A099392 -id:A099392 - cp's OEIS frontend

A124222 Triangular array read by rows, with shape sequence A002865, generated from A123682.

1, 1, 3, 5, 0, 9, 2, 13, 7, 1, 0, 19, 18, 4, 2, 25, 36, 12, 10, 0, 2, 0, 33, 64, 26, 32, 2, 11, 2, 1, 41, 103, 49, 78, 7, 40, 13, 6, 1, 0, 3, 0

Offset: 1

Alford Arnold, Oct 19 2006

The shape sequence for A124222 is essentially A002865 since its columns are based on source partitions (1,22,33,222,44,332,2222,333,...). The row sum sequence for A124222 is essentially A001045.

			The array begins
1
1
3
5 0
9 2
13 7 1 0
19 18 4 2
25 36 12 10 0 2 0
33 64 26 32 2 11 2 1
41 103 49 78 7 40 13 6 1 0 3 0

Cf. A001045 (row sums), A002865, A099392, A123682, A123683.

A132892 Square array T(m,n) read by antidiagonals; T(m,n) is the number of equivalence classes in the set of sequences of n nonnegative integers that sum to m, generated by the equivalence relation defined in the following manner: we write a sequence in the form a[1]0a[2]0...0a[p], where each a[i] is a (possibly empty) sequence of positive integers; two sequences in this form, a[1]0a[2]0...0a[p] and b[1]0b[2]0...0b[q] are said to be equivalent if p=q and b[1],b[2],...,b[q] is a cyclic permutation of a[1],a[2],...a[p].

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 3, 1, 1, 5, 9, 7, 4, 1, 1, 6, 13, 14, 10, 4, 1, 1, 7, 19, 25, 22, 12, 5, 1, 1, 8, 25, 41, 42, 30, 15, 5, 1, 1, 9, 33, 63, 79, 66, 43, 19, 6, 1, 1, 10, 41, 92, 131, 132, 99, 55, 22, 6, 1, 1, 11, 51, 129, 213, 245, 217, 143, 73, 26, 7, 1, 1, 12, 61, 175, 325, 428, 429, 335, 201, 91, 31, 7, 1

Offset: 1

Emeric Deutsch and Ira M. Gessel, Oct 02 2007

T(n,n) = A000108(n) (the Catalan numbers; see R. P. Stanley, Catalan addendum, problem starting "Equivalence classes of the equivalence relation ..."). T(m,m+1) = A007595(m+1); T(m,m+2) = A003441(m+1); T(m,m+3) = A003444(m+3); T(n+2,n) = A001453(n+1) (Catalan numbers - 1); T(m,1)=1; T(m,2)=m; T(m,3) = A080827(m) = A099392(m+1); T(m,4) = A004006(m).

			T(2,4) = 3 because we have {2000, 0200, 0020, 0002}, {1100, 0110, 0011} and {1010, 0101, 1001}.
T(4,2) = 4 because we have {40, 04}, {31}, {13} and {22}.
The square array starts:
  1....1.....1.....1......1.....1.....1...
  1....2.....3.....3......4.....4.....5...
  1....3.....5.....7.....10....12....15...
  1....4.....9....14.....22....30....43...
  1....5....13....25.....42....66....99...

Alois P. Heinz, antidiagonals n = 1..200, flattened
Emeric Deutsch and Ira Gessel, Equivalence Classes and Cyclic Arrangements:Problem 10525, Amer. Math. Monthly, 105, No. 8, 1998, 774-775 (published solution by D. Beckwith).
R. P. Stanley, Catalan addendum. See the interpretation (www, "Vertices of height n-1 of the tree T ...").

Cf. A000108, A007595, A003441, A003444, A001453, A080827, A099392, A004006.

with(numtheory): T:=proc(m,n) local r, div, N: r:=igcd(m,n+1): div:=divisors(r): N:=nops(div): (sum(phi(div[j])*(binomial((m+n+1)/div[j]-1,(n+1)/div[j]-1) -binomial(m/div[j]-1,(n+1)/div[j]-1)),j=1..N))/(n+1) end proc: for m to 12 do seq(T(m, n),n=1..12) end do; # yields the upper left 12 by 12 block of the infinite matrix T(m,n)
# second Maple program:
T:= proc(m, n) uses numtheory; (C-> add(phi(d)*(C((m+n+1)/d-1, (n+1)/d-1)
      -C(m/d-1, (n+1)/d-1))/(n+1), d=divisors(igcd(m, n+1))))(binomial)
    end:
seq(seq(T(1+d-n, n), n=1..d), d=1..14);  # Alois P. Heinz, Jan 28 2025

T[m_, n_] := Module[{r, div, N}, r = GCD[m, n + 1]; div = Divisors[r]; N = Length[div]; (Sum[EulerPhi[div[[j]]]*(Binomial[(m + n + 1)/div[[j]] - 1, (n + 1)/div[[j]] - 1] - Binomial[m/div[[j]] - 1, (n + 1)/div[[j]] - 1]), {j, 1, N}])/(n + 1)];
Table[T[m - n + 1, n], {m, 1, 13}, {n, 1, m}] // Flatten (* Jean-François Alcover, Sep 01 2024, after Maple program *)

T(m,n) = Sum_{d | gcd(m,n+1)} phi(d)*(C((m+n+1)/d-1, (n+1)/d-1) - C(m/d-1, (n+1)/d-1))/(n+1). [corrected by Jason Yuen, Jan 28 2025]

A274323 Number of partitions of n^4 into at most two parts.

Original entry on oeis.org

1, 1, 9, 41, 129, 313, 649, 1201, 2049, 3281, 5001, 7321, 10369, 14281, 19209, 25313, 32769, 41761, 52489, 65161, 80001, 97241, 117129, 139921, 165889, 195313, 228489, 265721, 307329, 353641, 405001, 461761, 524289, 592961, 668169, 750313, 839809, 937081

Offset: 0

Colin Barker, Oct 13 2016

Coefficient of x^(n^4) in 1/((1-x)*(1-x^2)).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A099392 (n^2), A274324 (n^3), A274325 (n^5).

Cf. A008619.

PARI
```
a(n) = (3+(-1)^n+2*n^4)/4
```

b(n) = (3+(-1)^n+2*n)/4 \\ the coefficient of x^n in the g.f. 1/((1-x)*(1-x^2))
vector(50, n, n--; b(n^4))

G.f.: (1 - 3*x + 10*x^2 + 10*x^3 + 5*x^4 + x^5) / ((1-x)^5*(1+x)).
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6) for n > 5.
a(n) = (3 + (-1)^n + 2*n^4)/4.
a(n) = A008619(n^4).
a(n) = 1 + floor(n^4/2). - Alois P. Heinz, Oct 13 2016
E.g.f.: ((2 + x + 7*x^2 + 6*x^3 + x^4)*cosh(x) + (1 + x + 7*x^2 + 6*x^3 + x^4)*sinh(x))/2. - Stefano Spezia, Mar 17 2024

A288797 Square array a(p,q) = p^2 + q^2 - 2p - 2q + 2*gcd(p,q), p >= 1, q >= 1, read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 4, 4, 4, 9, 5, 5, 9, 16, 12, 12, 12, 16, 25, 17, 13, 13, 17, 25, 36, 28, 20, 24, 20, 28, 36, 49, 37, 33, 25, 25, 33, 37, 49, 64, 52, 40, 36, 40, 36, 40, 52, 64, 81, 65, 53, 45, 41, 41, 45, 53, 65, 81, 100, 84, 72, 64, 52, 60, 52, 64, 72, 84, 100

Offset: 1

Luc Rousseau, Jun 16 2017

In the Cartesian plane, let r(p,q) denote the rotation with center origin and angle associated to slope p/q (p: number of units upwards, q: number of units towards the right).
Let R(p,q) be the square of area p^2 + q^2 = R^2, with vertices (0,0), (0,R), (R,R), (R,0).
The natural unit squares (i.e., the (a,a+1) X (b,b+1) Cartesian products, with a and b integers) are transformed by r(p,q) into rotated unit squares.
a(p,q) is the number of rotated unit squares that fully land inside R(p,q).

			Table begins:
    0   1   4   9  16  25 ...
    1   4   5  12  17  28 ...
    4   5  12  13  20  33 ...
    9  12  13  24  25  36 ...
   16  17  20  25  40  41 ...
   25  28  33  36  41  60 ...
  ... ... ... ... ... ... ...

Luc Rousseau, Illustration for p=6, q=4

Cf. A000217, A000290, A000566, A001844, A003989, A051624, A080827, A099392, A143101.

A[p_, q_] := p^2 + q^2 - 2*p - 2*q + 2*GCD[p, q];
(* or, checking without the formula: *)
okQ[{a_, b_}, p_, q_] := Module[{r2 = p^2 + q^2}, 0 <= a*q - b*p <= r2 && 0 <= a*p + b*q <= r2 && 0 <= a*q - b*p + q <= r2 && 0 <= a*p + b*q + p <= r2 && 0 <= a*q - (b + 1)*p <= r2 && 0 <= a*p + b*q + q <= r2 && 0 <= (a + 1)*q - (b + 1)*p <= r2 && 0 <= a*p + b*q + p + q <= r2];
A[p_, q_] := Module[{r}, r = Reduce[okQ[{a, b}, p, q], {a, b}, Integers]; If[r[[0]] === And, 1, Length[r]]];
Flatten[Table[A[p - q + 1, q], {p, 1, 11}, {q, 1, p}]] (* Jean-François Alcover, Jun 17 2017 *)

a(p,q)=p^2+q^2-2*p-2*q+2*gcd(p,q)
for(n=2,12,for(p=1,n-1,{q=n-p;print(a(p,q))}))

a(p,q) = p^2 + q^2 - 2*p - 2*q + 2*gcd(p,q).
a(p,1) = a(1,p) = (p-1)^2 = A000290(p-1).
a(p,p) = 2*p*(p-1) = 4*A000217(p-1).
a(p,p+1) = 2*p*(p-1)+1 = A001844(p-1).
a(p,p+2) = 2*p^2+2*gcd(2,p) = 2*p^2+3+(-1)^(p) = 4*A099392(p-1) = 4*A080827(p).
a(p,p+3) = 2*p^2+2*p+5+4*A079978(p) = 1+4*(1+A143101(p)).
a(p,2*p) = p*(5*p-4) = A051624(p).
a(p,3*p) = 2*p*(5*p-3) = 4*A000566(p).

A328005 Number of distinct coefficients in functional composition of 1 + x + ... + x^(n-1) with itself.

Original entry on oeis.org

0, 1, 2, 4, 8, 13, 19, 25, 33, 41, 51, 61, 73, 85, 99, 113, 129, 145, 163, 181, 201, 221, 243, 265, 289, 313, 339, 365, 393, 421, 451, 481, 513, 545, 579, 613, 649, 685, 723, 761, 801, 841, 883, 925, 969, 1013, 1059, 1105, 1153, 1201, 1251, 1301, 1353, 1405, 1459

Offset: 0

Vladimir Reshetnikov, Oct 01 2019

Sum_{i=0..n-1} x^i = (x^n - 1)/(x - 1).

			For n = 4, the composition of 1 + x + x^2 + x^3 with itself is 1 + (1 + x + x^2 + x^3) + (1 + x + x^2 + x^3)^2 + (1 + x + x^2 + x^3)^3 = 4 + 6 x + 10 x^2 + 15 x^3 + 15 x^4 + 14 x^5 + 11 x^6 + 6 x^7 + 3 x^8 + x^9 that has 8 distinct coefficients [1, 3, 4, 6, 10, 11, 14, 15], so a(4) = 8.
The first few polynomials p_n(x) are 0, 1, x + 2, x^4 + 2*x^3 + 4*x^2 + 3*x + 3, ... with p_n(1) = A023037(n), n >= 0.

Wikipedia, Function composition

Cf. A099392, A080827, A023037.

f:= n-> unapply(add(x^j, j=0..n-1), x):
a:= n-> nops({coeffs(expand((f(n)@@2)(x)))} minus {0}):
seq(a(n), n=0..60);  # Alois P. Heinz, Oct 01 2019

Table[With[{s = Sum[x^k, {k, 0, n - 1}]}, Length[Union[CoefficientList[Expand[s /. x -> s], x]]]], {n, 0, 53}]

a(n)={my(p=(1-x^n)/(1-x)); #Set(Vec(subst(p,x,p)))} \\ Andrew Howroyd, Oct 01 2019

def A328005(n):
    R. = PolynomialRing(ZZ)
    q = R(sum(x^k for k in range(n)))
    return len(Set(q.substitute(x=q).list()))
print([A328005(n) for n in range(55)]) # Peter Luschny, Oct 02 2019

It appears that a(n) = (2*n^2 + (-1)^n + 3)/4 for n >= 5.

Conjectured g.f.: (x^7 - x^6 - x^5 + 2*x^3 + 1)*x/((x + 1)*(1 - x)^3).

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A124222 Triangular array read by rows, with shape sequence A002865, generated from A123682.

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A274323 Number of partitions of n^4 into at most two parts.

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A288797 Square array a(p,q) = p^2 + q^2 - 2*p - 2*q + 2*gcd(p,q), p >= 1, q >= 1, read by antidiagonals.

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A328005 Number of distinct coefficients in functional composition of 1 + x + ... + x^(n-1) with itself.

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A288797 Square array a(p,q) = p^2 + q^2 - 2p - 2q + 2*gcd(p,q), p >= 1, q >= 1, read by antidiagonals.