A307752 Number of n-digit palindromic pentagonal numbers.

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

Offset: 1

Author

Robert Price, Apr 26 2019

Keywords

• nonn
• base
• more

Comments

Number of n-digit terms in A002069.

Examples

			There are only two 4-digit pentagonal number that are palindromic, 1001 and 2882. Thus, a(4)=2.
		

Links

• G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy] See page 95.

Crossrefs

Cf. A000326, A002069, A028386, A059868, A263618, A307717.

Programs

Mathematica

A002069 = {0, 1, 5, 22, 1001, 2882, 15251, 720027, 7081807, 7451547, 26811862, 54177145, 1050660501, 1085885801, 1528888251, 2911771192, 2376574756732, 5792526252975, 5875432345785, 10810300301801, 264571020175462, 5292834004382925, 10808388588380801, 15017579397571051, 76318361016381367, 150621384483126051, 735960334433069537, 1003806742476083001, 1087959810189597801, 2716280733370826172};
Table[Length[Select[A002069, IntegerLength[#] == n  || (n == 1 && # == 0) &]], {n, 18}] (* Robert Price, Apr 26 2019 *)

Python

def afind(terms):
  m, n, c = 0, 1, 0
  while n <= terms:
    p = m*(3*m-1)//2
    s = str(p)
    if len(s) == n:
       if s == s[::-1]: c += 1
    else:
      print(c, end=", ")
      n, c = n+1, int(s == s[::-1])
    m += 1
afind(14) # Michael S. Branicky, Mar 01 2021

Extensions

a(19)-a(22) from Michael S. Branicky, Mar 01 2021

a(23)-a(40) from Bert Dobbelaere, Apr 15 2025