Blaine J. Deal - cp's OEIS frontend

User: Blaine J. Deal

Blaine J. Deal has authored 2 sequences.

A114332 English spelling of n ends with a(n)-th letter of the alphabet.

15, 5, 15, 5, 18, 5, 24, 14, 20, 5, 14, 14, 5, 14, 14, 14, 14, 14, 14, 14, 25, 5, 15, 5, 18, 5, 24, 14, 20, 5, 25, 5, 15, 5, 18, 5, 24, 14, 20, 5, 25, 5, 15, 5, 18, 5, 24, 14, 20, 5, 25, 5, 15, 5, 18, 5, 24, 14, 20, 5, 25, 5, 15, 5, 18, 5, 24, 14, 20, 5, 25, 5, 15, 5, 18, 5, 24, 14, 20, 5

Offset: 0

Blaine J. Deal, Feb 06 2006

			'One' ends with 'e', which is the 5th letter of the alphabet, hence a(1)=5.
'Two' ends with 'o', which is the 15th letter of the alphabet, hence a(2)=15.

Tanar Ulric, Table of n, a(n) for n = 0..10000
B. Deal, Blaine's Puzzle Blog.
Chuck Gaydos, Sequence puzzle posting at Clifford Pickover forum.

Cf. A005606, A052360, A005589.

def a(n):
    if n == 0: return 15 # zerO
    if n%1000000 == 0: return 14 # millioN, billioN, ...
    r = n%100
    if r == 0: return 4 # hundreD, thousanD
    if r == 12: return 5 # twelvE
    if 10 <= r < 20: return 14 # teN, eleveN, thirteeN, ..., nineteeN
    return [25, 5, 15, 5, 18, 5, 24, 14, 20, 5][n%10] # *Y, *onE, ..., *ninE
print([a(n) for n in range(101)]) # Michael S. Branicky, Jan 19 2022

a(0)=15 prepended by Tanar Ulric, Jan 20 2022

A110883 Sum of consecutive digits in the decimal expansion of Pi.

Original entry on oeis.org

4, 5, 5, 6, 14, 11, 8, 11, 8, 8, 13, 17, 16, 16, 12, 5, 5, 11, 12, 10, 8, 8, 10, 7, 6, 11, 11, 5, 9, 16, 14, 5, 2, 10, 16, 12, 5, 10, 16, 8, 7, 15, 12, 12, 18, 12, 10, 12, 6, 1, 5, 13, 10, 2, 9, 16, 11, 13, 13, 8, 9, 14, 11, 5, 3, 7, 15, 9, 7, 10, 4, 6, 8, 10, 14, 8, 2, 8, 17, 18, 17, 14, 8

Offset: 1

Blaine J. Deal and Mark Nandor, Sep 19 2005

			a(1)=3+1 = 4, a(2)=1+4 = 5, a(3)=4+1 = 5, a(4)=1+5 = 6, a(5)=5+9 = 14

Paolo Xausa, Table of n, a(n) for n = 1..10000
Blaine Deal, Sequence puzzle posting at Clifford Pickover forum. [Broken link]
Blaine Deal, Home Page.
Albert Frank and Philippe Jacqueroux, International Contest, 2001. Item 25.

Cf. A000796.

listlength = 100; Table[IntegerDigits[IntegerPart[10^listlength Pi]][[i]] + IntegerDigits[IntegerPart[10^listlength (Pi - 3.0`100)]][[i]], {i, 1, listlength}]
(* Second program: *)
Total[Partition[First[RealDigits[Pi, 10, 100]], 2, 1], {2}] (* Paolo Xausa, Aug 22 2025 *)

a(n) = p(n) + p(n+1) where p(n) is the n-th digit of Pi (see sequence A000796).

a(n) = A000796(n) + A000796(n+1).

Typo in formula fixed by Zak Seidov, Jan 02 2014

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User: Blaine J. Deal

A114332 English spelling of n ends with a(n)-th letter of the alphabet.

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