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A240690
Number of partitions p of n such that p contains fewer 1s than its conjugate.
4
0, 1, 1, 2, 3, 4, 7, 8, 14, 16, 26, 30, 47, 54, 81, 95, 136, 161, 224, 266, 361, 431, 571, 684, 891, 1067, 1369, 1641, 2077, 2488, 3116, 3726, 4623, 5520, 6790, 8093, 9884, 11753, 14262, 16923, 20415, 24168, 29006, 34255, 40920, 48214, 57344, 67410, 79863
OFFSET
1,4
COMMENTS
a(n+1) = number of partitions p of n such that (# 1s in p) <= (#1s in conjugate(p)).
LINKS
FORMULA
2*a(n) + A240691(n) = A000041(n) for n >= 1.
a(n) + a(n+1) = A000041(n). - Omar E. Pol, Mar 07 2015
G.f.: (-1 + Product_{k>0} (1 - x^k)^(-1)) * x / (1 + x). - Michael Somos, Mar 16 2015
a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*n*sqrt(3)). - Vaclav Kotesovec, Jun 02 2018
EXAMPLE
a(6) counts these 4 partitions: 6, 51, 42, 411, of which the respective conjugates are 111111, 21111, 2211, 3111.
G.f. = x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 7*x^7 + 8*x^8 + 14*x^9 + 16*x^10 + ...
MATHEMATICA
z = 53; f[n_] := f[n] = IntegerPartitions[n]; c[p_] := Table[Count[#, _?(# >= i &)], {i, First[#]}] &[p]; (* conjugate of partition p *)
Table[Count[f[n], p_ /; Count[p, 1] < Count[c[p], 1]], {n, 1, z}] (* A240690 *)
Table[Count[f[n], p_ /; Count[p, 1] <= Count[c[p], 1]], {n, 1, z}] (* A240690(n+1) *)
Table[Count[f[n], p_ /; Count[p, 1] == Count[c[p], 1]], {n, 1, z}] (* A240691 *)
a[ n_] := SeriesCoefficient[ (-1 + 1 / QPochhammer[ x]) x / (1 + x), {x, 0, n}]; (* Michael Somos, Mar 16 2015 *)
PROG
(PARI) q='q+O('q^60); concat([0], Vec((-1 + 1/eta(q))*q/(1+q))) \\ G. C. Greubel, Aug 07 2018
CROSSREFS
Sequence in context: A215914 A006049 A084541 * A339593 A113050 A278180
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 11 2014
STATUS
approved