A264998 Number of partitions of n into distinct parts of the form 3^a*5^b or 2.

1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 3, 2, 3, 3, 2, 3, 2, 1, 2, 2, 3, 3, 4, 4, 4, 6, 4, 5, 5, 4, 5, 4, 4, 3, 4, 4, 4, 6, 5, 5, 7, 5, 5, 6, 4, 6, 6, 6, 6, 7, 7, 6, 8, 5, 6, 7, 5, 6, 5, 4, 4, 4, 4, 4, 5, 6, 5, 7, 6, 5, 9, 7, 8, 9, 7, 8, 9, 8, 7, 10, 8, 9, 11

Offset: 0

Author

Joseph Myers, Nov 29 2015

Keywords

• easy
• nonn

Examples

			15 = 15 = 9 + 5 + 1 = 9 + 3 + 2 + 1, so a(15) = 3.
		

Links

• Joseph Myers and Reinhard Zumkeller, Table of n, a(n) for n = 0..20000 (first 1000 terms from Joseph Myers)
• British Mathematical Olympiad 2015/16, Olympiad Round 1, Problem 6, Friday, 27 November 2015.

Crossrefs

Cf. A003593, A264997.

Programs

Haskell

import Data.MemoCombinators (memo2, list, integral)
a264998 n = a264998_list !! (n-1)
a264998_list = f 0 [] (1 : 2 : tail a003593_list) where
   f u vs ws'@(w:ws) | u < w = (p' vs u) : f (u + 1) vs ws'
                     | otherwise = f u (vs ++ [w]) ws
   p' = memo2 (list integral) integral p
   p _  0 = 1
   p [] _ = 0
   p (k:ks) m = if m < k then 0 else p' ks (m - k) + p' ks m
-- Reinhard Zumkeller, Dec 18 2015

Mathematica

nmax = 100; A003593 = Select[Range[nmax], PowerMod[15, #, #] == 0 &]; CoefficientList[Series[(1 + x^2) * Product[(1 + x^(A003593[[k]])), {k, 1, Length[A003593]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 18 2015 *)

Formula

G.f.: (1+x)(1+x^2)(1+x^3)(1+x^5)(1+x^9)(1+x^15)....