Tadayoshi Kamegai - cp's OEIS frontend

User: Tadayoshi Kamegai

Tadayoshi Kamegai has authored 3 sequences.

A372142 a(n) is the smallest prime p such that there exist exactly n distinct primes q where q < p and the representation of p in base q is a palindrome.

2, 3, 31, 443, 23053, 86677, 11827763, 27362989, 755827199, 1306369439

Offset: 0

Tadayoshi Kamegai, Apr 21 2024

This is a special case of A372141.
It need not be the case that a(n) is a palindrome in base 2, as 23053 is a counterexample.
For p > 3, one only needs to check q such that q^2 + 1 <= p else p = cc_q = c*(q+1), not prime for c != 1 and q != 2. A similar argument shows that p cannot have an even number of digits in base q, else it would be divisible by (q+1). - Michael S. Branicky, Apr 21 2024

			a(5) = 86677, as it is palindromic in base 2, 107, 113, 151, and 233, and no smaller number satisfies the property.

Cf. A372141, A002385, A002113.

Cf. A016041, A007500, A077798.

from math import isqrt
from sympy import sieve
from sympy.ntheory import digits
from itertools import islice
def ispal(v): return v == v[::-1]
def f(p): return sum(1 for q in sieve.primerange(1, isqrt(p-1)+1) if ispal(digits(p, q)[1:]))
def agen():
    adict, n = {0:2, 1:3}, 0
    for p in sieve:
        v = f(p)
        if v >= n and v not in adict:
            adict[v] = p
            while n in adict:
                yield adict[n]; del adict[n]; n += 1
print(list(islice(agen(), 6))) # Michael S. Branicky, Apr 21 2024

a(6) from Jon E. Schoenfield, Apr 21 2024
a(7) from Michael S. Branicky, Apr 21 2024
a(8) from Michael S. Branicky, Apr 22 2024
a(9) from Michael S. Branicky, Apr 24 2024

A372141 Primes p that are palindromic in some prime base q, where q < p.

Original entry on oeis.org

3, 5, 7, 13, 17, 23, 31, 41, 67, 71, 73, 83, 107, 109, 127, 151, 157, 173, 199, 233, 257, 271, 277, 307, 313, 353, 379, 409, 419, 421, 431, 443, 457, 499, 521, 523, 571, 587, 599, 601, 631, 643, 647, 653, 691, 701, 709, 719, 733, 743, 757, 787, 797, 809, 823, 829, 857, 863, 887

Offset: 1

Tadayoshi Kamegai, Apr 20 2024

If we remove either constraint of q < p or q being prime, then the sequence would be all prime numbers (A000040).
By definition it is a superset of A016041, and is a proper superset by construction (e.g., 13 is in the sequence).
Some terms have multiple bases that yield palindromic representations, the first being 31 (which is palindromic in both base 2 and base 5). The smallest prime p such that there exist n distinct primes less than p that give palindromic representations of p is A372142(n).

			11 is not in this sequence as its representation in base 2 is 1011, in base 3 is 102, in base 5 is 21, in base 7 is 14, none of which are palindromic.
1483 is in this sequence as its representation in base 37 is 131, which is palindromic.

Chai Wah Wu, Table of n, a(n) for n = 1..11371 (terms 1..1000 from Tadayoshi Kamegai)

Cf. A372142, A002385, A002113.

Cf. A007500, A016041, A077798.

a={}; For[i=1, i<=155, i++, flag=0; For[j=1, Prime[j] < Prime[i] && flag==0, j++, If[PalindromeQ[IntegerDigits[Prime[i], Prime[j]]], flag=1; AppendTo[a, Prime[i]]]]]; a (* Stefano Spezia, Apr 22 2024 *)

from sympy import sieve
from sympy.ntheory import digits
from itertools import islice
def ispal(v): return v == v[::-1]
def agen(): yield from (p for p in sieve if any(ispal(digits(p, q)[1:]) for q in sieve.primerange(1, p)))
print(list(islice(agen(), 60))) # Michael S. Branicky, Apr 20 2024

A369580 a(n) := f(n, n), where f(0,0) = 1/3, f(0,k) = 0 and f(k,0) = 3^(k-1) if k > 0, and f(n, m) = f(n, m-1) + f(n-1, m) + 3*f(n-1, m-1) otherwise.

Original entry on oeis.org

2, 16, 138, 1216, 10802, 96336, 861114, 7708416, 69072354, 619380496, 5557080938, 49879087296, 447852531986, 4022246329936, 36132550233498, 324645166734336, 2917340834679234, 26219438520320016, 235672871308226634, 2118552629658530496, 19046140604787232242, 171241206828437556816

Offset: 1

Tadayoshi Kamegai, Jan 26 2024

nonn

Take turns flipping a fair coin. The first to n heads wins. Sequence gives numerator of probability of first player winning. The denominator is .3^(2n-1).

It appears that a(n) for any n is divisible by 2^(A001511(n)).

Tadayoshi Kamegai, Table of n, a(n) for n = 1..100
Bartosz Sobolewski and Lukas Spiegelhofer, Decomposing the sum-of-digits correlation measure, arXiv:2411.07779 [math.NT], 2024. See p. 16.

Cf. A344576, A050231, A001850.

Cf. A001511 (see comments), A162326 (see formula).

def lis(n):
    table = [[0]*(n+1) for _ in range(n+1)]
    table[1][1] = 2
    for i in range(1, n+1) :
        table[i][0] = 3**(i-1)
    for i in range(1, n+1) :
        for j in range(1, n+1) :
            if (i == 1 and j == 1) :
                continue
            table[i][j] = table[i][j-1] + table[i-1][j] + 3*table[i-1][j-1]
    return [int(table[i][i]) for i in range(1, n+1)]

Limit_{n->oo} a(n)/3^(2n-1) = 1/2.
a(n) = Sum_{i>=n} Sum_{j=0..n-1} binomial(i-1,n-1)*binomial(i-1,j)*3^(2n-1)/2^(2i-1).
9*a(n) - a(n+1) = 2*A162326(n) (conjectured).
a(n) = 3^(2n-1)*A(n, n) where A(0, k) = 0 for k > 0, A(k, 0) = 1 for k >= 0 and A(n, m) = (A(n-1, m) + A(n, m-1) + A(n-1, m-1))/3.

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User: Tadayoshi Kamegai

A372142 a(n) is the smallest prime p such that there exist exactly n distinct primes q where q < p and the representation of p in base q is a palindrome.

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A372141 Primes p that are palindromic in some prime base q, where q < p.

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A369580 a(n) := f(n, n), where f(0,0) = 1/3, f(0,k) = 0 and f(k,0) = 3^(k-1) if k > 0, and f(n, m) = f(n, m-1) + f(n-1, m) + 3*f(n-1, m-1) otherwise.

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