A057569 -id:A057569 - cp's OEIS frontend

A133101 Expansion of f(x^2, x^3) in powers of x where f(, ) is Ramanujan's general theta function.

1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Michael Somos, Sep 11 2007

nonn

			G.f. = 1 + x^2 + x^3 + x^9 + x^11 + x^21 + x^24 + x^38 + x^42 + x^60 + x^65 + ...
G.f. = q + q^81 + q^121 + q^361 + q^441 + q^841 + q^961 + q^1521 + q^1681 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Index entries for characteristic functions

Cf. A057569, A113428.

a[ n_] := SeriesCoefficient[ QPochhammer[ -x^2, x^5] QPochhammer[ -x^3, x^5] QPochhammer[ x^5], {x, 0, n}]; (* Michael Somos, Oct 31 2015 *)

{a(n) = if( n<0, 0, polcoeff( prod(k=1,n, 1 + x^k*[ -1, 0, 1, 1, 0][k%5 + 1], 1 + x * O(x^n)), n))};

The characteristic function of A057569.
Euler transform of period 10 sequence [ 0, 1, 1, -1, -1, -1, 1, 1, 0, -1, ...].
G.f.: Prod_{k>0} (1 - x^(5*k)) * (1 + x^(5*k - 2)) * (1 + x^(5*k - 3)) = Sum_{k in Z} x^((5*k^2 + k) / 2).
a(n) = abs(A113428(n)).

A165720 Integers of the form k*(k+11)/10.

Original entry on oeis.org

6, 8, 18, 21, 35, 39, 57, 62, 84, 90, 116, 123, 153, 161, 195, 204, 242, 252, 294, 305, 351, 363, 413, 426, 480, 494, 552, 567, 629, 645, 711, 728, 798, 816, 890, 909, 987, 1007, 1089, 1110, 1196, 1218, 1308, 1331, 1425, 1449, 1547, 1572, 1674, 1700, 1806

Offset: 1

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Integers of the form k + k*(k+1)/10 = k + A000217(k)/5. For k see A047208, for A000217(k)/5 see A057569. - R. J. Mathar, Sep 25 2009
Are all terms composite numbers?
Yes. They are alternately of the form (h+2)*(5*h-1)/2 and h*(5*h+11)/2, with h>0. - Bruno Berselli, Dec 22 2016

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A047208, A057569, A165717, A165718, A165719, A279895.

Select[k = Range[0, 130]; k (k + 11)/10, IntegerQ] (* Bruno Berselli, Dec 22 2016 *)

From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = 5*(2*n^2 + 10*n + 3)/16 - 3*(-1)^n*(5 + 2*n)/16.
G.f.: x*(-6 - 2*x + 2*x^2 + x^3) / ((1 + x)^2*(x - 1)^3). (End)
Sum_{n>=1} 1/a(n) = 514/363 - 2*Pi*sqrt(1+2/sqrt(5))/11. - Amiram Eldar, Jul 26 2024

Definition simplified by R. J. Mathar, Sep 25 2009

Corrected A-number in my comment - R. J. Mathar, Oct 30 2009

A193470 Square array A(n,k) (n>=1, k>=0) read by antidiagonals: A(n,0) = 0 and A(n,k) is the least integer > A(n,k-1) that can be expressed as a triangular number divided by n.

Original entry on oeis.org

0, 0, 1, 0, 3, 3, 0, 1, 5, 6, 0, 7, 2, 14, 10, 0, 2, 9, 5, 18, 15, 0, 1, 3, 30, 7, 33, 21, 0, 3, 6, 9, 34, 12, 39, 28, 0, 15, 4, 11, 11, 69, 15, 60, 36, 0, 4, 17, 13, 13, 21, 75, 22, 68, 45, 0, 1, 5, 62, 15, 20, 24, 124, 26, 95, 55, 0, 5, 12, 17, 66, 30, 35, 38, 132, 35, 105, 66

Offset: 1

Peter Luschny, Jul 27 2011

			n\k  0   1   2    3    4     5     6     7
------------------------------------------
1 |  0   1   3    6   10    15    21    28    A000217
2 |  0   3   5   14   18    33    39    60    A074378
3 |  0   1   2    5    7    12    15    22    A001318
4 |  0   7   9   30   34    69    75   124    A154260
5 |  0   2   3    9   11    21    24    38    A057569
6 |  0   1   6   11   13    20    35    46    A154293
7 |  0   3   4   13   15    30    33    54    A057570
8 |  0  15  17   62   66   141   147   252    A157716

Cf. A061782, A011772.

A193470_rect := proc(n,k) local j,i,L; L := NULL; j := 0; while nops([L]) < k do add(i/n, i=1..j); if type(%,integer) then L := L,% fi; j := j+1 od; L end:
seq(print(A193470_rect(n, 12)),n = 1..8);

a[, 0] = 0; a[n, k_] := a[n, k] = For[j = a[n, k-1]+1, True, j++, If[Reduce[m > 0 && j == m(m+1)/(2n), m, Integers] =!= False, Return[j]]]; Table[a[n-k, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 07 2016 *)

A306690 Number of ways to write n as u^4 + (v(v+1)/2)^2 + (x(3x+1)/2)^2 + (y(5y+1)/2)^2 + (z(9z+1)/2)^2, where u and v are nonnegative integers and x,y,z are integers.

Original entry on oeis.org

1, 3, 3, 1, 2, 5, 4, 1, 1, 4, 6, 4, 1, 3, 5, 2, 2, 6, 6, 3, 5, 8, 6, 2, 2, 9, 14, 9, 2, 9, 14, 7, 2, 5, 10, 12, 9, 6, 8, 7, 5, 9, 10, 6, 4, 10, 10, 4, 1, 4, 12, 11, 5, 4, 10, 6, 5, 5, 5, 8, 8, 7, 8, 5, 1, 7, 11, 5, 3, 5, 8, 5, 3, 1, 6, 10, 4, 4, 6, 4, 1, 8, 8, 8, 6, 7, 11, 6, 1, 2, 10, 8, 3, 2, 7, 6, 1, 4, 8, 9, 4

Offset: 0

Zhi-Wei Sun, Mar 05 2019

nonn

Conjecture 1: a(n) > 0 for any nonnegative integer n.
Conjecture 2: Each n = 0,1,2,... can be written as f(u,v,x,y,z) with u,v,x,y,z integers, where f is any of the following polynomials: u^4 + (v*(v+1)/2)^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(5z+3)/2)^2, u^4 + (v*(v+1)/2)^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(3z+2))^2, (u*(u+1)/2)^2 + (v*(3v+1)/2)^2 + (x*(5x+1)/2)^2 + (y*(5y+3)/2)^2 + (z*(3z+2))^2, (u*(u+1)/2)^2 + (v*(3v+1)/2)^2 + (x*(5x+1)/2)^2 + (y*(5y+3)/2)^2 + (z*(4z+3))^2, (u*(u+1)/2)^2 + (v*(3v+1)/2)^2 + (x*(5x+1)/2)^2 + (y*(5y+3)/2)^2 + (z*(9z+7)/2)^2.
We have verified Conjectures 1 and 2 for n up to 2*10^6 and 10^6 respectively.

			a(8) = 1 with 8 = 0^4 + (0*(0+1)/2)^2 + (1*(3*1+1)/2)^2 + ((-1)*(5*(-1)+1)/2)^2 + (0*(9*0+1)/2)^2.
a(2953) = 1 with 2953 = 6^4 + (8*(8+1)/2)^2 + (0*(3*0+1)/2)^2 + (0*(5*0+1)/2)^2 + (2*(9*2+1)/2)^2.
a(8953) = 1 with 8953 = 2^4 + (7*(7+1)/2)^2 + (6*(3*6+1)/2)^2 + ((-1)*(5*(-1)+1)/2)^2 + ((-4)*(9*(-4)+1)/2)^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000118, A000217, A000290, A000583, A001318, A057569, A306606, A306614.

t[x_]:=t[x]=(x(x+1)/2)^2; f[x_]:=f[x]=(x(5x+1)/2)^2; g[x_]:=g[x]=(x(9x+1)/2)^2; SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; PQ[n_]:=PQ[n]=SQ[n]&&SQ[24*Sqrt[n]+1];
tab={};Do[r=0;Do[If[PQ[n-k^4-t[x]-f[y]-g[z]],r=r+1],{k,0,n^(1/4)},{x,0,(Sqrt[8*Sqrt[n-k^4]+1]-1)/2},{y,-Floor[(Sqrt[40*Sqrt[n-k^4-t[x]]+1]+1)/10],(Sqrt[40*Sqrt[n-k^4-t[x]]+1]-1)/10},{z,-Floor[(Sqrt[72*Sqrt[n-k^4-t[x]-f[y]]+1]+1)/18],(Sqrt[72*Sqrt[n-k^4-t[x]-f[y]]+1]-1)/18}];tab=Append[tab,r],{n,0,100}];Print[tab]

A364554 a(n) = number of primes of the form T(k)/n, for some k, where T(k)=A000217(k) is a triangular number.

Original entry on oeis.org

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

Offset: 1

Lamine Ngom, Jul 28 2023

nonn

Implementing a suggestion in the comment section of sequences A154296, ..., A154304, this sequence computes the number of primes of the form T(k)/n.
Equivalently, number of primes p such that 8*n*p+1 is a perfect square.
Let's consider, for all primes p, the set of linear recurrences {b(m)} defined as follows:
If p = 2, then {b(m)} = A074378 (numbers of the form x*(4*x -+ 1)); otherwise, b(m) = b(m-1) + 2*b(m-2) - 2*b(m-3) - b(m-4) + b(m-5) with initial terms b(0) = 0, b(1) = (p-1)/2, b(2) = (p+1)/2, b(3) = 2*p-1 and b(4) = 2*p+1. Numbers of the form x*(p*x -+ 1)/2.
Then a(n) = number of sequences {b(m)} in which n is a term.
This implies that:
i) for any n, the largest prime of the form T(k)/n is at most 2*n+1;
ii) if n is prime, then a(n) < 4. (3 and 5 are the only primes p such that a(p) = 3; primes p such that a(p) = 0 are A109998.)
Deeper in the examination of these results, we notice that the set of primes p of the form T(k)/n arises from the factorization of n. This set is exactly all primes p of the form (2*r -+ 1)/d or (r -+ 1)/(2*d), where d is some divisor of n and r is the ratio n/d. (Proof is welcome.)
Indices k of corresponding triangular numbers T(k) such that T(k) = n*p are then:
  2*r     if p = (2*r + 1)/d,
  2*r - 1 if p = (2*r - 1)/d,
    r     if p = (r + 1)/(2*d),
    r - 1 if p = (r - 1)/(2*d).
And pluging the value of p in the equivalent definition, the expression 8*n*p+1 yields respectively to following perfect squares: (4*r+1)^2, (4*r-1)^2, (2*r+1)^2 and (2*r-1)^2.

			a(15) = 4 since there are exactly 4 triangular numbers T(k) such that T(k) = 15*p, with p prime.
T(9)/15 = 45/15 = 3, T(14)/15 = 105/15 = 7, T(29)/15 = 435/15 = 29 and T(30)/15 = 465/15 = 31.
a(17) = 0 since there is no triangular number T(k) such that T(k) = 17*p, with p prime.

Cf. A000217, A154296, ..., A154304, A074378, A001318, A057569, A057570, A109998.

Cf. A364555 (indices of 0's).

Conjecture: a(n) = number of primes in the union of sets {(2*r -+ 1)/d; (r -+ 1)/(2*d)}, with d divisor of n and r = n/d.

A377224 Number of ways to write n as x(5x+1) + y(5y+1)/2 + z(5z+1)/2, where x,y,z are integers with y(5y+1) <= z(5z+1).

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 3, 1, 2, 3, 2, 3, 2, 2, 1, 3, 1, 3, 4, 1, 3, 2, 4, 2, 6, 2, 4, 5, 4, 3, 5, 3, 3, 4, 2, 2, 4, 1, 3, 3, 3, 3, 7, 1, 6, 6, 6, 3, 8, 4, 3, 7, 3, 7, 4, 4, 2, 4, 1, 5, 6, 1, 6, 7, 4, 4, 9, 6, 5, 8, 3, 6, 5, 3, 4, 5, 3, 3, 4, 1, 9, 6, 5, 3, 9, 5, 6, 9, 6, 8, 10, 3, 3, 9, 4, 7, 7, 4, 7, 5, 4

Offset: 0

Zhi-Wei Sun, Nov 13 2024

nonn

Conjecture 1: a(n) = 0 only for n = 1. Also, a(n) = 1 only for n = 0, 2, 3, 5, 7, 14, 16, 19, 37, 43, 58, 61, 79.
This has been verified for n <= 2*10^6.
Conjecture 2: Let N be the set of all nonnegative integers. Then
{x*(5*x+1) + y*(5*y+1)/2 + 5*z*(5*z+1)/2: x,y,z are integers} = N\{1,5},
{x*(5*x+1) + y*(5*y+1)/2 + 3*z*(5*z+1)/2: x,y,z are integers} = N\{1,5,32},
{x*(5*x+1) + y*(5*y+1)/2 + 2*z*(5*z+1): x,y,z are integers} = N\{1,5,70},
and
{x*(5*x+1)/2 + y*(5*y+1)/2 + z*(5*z+1)/2: x,y,z are integers} = N\{1,10,19,94}.
Conjecture 3: We have
{x*(5*x+3) + y*(5*y+3)/2 + 3*z*(5*z+3)/2: x,y,z are integers} = N\{31,77},
{x*(5*x+3) + y*(5*y+3)/2 + 5*z*(5*z+3): x,y,z are integers} = N\{10,16},
and
{x*(5*x+3)/2 + y*(5*y+3)/2 + 5*z*(5*z+3)/2: x,y,z are integers} = N\{3,15,29,44}.

			a(14) = 1 with 14 = 0*(5*0+1) + 1*(5*1+1)/2 + 2*(5*2+1)/2.
a(37) = 1 with 37 = (-1)*(5*(-1)+1) + (-2)*(5*(-2)+1)/2 + 3*(5*3+1)/2.
a(58) = 1 with 58 = (-2)*(5*(-2)+1) + (-1)*(5*(-1)+1)/2 + (-4)*(5*(-4)+1)/2.
a(79) = 1 with 79 = -4*(5*(-4)+1) + 0*(5*0+1)/2 + 1*(5*1+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162 (2016), 190-211.
Zhi-Wei Sun, Universal sums of three quadratic polynomials, Sci. China Math. 63 (2020), 501-520.
Zhi-Wei Sun, New results similar to Lagrange's four-square theorem, arXiv:2411.14308 [math.NT], 2024.

Cf. A057569, A085787, A306383.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[40(n-x(5x+1)-y(5y+1)/2)+1],r=r+1],{x,-Floor[(Sqrt[20n+1]+1)/10],(Sqrt[20n+1]-1)/10},{y,-Floor[(Sqrt[20(n-x(5x+1))+1]+1)/10],Floor[(Sqrt[20(n-x(5x+1))+1]-1)/10]}];tab=Append[tab,r],{n,0,100}];Print[tab]

cp's OEIS Frontend

A133101 Expansion of f(x^2, x^3) in powers of x where f(, ) is Ramanujan's general theta function.

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A165720 Integers of the form k*(k+11)/10.

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Extensions

A193470 Square array A(n,k) (n>=1, k>=0) read by antidiagonals: A(n,0) = 0 and A(n,k) is the least integer > A(n,k-1) that can be expressed as a triangular number divided by n.

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A306690 Number of ways to write n as u^4 + (v*(v+1)/2)^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(9z+1)/2)^2, where u and v are nonnegative integers and x,y,z are integers.

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A364554 a(n) = number of primes of the form T(k)/n, for some k, where T(k)=A000217(k) is a triangular number.

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A377224 Number of ways to write n as x*(5*x+1) + y*(5*y+1)/2 + z*(5*z+1)/2, where x,y,z are integers with y*(5*y+1) <= z*(5*z+1).

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A306690 Number of ways to write n as u^4 + (v(v+1)/2)^2 + (x(3x+1)/2)^2 + (y(5y+1)/2)^2 + (z(9z+1)/2)^2, where u and v are nonnegative integers and x,y,z are integers.

A377224 Number of ways to write n as x(5x+1) + y(5y+1)/2 + z(5z+1)/2, where x,y,z are integers with y(5y+1) <= z(5z+1).