A307790 Number of palindromic heptagonal numbers with exactly n digits.

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

Offset: 1

Author

Robert Price, Apr 28 2019

Keywords

• nonn
• base
• more

Comments

Number of terms in A054910 with exactly n digits.

Examples

			There are only two 4-digit heptagonal numbers that are palindromic, 3553 and 4774. 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. A000566, A054910, A054971, A059869, A307765, A307766.

Programs

Mathematica

A054910 = {0, 1, 7, 55, 616, 3553, 4774, 60606, 848848, 4615164, 5400045, 6050506, 7165445617, 62786368726, 65331413356, 73665056637, 91120102119, 345546645543, 365139931563, 947927729749, 3646334336463, 7111015101117, 17685292586717, 19480809790808491, 615857222222758516, 1465393008003935641, 8282802468642082828, 15599378333387399551, 20316023422432061302}; Table[Length[Select[A054910, IntegerLength[#] == n || (n == 1 && # == 0) &]], {n, 19}]

Python

def afind(terms):
  m, n, c = 0, 1, 0
  while n <= terms:
    p = m*(5*m-3)//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(20)-a(22) from Michael S. Branicky, Mar 01 2021