login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A334461 a(n) is the number of partitions of n into consecutive parts that differ by 4. 10
1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 2, 3, 1, 3, 1, 3, 3, 2, 1, 4, 1, 3, 2, 3, 1, 3, 2, 3, 2, 2, 1, 5, 1, 2, 2, 3, 2, 4, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 1, 4, 1, 4, 2, 2, 1, 5, 2, 2, 2, 3, 1, 5, 2, 3, 2, 2, 2, 5, 1, 3, 2, 4, 1, 4, 1, 3, 4 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,6
LINKS
FORMULA
The g.f. for "consecutive parts that differ by d" is Sum_{k>=1} x^(k*(d*k-d+2)/2) / (1-x^k); cf. A117277. - Joerg Arndt, Nov 30 2020
EXAMPLE
For n = 28 there are three partitions of 28 into consecutive parts that differ by 4, including 28 as a valid partition. They are [28], [16, 12] and [13, 9, 5, 1]. So a(28) = 3.
MATHEMATICA
nmax = 105;
col[k_] := col[k] = CoefficientList[Sum[x^(n(k n - k + 2)/2 - 1)/(1 - x^n), {n, 1, nmax}] + O[x]^nmax, x];
a[n_] := col[4][[n]];
Array[a, nmax] (* Jean-François Alcover, Nov 30 2020 *)
PROG
(PARI) seq(N, d)=my(x='x+O('x^N)); Vec(sum(k=1, N, x^(k*(d*k-d+2)/2)/(1-x^k)));
seq(100, 4) \\ Joerg Arndt, May 05 2020
CROSSREFS
Row sums of A334460.
Column k=4 of A323345.
Sequences of this family whose consecutive parts differ by k are A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), this sequence (k=4), A334541 (k=5), A334948 (k=6).
Sequence in context: A086435 A266226 A099305 * A338652 A033109 A321469
KEYWORD
nonn
AUTHOR
Omar E. Pol, May 01 2020
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified July 15 00:37 EDT 2024. Contains 374323 sequences. (Running on oeis4.)