[go: up one dir, main page]

login
A323345
Square array read by ascending antidiagonals: T(n, k) is the number of partitions of n where parts, if sorted in ascending order, form an arithmetic progression (AP) with common difference of k; n >= 1, k >= 0.
6
1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 1, 4, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 4, 2, 2, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 6, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
OFFSET
1,2
COMMENTS
T(n, k) is the number of positive integers in the sequence defined, for all i >= 1, by x_1 = n and x_i = (i-1)*(x_(i-1)-k)/i; or defined equivalently by x_i=n/i-(k/2)*(i-1). An x_i positive and integer characterizes the AP-partition with smallest part x_i and number of parts i.
T(n, k) is the number of i, positive integers, such that n - P(k+2,i) is both nonnegative and divisible by i, where P(r,i) denotes the i-th r-gonal number (see A057145).
FORMULA
T(n, 0) = A000005(n), the number of divisors of n.
T(n, 1) = A001227(n), the number of odd divisors of n.
T(n, 2) = A038548(n), the number of divisors of n that are at most sqrt(n).
T(n, 3) = A117277(n).
The g.f. for column d is Sum_{k>=1} x^(k*(d*k-d+2)/2)/(1-x^k) [information taken from A117277]. - Joerg Arndt, May 05 2020
EXAMPLE
There are 4 partitions of 150 such that the parts form an arithmetic progression with common difference of 9:
150 = 150
150 = 41 + 50 + 59
150 = 24 + 33 + 42 + 51
150 = 12 + 21 + 30 + 39 + 48
Then, T(150,9) = 4.
Array begins:
k 0 1 2 3 4 5 6 7 8 9
n +--------------------
1 | 1 1 1 1 1 1 1 1 1 1
2 | 2 1 1 1 1 1 1 1 1 1
3 | 2 2 1 1 1 1 1 1 1 1
4 | 3 1 2 1 1 1 1 1 1 1
5 | 2 2 1 2 1 1 1 1 1 1
6 | 4 2 2 1 2 1 1 1 1 1
7 | 2 2 1 2 1 2 1 1 1 1
8 | 4 1 2 1 2 1 2 1 1 1
9 | 3 3 2 2 1 2 1 2 1 1
10 | 4 2 2 1 2 1 2 1 2 1
MATHEMATICA
T[n_, k_] :=
Module[{c = 0, i = 1, x = n},
While[x >= 1, If[IntegerQ[x], c++]; i++; x = (i-1)*(x-k)/i]; c]
A004736[n_] := Binomial[Floor[3/2 + Sqrt[2*n]], 2] - n + 1
A002260[n_] := n - Binomial[Floor[1/2 + Sqrt[2*n]], 2]
a[n_] := T[A004736[n], A002260[n] - 1]
Table[a[n], {n, 1, 91}]
(* Second program: *)
nmax = 14;
col[k_] := col[k] = CoefficientList[Sum[x^(n(k n - k + 2)/2 - 1)/(1 - x^n), {n, 1, nmax}] + O[x]^nmax, x];
T[n_, k_] := col[k][[n]];
Table[T[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 30 2020 *)
PROG
(PARI)
T(n, k)=c=0; i=1; x=n; while(x>=1, if(frac(x)==0, c++); i++; x=n/i-(k/2)*(i-1)); c
for(s=1, 13, for(k=0, s-1, n=s-k; print1(T(n, k), ", ")))
KEYWORD
nonn,tabl
AUTHOR
Luc Rousseau, Jan 11 2019
STATUS
approved