# [seqfan] Re: a simple Partitions question

Rob Pratt Rob.Pratt at sas.com
Sun Mar 21 19:00:50 CET 2010

```Confirmed the complementary sequence by expanding the infinite product

prod {k=1 to inf} (1+x^(3^k))
* prod {k=1 to inf} (1+x^(5^k))
* prod {k=1 to inf} (1+x^(7^k))
- 1

To get the terms up to n, the following finite product suffices:

prod {k=1 to floor(log(n)/log(3))} (1+x^(3^k))
* prod {k=1 to floor(log(n)/log(5))} (1+x^(5^k))
* prod {k=1 to floor(log(n)/log(7))} (1+x^(7^k))
- 1

-----Original Message-----
From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Richard Mathar
Sent: Sunday, March 21, 2010 1:04 PM
To: seqfan at seqfan.eu
Subject: [seqfan] Re: a simple Partitions question

> %I A078198
> %S A078198 1,2,4,6,11,13,18,20,22,23,26,29,31,38,45,47,50,53,72,75,78,80,87,
> %T A078198 94,99,103,107,112
> %N A078198 Numbers that cannot be partitioned into positive powers of 3, 5 and 7.
> %e A078198 2271 is a memeber because 2271 = 3^1 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 5^1 + 5^2 + 5^3 + 5^4 + 7^1 + 7^2 + 7^3.
> %O A078198 1,2
> %K A078198 nonn,more
> %A A078198 Wouter Meeussen, Mar 20 2010

Better: numbers that cannot be partitioned into DISTINCT strictly positive
powers of 3, 5 and 7, because 6=3^1+3^1, 11=3^1+3^1+5^1 etc are
representations with repeated positive powers which are apparently
And: 3^0=1 is a positive power (albeit the exponent isn't), so what is
actually meant (?) is

Numbers that cannot be partitioned into distinct powers of 3, 5 and 7, each part >1.

The example needs negation: 2271 is NOT a member because .. is such a partition.

The auxiliary complementary sequence to A078198 is
"Numbers that can be partitioned.." = 3,5,7,8,9,10,12,14,15,16,17,...
And on that one could define "Numbers of partitions of A.....(n) into forms..."

3 -> 1 (namely 3)
5 -> 1 (namely 5)
7 -> 1 (namely 7)
8 -> 1 (namely 3+5)
9 -> 1 (namely 3^2)
10 -> 1 (namely 3+7)
12 -> 2 (namely 5+7, 3+3^2)
14 -> 1 (namely 5+3^2)
15 -> 1 (namely 3+5+7)
16 -> 1 (namely 7+3^2)
17 -> 1 (namely 3+5+3^2
19 -> 1 (namely 3+7+3^2)
21 -> 1 (namely 5+7+3^2)
24 -> 1 (namely 3+5+7+3^2)
25 -> 1 (namely 5^2)
27 -> 1 (namely 3^3)
28 -> 1 (namely 3+5^2)
30 -> 2 (namely 5+5^2 = 3+3^3)
32 -> 2 (namely 7+5^2 = 5+3^3)
33 -> 1 (namely 3+5+5^2)
34 -> 2 (namely 3^2+5^2 = 7+3^3)

to yield a sequence (everybody please check)

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 1,
3, 3, 1, 2, 2, 1, 2,

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