A056768 Number of partitions of the n-th prime into parts that are all primes.
1, 1, 2, 3, 6, 9, 17, 23, 40, 87, 111, 219, 336, 413, 614, 1083, 1850, 2198, 3630, 5007, 5861, 9282, 12488, 19232, 33439, 43709, 49871, 64671, 73506, 94625, 221265, 279516, 394170, 441250, 766262, 853692, 1175344, 1608014, 1975108, 2675925
Offset: 1
Keywords
Examples
a(4) = 3 because the 4th prime is 7 which can be partitioned using primes in 3 ways: 7, 5 + 2, and 3 + 2 + 2. In connection with the 6th prime 13, for instance, we have the a(6) = 9 prime partitions: 13 = 2 + 2 + 2 + 2 + 2 + 3 = 2 + 2 + 2 + 2 + 5 = 2 + 2 + 2 + 7 = 2 + 2 + 3 + 3 + 3 = 2 + 3 + 3 + 5 = 2 + 11 = 3 + 3 + 7 = 3 + 5 + 5.
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000 (first 4000 terms from Alois P. Heinz)
Programs
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Haskell
a056768 = a000607 . a000040 -- Reinhard Zumkeller, Aug 05 2012
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Maple
b:= proc(n, i) option remember; `if`(n=0 or i=2 and n::even, 1, `if`(i=2 or n=1, 0, b(n, prevprime(i)))+`if`(i>n, 0, b(n-i, i))) end: a:= n-> b(ithprime(n)$2): seq(a(n), n=1..50); # Alois P. Heinz, Sep 15 2016 -
Mathematica
Table[Count[IntegerPartitions[n],?(AllTrue[#,PrimeQ]&)],{n,Prime[ Range[ 40]]}] (* The program uses the AllTrue function from Mathematica version 10 *) (* _Harvey P. Dale, Mar 07 2015 *) n=40;ser=Product[1/(1-x^Prime[i]),{i,1,n}];Table[SeriesCoefficient[ser,{x,0,Prime[i]}],{i,1,n}] (* Gus Wiseman, Sep 14 2016 *)
Formula
a(n) = A000607(prime(n)).
a(n) = A168470(n) + 1. - Alonso del Arte, Feb 15 2014, restating the corresponding formula given by R. J. Mathar for A168470.
a(n) = [x^prime(n)] Product_{k>=1} 1/(1 - x^prime(k)). - Ilya Gutkovskiy, Jun 05 2017
Extensions
More terms from James Sellers, Aug 25 2000