OFFSET
1,3
COMMENTS
Each row n contains A002061(n) terms and is palindromic.
Apart from the last term, each term on row n gives the largest summand in the partitions encountered on the main trunk of the Bulgarian solitaire tree computed for the deck of n(n+1)/2 cards; from row 2 onward, the last term of row k is one less than the largest summand in the unordered triangular partition {1+2+...+k} that is at the root of each game tree of the deck of the same size. The function f(n) = A227185(A227452(n)) would correctly give the largest summand sizes also for those cases.
REFERENCES
Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.
LINKS
Antti Karttunen, The rows 1..31 of the table, flattened
Ethan Akin and Morton Davis, "Bulgarian solitaire", American Mathematical Monthly 92 (4): 237-250. (1985).
FORMULA
a(n) = A227141(A227177(n),A227181(n)). [As a sequence. Each row n is a subsequence from the section [n,n^2] of the n-th row of ordinary table A227141.]
;; The following two formulas use the table A227452:
a(n) = A227185(A227452(n)) - ([n>1] * (A227177(n+1) - A227177(n))). [Where the expression [n>1] is an instance of Iverson brackets]
For all n, a(n) = a(A227182(n)). [This is just a claim that each row is symmetric.]
EXAMPLE
Rows 1-6 of the table are:
1
1, 3, 1
2, 4, 3, 2, 3, 4, 2
3, 5, 4, 4, 3, 4, 5, 4, 3, 4, 4, 5, 3
4, 6, 5, 5, 5, 4, 5, 6, 5, 5, 4, 5, 5, 6, 5, 4, 5, 5, 5, 6, 4
5, 7, 6, 6, 6, 6, 5, 6, 7, 6, 6, 6, 5, 6, 6, 7, 6, 6, 5, 6, 6, 6, 7, 6, 5, 6, 6, 6, 6, 7, 5
PROG
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Antti Karttunen, Jul 03 2013
STATUS
approved