A362448 Triangle T(n,k) (n >= 0, 0 <= k <= n) read by rows: T(n,k) = 1 if the English names for n and k have a letter in common, otherwise 0.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1

Offset

0

Comments

Inspired by a problem in the GCHQ Puzzle Book.

References

GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See Problem 22, page 129.

Links

Michael S. Branicky, Table of n, a(n) for n = 0..11324 (150 rows)

Example

The triangle begins:
  1,
  1,1,
  1,1,1,
  1,1,1,1,
  1,1,1,1,1,
  1,1,0,1,1,1,
  0,0,0,0,0,1,1,
  1,1,0,1,0,1,1,1,
  1,1,1,1,0,1,1,1,1,
  1,1,0,1,0,1,1,1,1,1,
  1,1,1,1,0,1,0,1,1,1,1,
  ...

Mathematica

iName[n_]:=iName[n]=StringDelete[IntegerName[n, "Words"], Except[LetterCharacter]];
A362448row[n_]:=Table[Boole[StringContainsQ[iName[n], Characters[iName[k]]]], {k, 0, n}];
Array[A362448row, 15, 0] (* Paolo Xausa, Oct 17 2023 *)

Prog

(Python)
from num2words import num2words as n2w
def w(n): return [c for c in n2w(n).replace(" and", "") if c.isalpha()]
def T(n, k): return int(set(w(n)) & set(w(k)) != set())
print([T(n, k) for n in range(13) for k in range(n+1)]) # Michael S. Branicky, Apr 23 2023

Crossrefs

Cf. A362447.

Sequence in context: A014791 A014332 A014655 * A111024 A015586 A015700

Adjacent sequences: A362445 A362446 A362447 * A362449 A362450 A362451

Keyword

nonn,tabl,word

Author

N. J. A. Sloane, Apr 23 2023

Extensions

a(66) and beyond from Michael S. Branicky, Apr 23 2023

Status

approved