A254436 A component sequence of A254296.

0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 4, 3, 6, 3, 6, 5, 8, 7, 10, 7, 12, 9, 14, 11, 16, 14, 19, 17, 22, 20, 28, 23, 31, 26, 34, 32, 40, 35, 43, 38, 51, 46, 59, 51, 64, 61, 74, 71, 84, 76, 94, 86, 104, 96, 114, 108, 126, 120, 138, 132, 157, 146, 171

Offset: 1

Author

Md. Towhidul Islam, Feb 28 2015

Keywords

• nonn

Comments

This sequence is a component of the formula for counting A254296.

If m=ceiling(log_3(2k)), define n=(3^(m-1)+1)/2+(3^(m-2))-k for k in the range (3^(m-1)+1)/2<=k<=(3^(m-1)-1)/2+(3^(m-2)). Then this sequence gives the first 3^(m-2) terms.

Links

• Md. Towhidul Islam, Table of n, a(n) for n = 1..6561
• Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Crossrefs

Cf. A254296, A254430, A254431, A254432, A254433, A254435, A254437, A254438, A254439, A254440.

Formula

If m=ceiling(log_3(2k)), define n=(3^(m-1)+1)/2+(3^(m-2))-k for k in the range (3^(m-1)+1)/2<=k<=(3^(m-1)-1)/2+(3^(m-2)).

Then a(n)=Sum_{d=ceiling((3k+2)/5)..(3^(m-1)-1)/2} Sum_{p=ceiling((d-1)/3..2d-k-1} A254296(p).