A027568 -id:A027568 - cp's OEIS frontend

A088126 Let f(n, x) = x+3x^2+6x^3+...+(n(n+1)/2)x^n; then a(n) = least x such that f(n, x) is a triangular number, or 0 if no such x exists.

1, 18, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Amarnath Murthy, Sep 26 2003

nonn

If k is a member of A027568 (both triangular and tetrahedral) then k = A000292(n-1) for some n and a(n) = 1.

Zero values are conjectures. I have searched for a(4) up to x = 10^7, a(5) up to x = 10^6 and the rest up to x = 10^4. (Wasserman)

Cf. A087702.

Edited and extended by David Wasserman, Jun 16 2005

A303615 Complete list of solutions to y^2 + y = x^3 - 525x + 10156; sequence gives x values.

Original entry on oeis.org

-29, -25, -20, -14, -5, 5, 14, 16, 20, 25, 49, 70, 79, 130, 250, 305, 400, 695, 1555, 1645, 18895

Offset: 1

Tomohiro Yamada, May 29 2018

This equation gives the elliptic curve (W46) studied by Stroeker and de Weger. This curve has rank 3 with generators P1 = (25, 112), P2 = (-20, 112) and P3 = (70, 562). The list gives all integer points in this curve.

This equation can be transformed to A000332(n) = A000579(m) by x = (15/2)m^2 - (75/2)m + 25 and y = (225/2)n^2 - (675/2)n + 112. Hence, A000332(n) = A000579(m) (n >= 4, m >= 6) has no integer solutions other than (n, m)= (4, 6) and (10, 10).

			a(6) = 5: 5^3 - 525 * 5 + 10156 = 7656 = 88 * 87.

Roelof J. Stroeker and Benjamin M. M. de Weger, Elliptic binomial diophantine equations, Math. Comp. 68 (1999), 1257-1281.

Cf. A029728 (the complete list of solutions x to y^2=x^3+17), A102461 (the complete list of solutions n to A000217(n) = A027568(m)).

A307174 Numbers that are both triangular and icosahedral.

Original entry on oeis.org

0, 1, 1128, 87153

Offset: 1

William C. Laursen, Jul 09 2019

Similar to the list of triangular and tetrahedral numbers (A027568). It would appear that the similar sequence of pentagonal-dodecahedral numbers contains only the trivial cases of 0 and 1.

Terms n*(n+1)/2 = m*(5*m^2-5*m+2)/2 corresponds to integral points (X,Y) = (5*m,5*n) on the elliptic curve Y^2 + Y = X^3 - 5*X^2 + 10*X, which can be computed efficiently. There are none besides those already listed. - Max Alekseyev, Feb 12 2024

Intersection of A000217 and A006564. - A.H.M. Smeets, Aug 19 2019

Cf. A000217, A006564, A027568, A309009.

Intersection[Accumulate[Range[0, 199]], Table[n (5n^2 - 5n + 2)/2, {n, 0, 99}]] (* Alonso del Arte, Jul 10 2019 *)

lista(nn) = for (n=0, nn, if (ispolygonal(k=n*(5*n^2 - 5*n + 2)/2, 3), print1(k, ", "))); \\ Michel Marcus, Jul 10 2019

Keyword "full" added by Max Alekseyev, Feb 12 2024

A307491 Numbers that are both centered triangular and tetrahedral.

Original entry on oeis.org

1, 4, 10, 4960, 428536

Offset: 1

Ilya Gutkovskiy, Apr 10 2019

If it exists, a(6) > 10^29. - Bert Dobbelaere, Apr 12 2019

Eric Weisstein's World of Mathematics, Centered Triangular Number
Eric Weisstein's World of Mathematics, Tetrahedral Number

Intersection of A000292 and A005448.

Cf. A003556, A027568, A128862, A131750, A307492.

A309009 Numbers that are both triangular and octahedral.

Original entry on oeis.org

0, 1, 6, 231

Offset: 1

William C. Laursen, Jul 05 2019

Similar to the list of triangular and tetrahedral numbers (A027568). It would appear that the similar sequence of pentagonal-dodecahedral numbers contains only the trivial cases 0 and 1.
Terms n*(n+1)/2 = m*(2*m^2+1)/3 corresponds to integral points (X,Y) = (12*m,36*n+18) on the elliptic curve Y^2 = X^3 + 72*X + 324, which can be computed efficiently. There are none besides those already listed. - Max Alekseyev, Feb 12 2024
Is 1156 the only nontrivial number that is both octahedral and square? - Kelvin Voskuijl, Jan 07 2025

Intersection of A000217 and A005900. - A.H.M. Smeets, Aug 19 2019

Cf. A000217, A005900, A027568, A307174.

triQ[n_] := IntegerQ[Sqrt[8n + 1]]; oct[n_] := n(2n^2 + 1)/3 ; Select[oct@Range[0, 10], triQ] (* Amiram Eldar, Jul 15 2019 *)

lista(nn) = for (n=0, nn, if (ispolygonal(k=n*(2*n^2 + 1)/3, 3), print1(k, ", "))); \\ Michel Marcus, Jul 06 2019

Keyword "full" added by Max Alekseyev, Feb 12 2024

A344280 Numbers that are both 10-gonal numbers (A001107) and 10-gonal pyramidal numbers (A007585).

Original entry on oeis.org

0, 1, 175, 368050005576

Offset: 1

Seiichi Manyama, May 17 2021

Intersection of A001107 and A007585.

Numberphile, Cannon Ball Numbers.

Cf. A001107, A007585, A027568, A344376, A344377.

for(k=0, 1e4, if(ispolygonal(m=k*(k+1)*(8*k-5)/6, 10), print1(m", ")))

A385584 a(n) is the number of pairs (p, t) such that p is a pyramidal number, t is a triangular number, p + t <= n and t <= p.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 7, 8, 8, 9, 9, 9, 10, 10, 10, 10, 12, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 19, 19, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 25, 26, 26, 27, 27, 27, 28, 29, 29, 29, 30, 30, 30, 30, 30, 31, 31, 31

Offset: 0

Robert G. Wilson v, Jul 03 2025

nonn

This sequence is nondecreasing.

			[n] #  solutions
----------------------------------------------------
[0] 1 [(0, 0)]
[1] 2 [(0, 0), (1, 0)]
[2] 3 [(0, 0), (1, 0), (1, 1)]
[3] 3 [(0, 0), (1, 0), (1, 1)]
[4] 4 [(0, 0), (1, 0), (1, 1), (4, 0)]
[5] 5 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1)]
[6] 5 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1)]
[7] 6 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1), (4, 3)]
[8] 6 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1), (4, 3)]
[9] 6 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1), (4, 3)]

David A. Corneth, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Cf. A000217, A000292, A001399, A001477, A027568.

def a(n: int) -> int:
    count = 0
    for p in range(n + 1):
        pv = p * (p + 1) * (p + 2) // 6
        if pv > n: break
        for t in range(n - p + 1):
            tv = t * (t + 1) // 2
            if pv + tv <= n and tv <= pv:
                count += 1
    return count
print([a(n) for n in range(74)])  # Peter Luschny, Jul 10 2025

a(n) = card({(t in A000217, p in A000292) : t <= p, t + p <= n}). - Peter Luschny, Jul 10 2025

New name and two terms (n=4 and n=20) corrected by Peter Luschny, Jul 10 2025

A216267 Numbers that are both tetrahedral and pronic.

Original entry on oeis.org

0, 20, 56, 7140, 1414910

Offset: 1

Alex Ratushnyak, Mar 15 2013

Intersection of A000292 and A002378.

The equation y*(y+1) = x*(x+1)*(x+2)/6 leads to an elliptic curve, which has a finite number of solutions, all of which are already listed. - Max Alekseyev, Dec 28 2024

Cf. A000292, A002378, A027568, A029549, A003556.

t = {}; Do[tet = n (n + 1) (n + 2)/6; s = Floor[Sqrt[tet]]; If[s^2 + s == tet, AppendTo[t, tet]], {n, 0, 1000}]; t (* T. D. Noe, Mar 18 2013 *)
With[{nn=50000},Intersection[Binomial[Range[0,nn]+2,3],Table[n(n+1),{n,nn}]]] (* Harvey P. Dale, Apr 04 2016 *)

def rootPronic(a):
    sr = 1<<33
    while a < sr*(sr+1):
      sr>>=1
    b = sr>>1
    while b:
        s = sr+b
        if a >= s*(s+1):
          sr = s
        b>>=1
    return sr
for i in range(1<<20):
      a = i*(i+1)*(i+2)//6
      t = rootPronic(a)
      if a == t*(t+1):
        print(a)

fini, full keywords added by Max Alekseyev, Dec 28 2024

A317053 Complete list of solutions to y^2 + y = x^3 - 525x + 10156 with y > 0; sequence gives positive y values.

Original entry on oeis.org

31, 74, 76, 87, 112, 121, 319, 562, 679, 1462, 3937, 5312, 7987, 18312, 61312, 66712, 2597287

Offset: 1

Tomohiro Yamada, Jul 20 2018

This equation gives the elliptic curve (W46) studied by Stroeker and de Weger. This curve has rank 3 with generators P1 = (25, 112), P2 = (-20, 112) and P3 = (70, 562). The list gives all integer points with y > 0 in this curve.
Each positive y corresponds to a negative solution -y - 1, so that the sequence gives all y values of solutions.
Some y values corresponds to three solutions. For y = 87, we have x = -25, 5 or 20. For y = 112, we have x = -20, -5 or 25. Any other value of y corresponds to a unique solution.
This equation can be transformed to A000332(n) = A000579(m) by x = (15/2)m^2 - (75/2)m + 25 and y = (225/2)n^2 - (675/2)n + 112. Hence, A000332(n) = A000579(m) (n >= 4, m >= 6) has no integer solutions other than (n, m)= (4, 6) and (10, 10).

			a(1) = 31: (-29)^3 - 525 * (-29) + 10156 = 996 = 31 * 32.

Roelof J. Stroeker and Benjamin M. M. de Weger, Elliptic binomial diophantine equations, Math. Comp. 68 (1999), 1257-1281.

Cf. A303615 (x values)

Cf. A029728 (the complete list of solutions x to y^2 = x^3 + 17), A102461 (the complete list of solutions n to A000217(n) = A027568(m)).

EllipticCurve([0, 0, 1, -525, 10156]).integral_points()

cp's OEIS Frontend

A088126 Let f(n, x) = x+3x^2+6x^3+...+(n(n+1)/2)x^n; then a(n) = least x such that f(n, x) is a triangular number, or 0 if no such x exists.

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A303615 Complete list of solutions to y^2 + y = x^3 - 525x + 10156; sequence gives x values.

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A307174 Numbers that are both triangular and icosahedral.

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A307491 Numbers that are both centered triangular and tetrahedral.

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A309009 Numbers that are both triangular and octahedral.

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A344280 Numbers that are both 10-gonal numbers (A001107) and 10-gonal pyramidal numbers (A007585).

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A385584 a(n) is the number of pairs (p, t) such that p is a pyramidal number, t is a triangular number, p + t <= n and t <= p.

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A216267 Numbers that are both tetrahedral and pronic.

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Python

Extensions

A317053 Complete list of solutions to y^2 + y = x^3 - 525x + 10156 with y > 0; sequence gives positive y values.

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