reviewed
approved
reviewed
approved
proposed
reviewed
editing
proposed
allocated for Gus WisemanNumber of knapsack factorizations whose factors sum to n.
1, 1, 1, 2, 2, 4, 4, 6, 8, 11, 12, 19, 21, 27, 34, 45, 51, 69, 77, 100, 117, 146
1,4
A knapsack factorization is a finite multiset of positive integers greater than one such that every distinct submultiset has a different product.
The a(12) = 19 partitions are:
(12),
(10 2), (9 3), (8 4), (7 5), (6 6),
(8 2 2), (7 3 2), (6 4 2), (6 3 3), (5 5 2), (5 4 3), (4 4 4),
(6 2 2 2), (5 3 2 2), (4 3 3 2), (3 3 3 3),
(3 3 2 2 2),
(2 2 2 2 2 2).
nn=22;
apsQ[y_]:=UnsameQ@@Times@@@Union[Rest@Subsets[y]];
Table[Length@Select[IntegerPartitions[n], apsQ], {n, nn}]
allocated
nonn,more
Gus Wiseman, Oct 23 2017
approved
editing
allocated for Gus Wiseman
recycled
allocated
proposed
approved
editing
proposed
Take an integer k and sum the numbers obtained deleting only one of the digit of k, repeating the process for any digit of k. Sequence lists numbers such that the sum is equal to the reverse of k.
10, 20, 30, 40, 50, 60, 70, 80, 90, 531, 27681, 67743, 70692, 4812363, 5741514, 34999992, 36600123, 63690684, 65523315, 94444650, 261400662, 315000072, 385601193, 906400737
1,1
With 70692 we have 692 + 7692 + 7092 + 7062 + 7069 = 29607 that is the reverse of 70692;
With 36600123 we have 6600123 + 3600123 + 3600123 + 3660123 + 3660123 + 3660023 + 3660013 + 3660012 = 32100663 that is the reverse of 36600123.
T:=proc(w) local x, y, z; x:=0; y:=w; for z from 1 to ilog10(w)+1 do
x:=10*x+(y mod 10); y:=trunc(y/10); od; x; end:
P:=proc(q) local c, d, k, n; for n from 1 to q do d:=ilog10(n)+1; c:=0;
for k from 0 to d-1 do c:=n mod 10^k+trunc(n/10^(k+1))*10^k+c; od;
if c=T(n) then print(n); fi; od; end: P(10^9);
nonn,base,more,changed
recycled
Paolo P. Lava, Oct 13 2017
proposed
editing
editing
proposed