Gerry Leversha - cp's OEIS frontend

User: Gerry Leversha

Gerry Leversha has authored 2 sequences.

A235163 Number of positive integers with n digits in which adjacent digits differ by at most 1.

9, 26, 75, 217, 629, 1826, 5307, 15438, 44941, 130900, 381444, 1111926, 3242224, 9455987, 27583372, 80472698, 234799873, 685149328, 1999414181, 5835044495, 17029601028, 49702671494, 145066398937, 423412132499, 1235854038791, 3607255734629, 10529101874491

Offset: 1

Gerry Leversha, Jan 04 2014

			a(2) = 26: 10, 11, 12, 21, 22, 23, 32, 33, 34, 43, 44, 45, 54, 55, 56, 65, 66, 67, 76, 77, 78, 87, 88, 89, 98, 99.

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-10,1,6,-1).

Cf. A032981, A090994, A126364 (allowing leading zeros).

u:= proc(n, r) option remember; `if`(n=1, `if`(r=0, 0, 1),
      add(`if`(r+i in [$0..9], u(n-1, r+i), 0), i=-1..1))
    end:
a:= n-> add(u(n, r), r = 0..9):
seq(a(n), n=1..30);  # Alois P. Heinz, Jan 12 2014

CoefficientList[Series[-x*(3*x^4-18*x^3-9*x^2+28*x-9)/(x^5-6*x^4-x^3+10*x^2-6*x+1),{x,0,30}],x]//Rest (* Harvey P. Dale, Aug 13 2019 *)

from functools import cache
@cache
def u(n, r):
    if r < 0 or r > 9: return 0
    if n == 1: return (r > 0)
    return u(n-1, r-1) + u(n-1, r) + u(n-1, r+1)
def a(n): return sum(u(n, r) for r in range(10))
print([a(n) for n in range(1, 28)]) # Michael S. Branicky, Sep 26 2021

a(n) = Sum_{r=0..9} u(n,r) where u(n,r) = 0 if r<0 or r>9, u(1,0) = 0, u(1,r) = 1 for 1<=r<=9, and otherwise u(n,r) = u(n-1,r-1) + u(n-1,r) + u(n-1,r+1).

G.f.: -x*(3*x^4-18*x^3-9*x^2+28*x-9)/(x^5-6*x^4-x^3+10*x^2-6*x+1). - Alois P. Heinz, Jan 12 2014

A126933 Quotients arising from sequence A053312.

Original entry on oeis.org

1, 3, 14, 132, 691, 1908, 16579, 47352, 414301, 1183713, 5474669, 27151397, 135646011, 678174568, 6442602909, 18480090517, 85533990571, 424236721848, 4026815626549, 11550150977337, 53458791308981, 265147974756053, 1324666882885839, 6622797918981982, 62916043734881616, 329481245744393933

Offset: 1

Gerry Leversha, Mar 18 2007

Take the decimal number formed by the first n digits of A023396 in reverse order and divide by 2^n.
The sequence A053312 gives n-digit numbers consisting entirely of 1s and 2s which are divisible by 2^n. The quotients upon division form the present sequence. The parity of the n-th term here determines the next term in A023396; if odd, it is a 1 and if even, a 2.
This was set as a problem in the All Union Mathematical Olympiad of 1971 and can be found in the reference cited here.

			a(4) = A053312(4) / 2^4 = 2112 / 16 = 132. - _David A. Corneth_, Jun 11 2020

J. B. Tabov and P. J. Taylor, Methods of Problem Solving, Book 1, Australian Mathematics Trust, 1996.

David A. Corneth, Table of n, a(n) for n = 1..1431 (terms <= 10^1000)
Index to sequences related to Olympiads.

Cf. A023396, A053312.

from itertools import count, islice
def A126933_gen(): # generator of terms
    a, b = 2, 10
    for n in count(1):
        a+=b if (c:=a>>n)&1 else b<<1
        b *= 10
        yield c
A126933_list = list(islice(A126933_gen(),20)) # Chai Wah Wu, Mar 16 2023

a(n) < 0.3 * 5^n. - David A. Corneth, Jun 11 2020

Name changed and other minor edits by Ray Chandler, Jun 17 2020

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User: Gerry Leversha

A235163 Number of positive integers with n digits in which adjacent digits differ by at most 1.

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