A270270 The number of n-digit numbers in A270048.

4, 6, 17, 45, 131, 381, 1123, 3334, 9973, 29991, 90601, 274746, 835844, 2549874, 7797469, 23894630, 73358721, 225589420, 694745922, 2142444490, 6614766985, 20445300258, 63256499281, 195890524486, 607136782567, 1883199766658, 5845450449249, 18156369461770, 56429925440218

Offset

1,1

Comments

Conjecture: lim_{n -> infinity} a(n)/a(n-1) = sqrt(10).

(Similar to A265108, where we count the n-digit numbers of A264847, pluritriangular numbers.)

It is not possible to count some hundred-digit numbers without a "climbing algorithm" (see also Program and Links).

Links

Francesco Di Matteo, Table of n, a(n) for n = 1..300

Francesco Di Matteo, Calculating the A270270 terms

Example

a(1) = 4 because in A270048 there are 4 numbers with 1 digit (0, 1, 3, 6).
a(2) = 6 because in A270048 there are 8 numbers with 2 digits (10, 20, 32, 46, 62, 80).

Prog

(Python)
# init values
seq = [4]   # the output list
somme = [4] # the n-value list after the adding of the last seq term
last = [10] # last a(n) term, or the first k-digit number (10 with k=2)
# CLIMBING loop, put a bigger value if you want
for n in range (1, 30):
  k = (len(somme)+1)     # the digits number
  limit = 10**k          # the newest value to achieve
  base = (somme[-1]+1)*k # this is the (n+1)*k value
  hypo = seq[-1]*3       # to obtain rapidly the limite value
  rid = 10**(len(str(hypo))-1) # the reduction factor
  # Adjustment LOOP #
  s, p, m = 0, 0, 0
  while s < 1:
    diff_1 = (base + (base + (k*(hypo-1))))*float(hypo)/2
    tot = last[-1] + diff_1
    if tot < limit:
      p = 1
      if m == 1 and rid > 1:
        m = 0; rid = rid/10
      hypo = hypo + rid
    else:
      diff_2 = (base + (base + (k*(hypo-2))))*float(hypo-1)/2
      tot = last[-1] + diff_2
      if tot > limit:
        m = 1
        if p == 1 and rid > 1:
          p = 0; rid = rid/10
        hypo = hypo - rid
      else:
        s = 1    # escape value
  # lists updating
  seq.append(hypo)
  somme.append(somme[-1]+ hypo)
  last.append(last[-1]+ diff_1)
# if you want to prove the conjecture values, uncomment next line
#print(seq[-1], float(seq[-1])/seq[-2])
print(seq)

Crossrefs

Cf. A264847, A265108, A270048.

Sequence in context: A034492 A125691 A302878 * A211949 A079803 A061361

Adjacent sequences: A270267 A270268 A270269 * A270271 A270272 A270273

Keyword

nonn,base

Author

Francesco Di Matteo, Mar 14 2016

Status

approved