A307765 Number of palindromic hexagonal numbers with exactly n digits.

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

Offset

1,1

Comments

Number of terms in A054969 with exactly n digits.

Links

Table of n, a(n) for n=1..22.

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

Example

There are only two 4-digit hexagonal numbers that are palindromic, 3003 and 5995. Thus, a(4)=2.

Mathematica

A054969 = {0, 1, 6, 66, 3003, 5995, 15051, 66066, 617716, 828828, 1269621, 1680861, 5073705, 5676765, 1264114621, 5289009825, 6172882716, 13953435931, 1313207023131, 5250178710525, 6874200024786, 61728399382716, 602224464422206, 636188414881636, 1250444114440521, 16588189498188561, 58183932923938185, 66056806460865066, 67898244444289876, 514816979979618415, 3075488771778845703, 6364000440440004636, 15199896744769899151}; Table[Length[ Select[A054969, IntegerLength[#] == n || (n == 1 && # == 0) &]], {n, 19}]

Prog

(Python)
def afind(terms):
  m, n, c = 0, 1, 0
  while n <= terms:
    p = m*(2*m-1)
    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

Crossrefs

Cf. A000384, A054969, A054970, A082721, A263618, A307752.

Sequence in context: A002726 A119734 A073200 * A308077 A336344 A352609

Adjacent sequences: A307762 A307763 A307764 * A307766 A307767 A307768

Keyword

nonn,base,more

Author

Robert Price, Apr 27 2019

Extensions

a(20)-a(22) from Michael S. Branicky, Mar 01 2021

Status

approved