This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(4) = 15 = 0+1+2+2+3+3+4, the total number of 1's in 0000, 1000, 1010, 1100, 1101, 1110, 1111.
b:= proc(n) option remember; `if`(n=0, 1, (g-> (f-> expand(
1+x*b(f)*b(n-1-f)))(min(g-1, n-g/2)))(2^ilog2(n)))
end:
a:= n-> subs(x=1, diff(b(n), x)):
seq(a(n), n=0..40);
b[n_][x_] := b[n][x] = If[n == 0, 1, Function[g, Function[f, Expand[1 + x b[f][x] b[n - 1 - f][x]]][Min[g - 1, n - g/2]]][2^(Length[IntegerDigits[ n, 2]] - 1)]]; a[n_] := b[n]'[1]; a /@ Range[0, 40] (* Jean-François Alcover, Apr 22 2021, after Alois P. Heinz *)
a(2) = 1: 12. a(3) = 2: 213, 231. a(4) = 9: 2413, 3124, 3214, 3241, 3412, 3421, 4123, 4132, 4213. a(5) = 28: 25134, 25143, 35124, 35142, 35214, 35241, 42315, 42351, 43125, 43152, 43215, 43251, 43512, 43521, 45123, 45132, 45213, 45231, 45312, 45321, 52314, 52341, 52413, 52431, 53124, 53142, 53214, 53241. a(6) = 90: 362451, 362541, 436125, 436215, ..., 652314, 652413, 653124, 653214. (The examples use max-heaps.)
b:= proc(u, o) option remember; local n, g, l; n:= u+o;
if n=0 then 1
else g:= 2^ilog2(n); l:= min(g-1, n-g/2); expand(
add(add(binomial(j-1, i)*binomial(n-j, l-i)*
b(i, l-i)*b(j-1-i, n-l-j+i), i=0..min(j-1, l)), j=1..u)+
add(add(binomial(j-1, i)*binomial(n-j, l-i)*
b(l-i, i)*b(n-l-j+i, j-1-i), i=0..min(j-1, l)), j=1..o)*x)
fi
end:
a:= n-> coeff(b(n, 0), x, 1):
seq(a(n), n=1..25);
b[u_, o_] := b[u, o] = Module[{n = u+o, g, l}, If[n == 0, 1,
g = 2^(Length[IntegerDigits[n, 2]] - 1);
l = Min[g - 1, n - g/2]; Expand[
Sum[ Sum[Binomial[j - 1, i]*Binomial[n - j, l - i]*
b[i, l-i]*b[j-1-i, n-l-j+i], {i, 0, Min[j-1, l]}], {j, 1, u}] +
Sum[Sum[Binomial[j - 1, i]*Binomial[n - j, l - i]*
b[l-i, i]*b[n-l-j+i, j-1-i], {i, 0, Min[j-1, l]}], {j, 1, o}]*x]]];
a[n_] := Coefficient[b[n, 0], x, 1];
Array[a, 25] (* Jean-François Alcover, Apr 22 2021, after Alois P. Heinz *)
a(0) = 1: the empty heap. a(1) = 1: 1. a(2) = 2: 11, 21. a(3) = 6: 111, 211, 212, 221, 312, 321. a(4) = 16: 1111, 2111, 2121, 2211, 2212, 2221, 3121, 3211, 3212, 3221, 3231, 3312, 3321, 4231, 4312, 4321. (The examples use max-heaps.)
b:= proc(n, k) option remember; `if`(n=0, 1,
(g-> (f-> add(b(f, j)*b(n-1-f, j), j=1..k)
)(min(g-1, n-g/2)))(2^ilog2(n)))
end:
a:= n-> add(add(binomial(k, j)*(-1)^j*b(n, k-j), j=0..k), k=0..n):
seq(a(n), n=0..24);
b[n_, k_] := b[n, k] = If[n == 0, 1, Function[g, Function[f, Sum[b[f, j]*b[n - 1 - f, j], {j, 1, k}]][Min[g - 1, n - g/2]]][2^(Length@IntegerDigits[n, 2] - 1)]];
T[n_, k_] := Sum[Binomial[k, j]*(-1)^j*b[n, k - j], {j, 0, k}];
a[n_] := Sum[T[n, k], {k, 0, n}];
Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Sep 24 2024, after Alois P. Heinz *)
a(4) = 3 because 3 = 24/8 and there are 4! = 24 permutations on 4 elements and 8 min-heaps on 5 elements, namely (0,1,2,3,4), (0,1,2,4,3), (0,1,3,2,4), (0,1,3,4,2), (0,1,4,2,3), (0,1,4,3,2), (0,2,1,3,4), and (0,2,1,4,3). In every (min-) heap, the element at position i has to be larger than the element at position floor(i/2) for all i=2..n. The minimum is always found at position 1.
h:= proc(n) option remember; `if`(n=0, 1, (b-> (f->
h(f)*n*h(n-1-f))(min(b-1, n-b/2)))(2^ilog2(n)))
end:
a:= n-> h(n+1)/(n+1):
seq(a(n), n=0..50);
aa[n_] := aa[n] = Module[{b, nl}, If[n<2, 1, b = 2^Floor[Log[2, n]]; nl = Min[b-1, n-b/2]; n*aa[nl]*aa[n-1-nl]]]; a[n_] := aa[n+1]/(n+1); Table[a[i], {i, 0, 50}] (* Jean-François Alcover, Mar 05 2014, after Alois P. Heinz *)
a(4) = 13 = 4+3+2+2+1+1+0, the total number of 0's in 0000, 1000, 1010, 1100, 1101, 1110, 1111.
b:= proc(n) option remember; `if`(n=0, 1, (g-> (f-> expand(
x^n+b(f)*b(n-1-f)))(min(g-1, n-g/2)))(2^ilog2(n)))
end:
a:= n-> subs(x=1, diff(b(n), x)):
seq(a(n), n=0..40);
b[n_][x_] := b[n][x] = If[n == 0, 1, Function[g, Function[f, Expand[x^n + b[f][x] b[n - 1 - f][x]]][Min[g - 1, n - g/2]]][2^(Length[IntegerDigits[ n, 2]] - 1)]]; a[n_] := b[n]'[1]; a /@ Range[0, 40] (* Jean-François Alcover, Apr 22 2021, after Alois P. Heinz *)
a(0) = 1: the empty heap. a(1) = 1: 1. a(2) = 3: 11, 21, 22. a(3) = 14: 111, 211, 212, 221, 222, 311, 312, 313, 321, 322, 323, 331, 332, 333. (The examples use max-heaps.)
b:= proc(n, k) option remember; `if`(n=0, 1,
(g-> (f-> add(b(f, j)*b(n-1-f, j), j=1..k)
)(min(g-1, n-g/2)))(2^ilog2(n)))
end:
a:= n-> b(n$2):
seq(a(n), n=0..21);
A heap on 4 elements is pictured in the 2nd link, and has breadth first reading word abcd. Then for n = 4 the a(4) = 3 heaps have reading words 1234, 1243, and 1324.
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