A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3*x^2*y^2 + 5*y^2*z^2 + 8*z^2*x^2 is a square.

0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 4, 2, 5, 2, 1, 2, 2, 2, 2, 4, 2, 4, 3, 3, 3, 4, 5, 2, 3, 5, 5, 4, 4, 2, 4, 4, 5, 3, 4, 3, 6, 3, 2, 5, 2, 2, 7, 7, 1, 3, 6, 4, 4, 3, 3, 6, 2, 5, 5, 3, 6, 5, 4, 6, 6, 6, 3, 6, 6, 4, 5, 6, 2, 6, 3, 5, 4, 5, 3, 5, 12, 4, 4, 5, 1, 6, 6, 7, 9, 3, 3, 6, 5, 6, 7, 4

Offset

1,9

Comments

Conjecture 1: a(n) > 0 for all n > 2.

We have verified a(n) > 0 for all n = 3..10000. The conjecture holds if a(p) > 0 for every odd prime p. For any n > 0 we have a(3*n) > 0, since 3*n = n + n + n and 3 + 5 + 8 = 4^2.

It seems that a(n) = 1 only for n = 3..8, 10, 11, 19, 53, 89, 127, 178, 257, 461.

See also A343862 for similar conjectures.

Conjecture 1 holds for all n < 2^15. Note a(1823) = 1. - Martin Ehrenstein, May 03 2021

Links

Martin Ehrenstein, Table of n, a(n) for n = 1..32767 (first 1500 terms from Zhi-Wei Sun)

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190. See also arXiv version, arXiv:1604.06723 [math.NT], 2016-2017.

Example

a(4) = 1 with 4 = 2 + 1 + 1 and 3*2^2*1^2 + 5*1^2*1^2 + 8*1^2*2^2 = 7^2.
a(19) = 1 with 19 = 9 + 9 + 1 and 3*9^2*9^2 + 5*9^2*1^2 + 8*1^2*9^2 = 144^2.
a(53) = 1 with 53 = 23 + 7 + 23 and 3*23^2*7^2 + 5*7^2*23^2 + 8*23^2*23^2 = 1564^2.
a(89) = 1 with 89 = 2 + 58 + 29 and 3*2^2*58^2 + 5*58^2*29^2 + 8*29^2*2^2 = 3770^2.
a(257) = 1 with 257 = 11 + 164 + 82 and 3*11^2*164^2 + 5*164^2*82^2 + 8*82^2*11^2 = 30340^2.
a(461) = 1 with 461 = 186 + 165 + 110 and 3*186^2*165^2 + 5*165^2*110^2 + 8*110^2*186^2 = 88440^2.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[3x^2*y^2+(n-x-y)^2*(5*y^2+8*x^2)], r=r+1], {x, 1, n-2}, {y, 1, n-1-x}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

Crossrefs

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A343862.

Keyword

nonn

Author

Zhi-Wei Sun, Apr 24 2021

Status

approved

A099509 Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 7, 1, 1, 5, 10, 16, 6, 3, 1, 6, 15, 30, 19, 16, 1, 1, 7, 21, 50, 45, 51, 10, 4, 1, 8, 28, 77, 90, 126, 45, 30, 1, 1, 9, 36, 112, 161, 266, 141, 126, 15, 5, 1, 10, 45, 156, 266, 504, 357, 393, 90, 50, 1, 1, 11, 55, 210, 414, 882, 784, 1016, 357

Offset

0,5

Comments

Row sums form absolute values of A078039. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2).

Links

Table of n, a(n) for n=0..74.

Formula

G.f.: (1-x+x*y-x^2*y^2)/((1-x)^2-2*x^2*y^2+x^3*y^2+x^4*y^4).

Example

Rows begin:
[1],
[1,1],
[1,2,1],
[1,3,3,2],
[1,4,6,7,1],
[1,5,10,16,6,3],
[1,6,15,30,19,16,1],
[1,7,21,50,45,51,10,4],
[1,8,28,77,90,126,45,30,1],
[1,9,36,112,161,266,141,126,15,5],...
and can be derived from coefficients of (1+z+z^2)^n:
[1],
[1,1,1],
[1,2,3,2,1],
[1,3,6,7,6,3,1],
[1,4,10,16,19,16,10,4,1],
[1,5,15,30,45,51,45,30,15,5,1],...
by shifting each column k down by [k/2] rows.

Prog

(PARI) T(n, k)=if(n<k || k<0, 0, polcoeff((1+z+z^2+z*O(z^k))^(n-k\2), k, z))

Crossrefs

Cf. A027907, A078039, A099510.

Keyword

nonn,tabl

Author

Paul D. Hanna, Oct 20 2004

Status

approved

A368343 Square array T(n,k), n >= 3, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} k^(n-j) * floor(j/3).

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 7, 5, 2, 1, 5, 13, 16, 7, 2, 1, 6, 21, 41, 34, 9, 3, 1, 7, 31, 86, 125, 70, 12, 3, 1, 8, 43, 157, 346, 377, 143, 15, 3, 1, 9, 57, 260, 787, 1386, 1134, 289, 18, 4, 1, 10, 73, 401, 1562, 3937, 5547, 3405, 581, 22, 4

Offset

3,5

Links

Seiichi Manyama, Antidiagonals n = 3..142, flattened

Formula

T(n,k) = T(n-3,k) + Sum_{j=0..n-3} k^j.

T(n,k) = 1/(k-1) * Sum_{j=0..n} floor(k^j/(k^2+k+1)) = Sum_{j=0..n} floor(k^j/(k^3-1)) for k > 1.

T(n,k) = (k+1)*T(n-1,k) - k*T(n-2,k) + T(n-3,k) - (k+1)*T(n-4,k) + k*T(n-5,k).

G.f. of column k: x^3/((1-x) * (1-k*x) * (1-x^3)).

T(n,k) = 1/(k-1) * (floor(k^(n+1)/(k^3-1)) - floor((n+1)/3)) for k > 1.

Example

Square array begins:
  1,  1,   1,    1,    1,     1,     1, ...
  1,  2,   3,    4,    5,     6,     7, ...
  1,  3,   7,   13,   21,    31,    43, ...
  2,  5,  16,   41,   86,   157,   260, ...
  2,  7,  34,  125,  346,   787,  1562, ...
  2,  9,  70,  377, 1386,  3937,  9374, ...
  3, 12, 143, 1134, 5547, 19688, 56247, ...

Prog

(PARI) T(n, k) = sum(j=0, n, k^(n-j)*(j\3));

Crossrefs

Columns k=0..4 give A002264, A130518, A178455, A368344, A368345.

Cf. A055129, A368296.

Keyword

nonn,tabl

Author

Seiichi Manyama, Dec 22 2023

Status

approved

A247749 Number T(n,k) of lattice paths from (0,0) to (n,0) that do not go below the x-axis or above the diagonal x=y, consist of steps u=(1,1), U=(1,3), H=(1,0), d=(1,-1) and D=(1,-3) for which the area below the path is k; triangle T(n,k), n>=0, read by rows.

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 2, 0, 1, 1, 4, 6, 6, 6, 3, 4, 2, 1, 1, 1, 5, 10, 13, 15, 12, 14, 15, 9, 12, 5, 5, 1, 1, 1, 6, 15, 24, 32, 33, 37, 46, 40, 43, 34, 28, 23, 16, 10, 5, 2, 1, 1, 7, 21, 40, 61, 75, 88, 114, 122, 134, 137, 118, 127, 101, 99, 69, 68, 41, 38, 19, 17, 5, 5, 0, 1

Offset

0,6

Links

Alois P. Heinz, Rows n = 0..50, flattened

Formula

Sum_{k>=0} T(n,k) = A240904(n).

Sum_{k>=1} k * T(n,k) = A247748(n).

Example

Triangle T(n,k) begins:
1;
1;
1, 1;
1, 2,  1;
1, 3,  3,  2,  2,  0,  1;
1, 4,  6,  6,  6,  3,  4,  2,  1,  1;
1, 5, 10, 13, 15, 12, 14, 15,  9, 12,  5,  5,  1,  1;
1, 6, 15, 24, 32, 33, 37, 46, 40, 43, 34, 28, 23, 16, 10, 5, 2, 1;

Maple

b:= proc(x, y) option remember; `if`(y<0 or x<y, 0, `if`(x=0, 1,
      expand(add(z^(y+j/2)*b(x-1, y+j), j=[-1, -3, 0, 1, 3]))))
    end:
T:= n-> (p->seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..10);

Mathematica

b[x_, y_] := b[x, y] = If[y < 0 || x < y, 0, If[x == 0, 1,
     Expand[Sum[z^(y+j/2)*b[x-1, y+j], {j, {-1, -3, 0, 1, 3}}]]]];
T[n_] := CoefficientList[b[n, 0], z];
Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 29 2022, after Alois P. Heinz *)

Crossrefs

Cf. A240904, A247748.

Keyword

nonn,tabf

Author

Alois P. Heinz, Sep 23 2014

Status

approved

A034929 A triangle of Motzkin ballot numbers, read by rows.

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 6, 4, 1, 5, 10, 13, 13, 9, 1, 6, 15, 24, 30, 30, 21, 1, 7, 21, 40, 59, 72, 72, 51, 1, 8, 28, 62, 105, 148, 178, 178, 127, 1, 9, 36, 91, 174, 276, 378, 450, 450, 323, 1, 10, 45, 128, 273, 480, 730, 980, 1158, 1158, 835, 1, 11, 55, 174, 410, 791

Offset

1,5

Comments

Mirror image of A091836. Row sums are the Motzkin numbers (A001006). T(n,n-1)=A001006(n-2) (the Motzkin numbers). T(n,n-2)=A005554(n-1).

References

M. Aigner, Motzkin numbers, Europ. J. Comb. 19 (1998), 663-675.

Links

Table of n, a(n) for n=1..72.

Formula

G.f.= 2(1+tz)/[1-2z+tz-2tz^2+sqrt(1-2tz-3t^2*z^2)].

Example

Triangle begins:
[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 2],
[1, 4, 6, 6, 4],
[1, 5, 10, 13, 13, 9],
[1, 6, 15, 24, 30, 30, 21],
[1, 7, 21, 40, 59, 72, 72, 51]

Crossrefs

Cf. A001006, A091836, A005554.

Keyword

nonn,tabl

Author

N. J. A. Sloane.

Extensions

Edited by Emeric Deutsch, Mar 11 2004

Status

approved

A153859 Triangle read by rows, A007318 * (A007476 * 0 ^(n-k))

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 8, 4, 1, 5, 10, 20, 20, 9, 1, 6, 15, 40, 60, 54, 23, 1, 7, 21, 70, 140, 189, 161, 65, 1, 8, 28, 112, 280, 504, 644, 520, 199, 1, 9, 36, 168, 504, 1134, 1932, 2340, 1791, 654, 1, 10, 45, 240, 840, 2268, 4830, 7800, 8955, 6540, 2296

Offset

0,5

Comments

Row sums = A007476 starting (1, 2, 4, 9, 23, 65, 199,...).

Links

Table of n, a(n) for n=0..65.

Formula

Triangle read by rows A007318 * (A007476 * 0^(n-k)) = binomial transform of an infinite lower triangular matrix with A007476 as the main diagonal: (1, 1, 1, 2, 4, 9, 23, 65, 199,...) and the rest zeros.

Example

First few rows of the triangle =
1;
1, 1;
1, 2, 1;
1, 3, 3, 2;
1, 4, 6, 8, 4;
1, 5, 10, 20, 20, 9;
1, 6, 15, 40, 60, 54, 23;
1, 7, 21, 70, 140, 189, 161, 65;
1, 8, 28, 112, 280, 504, 644, 520, 199;
1, 9, 36, 168, 504, 1134, 1932, 2340, 1791, 654;
1, 10, 45, 240, 840, 2268, 4830, 7800, 8955, 6540, 2296;
...

Crossrefs

Cf. A007476

Keyword

nonn,tabl

Author

Gary W. Adamson, Jan 02 2009

Status

approved

A144401 Padovan ( A000931) version of A038137: expansion of polynomials as antidiagonal: p(x,n)=1/(1-x-x^3)^n.

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 6, 3, 1, 5, 10, 13, 11, 4, 1, 6, 15, 24, 27, 18, 6, 1, 7, 21, 40, 55, 51, 30, 9, 1, 8, 28, 62, 100, 116, 94, 50, 13, 1, 9, 36, 91, 168, 231, 234, 171, 81, 19, 1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28, 1, 11, 55, 174, 402, 714, 987, 1065

Offset

1,5

Comments

Row sums are: 1, 2, 4, 9, 20, 44, 97, 214, 472, 1041, 2296, 5064, 11169, 24634, 54332 (cf. A008998).

These polynomials are sort of pseudo-combinations with the last element Padovan instead of one.

If you subtract the binomial triangle sequence you get:

{0},

{0, 0},

{0, 0, 0},

{0, 0, 0, 1},

{0, 0, 0, 2, 2},

{0, 0, 0, 3, 6, 3},

{0, 0, 0, 4, 12, 12, 5},

{0, 0, 0, 5, 20, 30, 23, 8},

{0, 0, 0, 6, 30, 60, 66, 42, 12}

Links

Table of n, a(n) for n=1..74.

Formula

p(x,n)=1/(1-x-x^3)^n; t(n,m)=anti_diagonal_expansion(p(x,n)).

Example

{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 2},
{1, 4, 6, 6, 3},
{1, 5, 10, 13, 11, 4},
{1, 6, 15, 24, 27, 18, 6},
{1, 7, 21, 40, 55, 51, 30, 9},
{1, 8, 28, 62, 100, 116, 94, 50, 13},
{1, 9, 36, 91, 168, 231, 234, 171, 81, 19},
{1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28},
{1, 11, 55, 174, 402, 714, 987, 1065, 879, 527, 208, 41},
{1, 12, 66, 230, 585, 1152, 1792, 2220, 2175, 1640, 906, 330, 60},
{1, 13, 78, 297, 825, 1782, 3072, 4278, 4815, 4320, 3006, 1539, 520, 88},
{1, 14, 91, 376, 1133, 2662, 5028, 7752, 9807, 10122, 8391, 5424, 2586, 816, 129}

Mathematica

Clear[f, b, a, g, h, n, t]; f[t_, n_] = 1/(1 - t - t^3)^n; a = Table[Table[SeriesCoefficient[Series[f[t, m], {t, 0, 30}], n], {n, 0, 30}], {m, 1, 31}]; b = Table[Table[a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}]; Flatten[b]

Crossrefs

Cf. A000931, A038137.

Keyword

nonn,tabl,uned

Author

Roger L. Bagula and Gary W. Adamson, Oct 03 2008

Status

approved

A175424 a(n) is the number of steps of iterations of {(((D_k^D_(k-1))^D_(k-2))^...)^D_1, where D_k is the k-th digit D of number r and k is the number of digits of number r in decimal expansion of r (A055642)} needed to reach a single-digit number starting at r = n, or a(n) = -1 if a single-digit number is never reached.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, -1, -1, 3, 3, 2, 1, 1, 1, 4, 3, -1, -1, 3, 3, 3, 1, 1, 2, 2, 3, 3, 3, 2, 2, 2, 1, 1, 2, 3, 2, 4, 3, 2, 3, 2, 1, 1, 3, 3, 2, 3, 3, 3, 3, 2, 1, 1, 4, 3, 3, 3, 3, 3, 2, 3, 1, 1, 3, 2, 3, 2, 3, 2, 2, 2, 1, 1, 3, 3, 3, 3, 2, 2, 2, 2

Offset

0,25

Comments

Conjecture: max(a(n)) = 4.

Links

Table of n, a(n) for n=0..99.

Example

For n = 33: a(33) = 4 because for the number 33 there are 4 steps of defined iteration: {3^3 = 27}, {7^2 = 49}, {9^4 = 6561}, {((1^6)^5)^6 = 1}.

Prog

(PARI) f(n) = if (n, my(d=digits(n), r=d[#d]); if (!vecmin(d), return(0)); forstep (k=#d-1, 1, -1, r = r^d[k]; ); r); \\ A175420
findpos(n, list) = {forstep (k=#list, 1, -1, if (list[k] == n, return (k)); ); return (0); }
a(n) = {my(list = List(n), nb = 0); while (n >= 10, n = f(n); my(k=findpos(n, list)); nb++; if (k, if (k==#list-1, if (list[k]<10, return (nb), return(-1)), return(-1)); ); listput(list, n); ); return (nb); } \\ Michel Marcus, Jan 20 2022

Crossrefs

Cf. A175419, A175420, A175421, A175422, A175423, A175425, A175426, A175427.

Keyword

sign,base

Author

Jaroslav Krizek, May 09 2010

Status

approved

A363394 Triangle read by rows. T(n, k) = A081658(n, k) + A363393(n, k) for k > 0 and T(n, 0) = 1.

1, 1, 1, 1, 2, -1, 1, 3, -3, -2, 1, 4, -6, -8, 5, 1, 5, -10, -20, 25, 16, 1, 6, -15, -40, 75, 96, -61, 1, 7, -21, -70, 175, 336, -427, -272, 1, 8, -28, -112, 350, 896, -1708, -2176, 1385, 1, 9, -36, -168, 630, 2016, -5124, -9792, 12465, 7936

Offset

0,5

Links

Table of n, a(n) for n=0..54.

Richard P. Stanley, A survey of alternating permutations, arXiv:0912.4240 [math.CO], 2009.

Index entries for sequences related to Euler numbers.

Formula

|T(n, k)| = (-1)^(n - k) * n! * [x^(n - k)][y^n] (sec(y) + tan(y)) / exp(x*y).

T(n, k) = [x^(n - k)] -2^(k-(0^k))*(Euler(k, 0) + Euler(k, 1/2)) / (x-1)^(k + 1).

For a recursion see the Python program.

Example

The triangle T(n, k) begins:
  [0] 1;
  [1] 1, 1;
  [2] 1, 2,  -1;
  [3] 1, 3,  -3,   -2;
  [4] 1, 4,  -6,   -8,   5;
  [5] 1, 5, -10,  -20,  25,   16;
  [6] 1, 6, -15,  -40,  75,   96,   -61;
  [7] 1, 7, -21,  -70, 175,  336,  -427,  -272;
  [8] 1, 8, -28, -112, 350,  896, -1708, -2176,  1385;
  [9] 1, 9, -36, -168, 630, 2016, -5124, -9792, 12465, 7936;

Maple

# Variant, computes abs(T(n, k)):
P := n -> n!*coeff(series((sec(y) + tan(y))/exp(x*y), y, 24), y, n):
seq(print(seq((-1)^(n - k)*coeff(P(n), x, n - k), k = 0..n)), n = 0..9);

Prog

(Python)
from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k == 0: return 1
    if k == n:
        p = k % 2
        return p - sum(T(n, j) for j in range(p, n - 1, 2))
    return (T(n - 1, k) * n) // (n - k)
for n in range(10): print([T(n, k) for k in range(n + 1)])

Crossrefs

Variants (row reversed): A109449, A247453.

Cf. A081658 (signed secant part), A363393 (signed tangent part), A000111 (main diagonal), A122045, A155585 (aerated main diagonal), A000667, A062162 (row sums of signless variant).

Keyword

sign,tabl

Author

Peter Luschny, Jun 06 2023

Status

approved