A055997 -id:A055997 - cp's OEIS frontend

A158464 Number of distinct squares in row n of Pascal's triangle.

1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1

Offset: 0

Reinhard Zumkeller, Mar 19 2009

nonn

It seems that some subset of terms in A055997 (A115599) gives the positions of 3's. E.g., we have a(9) = a(50) = a(289) = a(9801) = 3, but on the other hand, a(1682) = a(57122) = 2. - Antti Karttunen, Nov 03 2017

			a(8) = #{1} = 1;
a(9) = #{1,9,36} = 3.

Antti Karttunen, Table of n, a(n) for n = 0..11111
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000290, A055997 (A115599).

A158464 := proc(n)
    local sqset,k ;
    sqset := {} ;
    for k from 0 to n do
        P := binomial(n,k) ;
        if issqr(P) then
            sqset := sqset union {P} ;
        end if;
    end do:
    nops(sqset) ;
end proc:
seq(A158464(n),n=0..120) ; # R. J. Mathar, Jul 09 2016

CountDistinct /@ Table[Sqrt@ Binomial[n, k] /. k_ /; ! IntegerQ@ k -> Nothing, {n, 0, 104}, {k, 0, n}] (* Michael De Vlieger, Nov 03 2017 *)

A158464(n) = sum(k=0,n\2,issquare(binomial(n,k))); \\ Antti Karttunen, Nov 03 2017

a(n) = Sum_{k=0..floor(n/2)} A010052(A007318(n,k));

a(A000290(n)) > 1 for n > 1.

More terms from Antti Karttunen, Nov 03 2017

A175492 Numbers m >= 3 such that binomial(m,3) + 1 is a square.

Original entry on oeis.org

7, 10, 24, 26, 65, 13777

Offset: 1

Ctibor O. Zizka, May 29 2010

Related sequences:
Numbers m such that binomial(m,2) is a square: A055997;
Numbers m such that binomial(m,2) + 1 is a square: A006451 + 1;
Numbers m such that binomial(m,2) - 1 is a square: A072221 + 1;
Numbers m >= 3 such that binomial(m,3) is a square: {3, 4, 50} (Proved by A. J. Meyl in 1878);
Numbers m >= 4 such that binomial(m,4) + 1 is a square: {6, 7, 45, 55, ...};
Numbers m >= 7 such that binomial(m,7) + 1 is a square: {8, 10, 21, 143, ...}.
No additional terms up to 10 million. - Harvey P. Dale, Apr 04 2017
No additional terms up to 10 billion. - Jon E. Schoenfield, Mar 18 2022
No additional terms up to 1 trillion. The sequence is finite by Siegel's theorem on integral points. - David Radcliffe, Jan 01 2024

Wikipedia, Tetrahedral number

Cf. A006451, A007318, A055997, A072221.

Cf. A216268 (values of binomial(m, 3)) and A216269 (square roots of binomial(m, 3) + 1).

lst = {}; k = 3; While[k < 10^6, If[ IntegerQ@ Sqrt[ Binomial[k, 3] + 1], AppendTo[lst, k]]; k++ ]; lst (* Robert G. Wilson v, Jun 11 2010 *)
Select[Range[3,14000],IntegerQ[Sqrt[Binomial[#,3]+1]]&] (* Harvey P. Dale, Apr 04 2017 *)

isok(m) = (m>=3) && issquare(binomial(m,3)+1); \\ Michel Marcus, Mar 15 2022

from sympy import binomial
from sympy.ntheory.primetest import is_square
for m in range(3, 10**6):
    if is_square(binomial(m,3)+1):
        print(m) # Mohammed Yaseen, Mar 18 2022

A270889 Integers n such that the circular graph C_n has a square size deficiency.

Original entry on oeis.org

3, 6, 27, 150, 867, 5046, 29403, 171366, 998787, 5821350, 33929307, 197754486, 1152597603, 6717831126, 39154389147, 228208503750, 1330096633347, 7752371296326, 45184131144603, 263352415571286, 1534930362283107, 8946229758127350, 52142448186480987, 303908459360758566, 1771308307978070403

Offset: 0

John Rafael M. Antalan, Mar 25 2016

Define the size deficiency of a graph G as the number of edges needed to complete G. If G is a cycle graph C_n, this sequence gives the values of n for which C_n has a size deficiency which is a perfect square.

Colin Barker, Table of n, a(n) for n = 0..1000
J. R. M. Antalan and I. F. Callano, On the Size Deficiency of Cycle Graphs and Some Integer Sequences, Asian Journal of Mathematics and Computer Research, 11 (3) (2016),192-200.
Mathematics Stack Exchange, Solving the nonlinear Diophantine equation x2-3x=2y^2
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

a[0] = 3; a[1] = 6; a[n_] := a[n] = 6 a[n - 1] - a[n - 2] - 6; Table[a@ n, {n, 0, 24}] (* Michael De Vlieger, Mar 25 2016 *)
LinearRecurrence[{7,-7,1},{3,6,27},30] (* Harvey P. Dale, Jan 23 2019 *)

is(n)=issquare(n*(n-3)/2) \\ Charles R Greathouse IV, Mar 25 2016

a(n)=([0,1,0;0,0,1;1,-7,7]^n*[3;6;27])[1,1] \\ Charles R Greathouse IV, Mar 25 2016

Vec(3*(1-5*x+2*x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, Apr 03 2016

a(n+2) = 6*a(n+1) - a(n) - 6; a(0) = 3 , a(1) = 6.
G.f.: 3*(1-5*x+2*x^2)/((1-x)*(1-6*x+x^2)). - Joerg Arndt, Mar 25 2016
a(n) = 3 * A055997(n+1). - Joerg Arndt, Mar 25 2016
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>2. - Colin Barker, Apr 03 2016
a(n) = 3*(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4. - Colin Barker, Apr 03 2016

A275151 a(1) = 8; a(n) = 3a(n-1) + 2sqrt(2a(n-1)(a(n-1)-7)) - 7 for n > 1.

Original entry on oeis.org

8, 25, 128, 729, 4232, 24649, 143648, 837225, 4879688, 28440889, 165765632, 966152889, 5631151688, 32820757225, 191293391648, 1114939592649, 6498344164232, 37875125392729, 220752408192128, 1286639323760025, 7499083534368008, 43707861882448009, 254748087760320032, 1484780664679472169

Offset: 1

Manuel López Holgueras, Jul 17 2016

nonn

Related to A055997.
If we solve X^2 + (X+7)^2 = (X+N)^2 over the positive integers we find that the solutions belong to three sequences:
1) The first is a(1) = 7; a(n) = 3*a(n-1) + 2*sqrt(2*a(n-1)*(a(n-1)-7)) - 7 for n > 1: 7, 14, 63, 350, 2023, 11774, 68607, 399854, 2330503, 13583150, 79168383, 461427134, ... We observe that a(n) = 7*A055997(n).
2) The second is this sequence.
3) The third is a(1) = 9; a(n) = 3*a(n-1) + 2*sqrt(2*a(n-1)*(a(n-1)-7))-7 for n > 1: 9, 32, 169, 968, 5625, 32768, 190969, 1113032, 6487209, 37810208, 220374025, 1284433928, 7486229529, 43632943232, 254311429849, 1482235635848, ...
There is a property of the formula:
If y = 3*x + 2*sqrt(2*x*(x-q)) - q then x = 3*y - 2*sqrt(2*y*(y-q)) - q.
Let F(X) = 3*x - 2*sqrt(2*x*(x-7)) - 7.
  Let us use this function:
With the 1st sequence:   With the 2nd:         With the 3rd:
 F(2023)=350              F(729)=128            F(968)=169
 F(350)=63                F(128)=25             F(169)=32
 F(63)=14                 F(25)=8               F(32)=9
 F(14)=7                  F(8)=9                F(9)=8
 F(7)=14                  F(9)=8                F(8)=9

G. C. Greubel, Table of n, a(n) for n = 1..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 54, 56.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A055997.

I:=[8]; [n le 1 select I[n] else Floor(3*Self(n-1) +2*Sqrt(2*Self(n-1)*(Self(n-1) - 7)) -7): n in [1..30]]; // G. C. Greubel, Oct 07 2018

a:= proc(n) option remember; `if`(n=1, 8,
      3*a(n-1)+2*isqrt(2*a(n-1)*(a(n-1)-7))-7)
    end:
seq(a(n), n=1..25);

NestList[3 # + 2 Sqrt[2 # (# - 7)] - 7 &, 8, 23] (* Michael De Vlieger, Jul 18 2016 *)

m=30; v=concat([8], vector(m-1)); for(n=2, m, v[n] = floor(3*v[n-1] +2*sqrt(2*v[n-1]*(v[n-1]-7))-7)); v \\ G. C. Greubel, Oct 07 2018

a(n) = 3*a(n-1) + 2*sqrt(2*a(n-1)*(a(n-1)-7)) - 7, for n > 1, with a(1)=8.
Conjectures from Colin Barker, Jul 19 2016: (Start)
a(n) = (14 + (11-6*sqrt(2))*(3+2*sqrt(2))^n + (3-2*sqrt(2))^n*(11+6*sqrt(2)))/4.
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) for n > 3.
G.f.: x*(8 - 31*x + 9*x^2) / ((1-x)*(1 - 6*x + x^2)). (End)

A308085 a(n) is the least positive number k such that n(n-1)/2 + k(k-1)/2 is a square.

Original entry on oeis.org

1, 1, 2, 3, 4, 2, 6, 7, 1, 9, 10, 6, 3, 13, 14, 2, 16, 17, 18, 4, 6, 21, 3, 23, 24, 9, 5, 27, 13, 4, 30, 31, 2, 6, 34, 35, 5, 37, 38, 16, 7, 30, 23, 6, 44, 20, 46, 8, 16, 1, 7, 51, 12, 53, 9, 42, 23, 8, 58, 59, 60, 10, 27, 63, 9, 65, 20, 67, 11, 69, 6, 10, 72, 3, 44, 12, 21, 77, 11, 34, 80, 46

Offset: 1

J. M. Bergot and Robert Israel, Jun 05 2019

a(n) <= n-1 if n > 1, because (n-1)*(n-2)/2 + n*(n-1)/2 = (n-1)^2.
a(7*k+2) <= k and a(7*k+6) <= k+2, because (7*k+2)*(7*k+1)/2 + k*(k-1)/2 = (5*k+1)^2 and (7*k+6)*(7*k+5)/2 + (k+2)*(k+1)/2 = (5*k+4)^2.
From Bernard Schott, Jun 27 2019: (Start)
a(m) = 1 iff the triangular number t(m-1) = (m-1)*m/2 is a square, so iff m-1 is in A001108, or m in A055997.
a(m) = 2 iff the triangular number t(m-1) + 1 is a square, so iff m-1 is in A006451. (End)

			a(5) = 4 because 4*3/2 + 5*4/2 = 4^2 and none of 1*0/2 + 5*4/2, 2*1/2 + 5*4/2, 3*2/2 + 5*4/2 are squares.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000217, A055997 (a(n)=1).

f:= proc(n) local k;
  for k from 1 do
    if issqr((k*(k-1)+n*(n-1))/2) then return k fi
  od
end proc:
map(f, [$1..100]);

a(n) = {my(k=1); while (!issquare(n*(n-1)/2 + k*(k-1)/2), k++); k;} \\ Michel Marcus, Jun 27 2019

A373504 Triangular array: row n gives the coefficients T(n,k) of powers x^(2k) in the series expansion of ((b^n + b^(-n))/2)^2, where b = x + sqrt(x^2 + 1).

Original entry on oeis.org

1, 1, 1, 1, 4, 4, 1, 9, 24, 16, 1, 16, 80, 128, 64, 1, 25, 200, 560, 640, 256, 1, 36, 420, 1792, 3456, 3072, 1024, 1, 49, 784, 4704, 13440, 19712, 14336, 4096, 1, 64, 1344, 10752, 42240, 90112, 106496, 65536, 16384, 1, 81, 2160, 22176, 114048, 329472, 559104, 552960, 294912, 65536

Offset: 0

Clark Kimberling, Aug 03 2024

Related to Chebyshev polynomials of the first kind; see A123588.

			First 8 rows:
  1
  1    1
  1    4     4
  1    9    24     16
  1   16    80    128     64
  1   25   200    560    640    256
  1   36   420   1792   3456   3072   1024
  1   49   784   4704  13440  19612  14336  4096
The 4th polynomial is 1 + 9 x^2 + 24 x^4 + 16 x^6.

Cf. A000012 (col 0), A000290 (col 1), A002415 ((1/4)*col(2)), A112742 (col 2), A000302 (T(n,n)), A123588, A008310.
Row sums give A055997(n+1).
Triangle without column 0 gives A334009.

p:= proc(n) option remember; (b-> series(
      ((b^n+b^(-n))/2)^2, x, 2*n+1))(x+sqrt(x^2+1))
    end:
T:= (n, k)-> coeff(p(n), x, 2*k):
seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Aug 03 2024

t[n_] := ((x + Sqrt[x^2 + 1])^n + (x + Sqrt[x^2 + 1])^(-n))/2
u = Expand[Table[FullSimplify[Expand[t[n]]], {n, 0, 10}]^2]
v = Column[CoefficientList[u, x^2]] (* array *)
Flatten[v] (* sequence *)
T[n_, k_] := If[k==0, 1, 4^(k - 1)*(2*Binomial[n + k, 2*k] - Binomial[n + k -1, 2*k -1])]; Flatten[Table[T[n,k],{n,0,9},{k,0,n}]] (* Detlef Meya, Aug 11 2024 *)

T(n, k) = if (k=0) then 1, otherwise 4^(k - 1)*(2*binomial(n + k, 2*k) - binomial(n + k - 1, 2*k - 1)). - Detlef Meya, Aug 11 2024

cp's OEIS Frontend

A158464 Number of distinct squares in row n of Pascal's triangle.

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A175492 Numbers m >= 3 such that binomial(m,3) + 1 is a square.

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A270889 Integers n such that the circular graph C_n has a square size deficiency.

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A275151 a(1) = 8; a(n) = 3*a(n-1) + 2*sqrt(2*a(n-1)*(a(n-1)-7)) - 7 for n > 1.

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A308085 a(n) is the least positive number k such that n*(n-1)/2 + k*(k-1)/2 is a square.

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A373504 Triangular array: row n gives the coefficients T(n,k) of powers x^(2k) in the series expansion of ((b^n + b^(-n))/2)^2, where b = x + sqrt(x^2 + 1).

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A275151 a(1) = 8; a(n) = 3a(n-1) + 2sqrt(2a(n-1)(a(n-1)-7)) - 7 for n > 1.

A308085 a(n) is the least positive number k such that n(n-1)/2 + k(k-1)/2 is a square.