[3,1],[4]]
		

Links

• Peter Luschny, Counting with Partitions.

Crossrefs

Cf. A157406

Programs

Maple

a := proc(n) local rev,P,R,Q,i,l,s,k,j;
rev := l -> [seq(l[nops(l)-j+1],j=1..nops(l))];
P := combinat[partition](n); R := NULL;
for i to nops(P) do Q := rev(P[i]);
l := convert(Q,base,n+1,10);
s := add(l[k]*10^(k-1), k=1..nops(l));
R:= R,s; od; R end: [0,seq(a(i),i=1..7)];