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 A307765 Number of palindromic hexagonal numbers with exactly n digits. 2
 3, 1, 0, 2, 2, 2, 4, 0, 0, 3, 1, 0, 3, 1, 2, 1, 4, 1, 2, 1, 2, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Number of terms in A054969 with exactly n digits. LINKS 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 A198345 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

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Last modified June 24 02:56 EDT 2021. Contains 345415 sequences. (Running on oeis4.)