A345038 Triangle T(n,k) read by rows of the smallest centered n-gonal number greater than 1 that is also centered k-gonal, or 0 if none exists, for 1 <= k <= n.

2, 7, 3, 4, 31, 4, 0, 13, 85, 5, 16, 31, 31, 181, 6, 7, 7, 19, 61, 331, 7, 22, 43, 316, 841, 106, 547, 8, 121, 0, 361, 25, 22801, 169, 841, 9, 0, 91, 10, 0, 1891, 91, 253, 1225, 10, 11, 31, 31, 61, 31, 61, 2101, 361, 1711, 11, 67, 111, 166, 8581, 1156, 397, 6931, 179479609, 496, 2311, 12

Offset

1,1

Comments

The i-th centered j-gonal number is j*i*(i-1)/2 + 1. Thus if the p-th centered n-gonal number is also the q-th centered k-gonal number, then n*p*(p-1) = k*q*(q-1). Therefore T(n,k) = n*p*(p-1)/2 + 1 = k*q*(q-1)/2 + 1 iff n*p*(p-1) = k*q*(q-1) has a nontrivial positive integer solution. Otherwise T(n,k) = 0. It also implies that when T(n,k) = 0, T(r*n,r*k) = 0 for any positive integer r.

Links

Table of n, a(n) for n=1..66.

Index entries for sequences related to centered polygonal numbers

Formula

T(n,n) = n+1.

Example

The triangle begins:
   2;
   7,   3;
   4,  31,   4;
   0,  13,  85,   5;
  16,  31,  31, 181,   6;
   7,   7,  19,  61, 331,   7;
  22,  43, 316, 841, 106, 547,   8;
  ...

Prog

(PARI)
iszero(n, k)={if(issquare(n) && issquare(k) && n<>k, my(t=n-k); fordiv(t, d, my(p=(d+t/d)/2/sqrtint(n), q=(d-t/d)/2/sqrtint(k)); if(abs(p)!=1 && !frac(p) && !frac(q) && p%2==1 && q%2==1, return(0))); 1, 0)}
T(n, k)={my(g=gcd(n, k)); n/=g; k/=g; if(iszero(n, k), 0, for(p=2, oo, my(t=n*p*(p-1)/2); if(t%k==0 && ispolygonal(t/k, 3), return(t*g+1))))} \\ Andrew Howroyd, Jun 08 2021

Crossrefs

Cf. A101321, A189216.

Sequence in context: A350505 A326661 A326662 * A345237 A011049 A372131

Adjacent sequences: A345035 A345036 A345037 * A345039 A345040 A345041

Keyword

nonn,tabl

Author

Mohammed Yaseen, Jun 06 2021

Status

approved