A018247 -id:A018247 - cp's OEIS frontend

A318140 The 10-adic integer f = ...6510474330 satisfying f^2 + 1 = a, a^2 + 1 = b, b^2 + 1 = c, c^2 + 1 = d, d^2 + 1 = e, and e^2 + 1 = f.

0, 3, 3, 4, 7, 4, 0, 1, 5, 6, 7, 8, 5, 7, 3, 3, 2, 6, 6, 6, 1, 6, 1, 5, 8, 4, 0, 3, 9, 4, 0, 2, 6, 0, 9, 4, 0, 4, 2, 2, 1, 7, 7, 0, 7, 8, 9, 7, 0, 6, 3, 0, 3, 5, 1, 6, 5, 2, 4, 3, 5, 6, 0, 6, 0, 4, 6, 4, 8, 1, 8, 6, 2, 8, 0, 7, 9, 4, 1, 2, 6, 2, 9, 1, 7, 5, 2, 0, 6, 4, 5, 4, 3, 7, 4, 4, 9, 8, 0, 7, 6, 7, 0, 1, 3, 2, 8, 4, 7, 5, 5, 6, 3, 9, 7, 0, 1, 8, 0, 8, 3, 7, 3, 5, 8, 6, 2, 4, 6, 6, 9, 5, 0, 7, 8, 6, 2, 7, 7, 4, 8, 2, 1, 7, 1, 4, 5, 5, 9, 3

Offset: 0

Patrick A. Thomas, Aug 19 2018

Data generated using MATLAB.

			330^2 + 1 == 901 (mod 10^3), 901^2 + 1 == 802 (mod 10^3), 802^2 + 1 == 205 (mod 10^3), 205^2 + 1 == 26 (mod 10^3), 26^2 + 1 == 677 (mod 10^3), and 677^2 + 1 == 330 (mod 10^3), so 0 3 3 comprise the sequence's first three terms.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A018247, A318135 (a), A318136 (b), A318137 (c), A318138 (d), A318139 (e).

A055620 Digits of an idempotent 6-adic number.

Original entry on oeis.org

4, 4, 3, 5, 0, 2, 4, 3, 3, 3, 0, 4, 0, 0, 4, 1, 4, 2, 4, 3, 0, 0, 0, 5, 0, 3, 0, 0, 0, 2, 4, 1, 2, 2, 5, 1, 3, 3, 1, 5, 4, 2, 2, 4, 1, 5, 3, 5, 4, 3, 0, 3, 1, 5, 3, 2, 2, 5, 2, 1, 0, 0, 3, 0, 0, 1, 2, 3, 2, 4, 0, 1, 0, 1, 5, 4, 4, 5, 1, 3, 5, 4, 2, 5, 4, 0, 5, 1, 2, 0, 5, 4, 5, 3, 1, 5, 2, 1, 3, 3, 2, 3, 3, 5, 3

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), Jun 04 2000

( a(0) + a(1)*6 + a(2)*6^2 + ... )^k = a(0) + a(1)*6 + a(2)*6^2 + ... for each k. Apart from 0 and 1 in base 6 there are only 2 numbers with this property. For the other see A054869.

			(a(0) + a(1)*6 + a(2)*6^2 + a(3)*6^3)^2 == (a(0) + a(1)*6 + a(2)*6^2 + a(3)*6^3) mod 6^4 because 1478656 == 1216 (mod 1296).

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Kenny Lau, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

first(p)=Vecrev(digits(lift(Mod(4,6^p)^3^p), 6)) \\ Charles R Greathouse IV, Nov 01 2022

n=10000;res=pow((3**n+1)//2,n,3**n)*2**n
for i in range(n):print(i,res%6);res//=6
# Kenny Lau, Jun 09 2018

If A is the 6-adic number, A == 4^(3^n) mod 6^n. - Robert Dawson, Oct 28 2022

A259468 Digits of an idempotent 12-adic number.

Original entry on oeis.org

4, 5, 8, 3, 11, 1, 6, 11, 1, 2, 11, 2, 5, 4, 3, 11, 7, 6, 2, 11, 6, 8, 9, 6, 1, 10, 3, 10, 1, 8, 10, 3, 3, 1, 10, 8, 6, 4, 6, 11, 2, 0, 4, 0, 3, 8, 9, 9, 5, 3, 1, 0, 9, 7, 11, 8, 7, 6, 5, 2, 5, 1, 10, 2, 11, 0, 2, 7, 6, 6, 5, 4, 7, 2, 0, 5, 11, 7, 8, 7, 2, 6

Offset: 0

N. J. A. Sloane, Jul 02 2015

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Eric M. Schmidt, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 No. 3, (1968), 173-179.

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

def a_list(n) : return crt(1, 0, 3^n, 4^n).digits(12) # Eric M. Schmidt, Jul 09 2015

More terms from Eric M. Schmidt, Jul 09 2015

A259469 Digits of an idempotent 12-adic number.

Original entry on oeis.org

9, 6, 3, 8, 0, 10, 5, 0, 10, 9, 0, 9, 6, 7, 8, 0, 4, 5, 9, 0, 5, 3, 2, 5, 10, 1, 8, 1, 10, 3, 1, 8, 8, 10, 1, 3, 5, 7, 5, 0, 9, 11, 7, 11, 8, 3, 2, 2, 6, 8, 10, 11, 2, 4, 0, 3, 4, 5, 6, 9, 6, 10, 1, 9, 0, 11, 9, 4, 5, 5, 6, 7, 4, 9, 11, 6, 0, 4, 3, 4, 9, 5, 11

Offset: 0

N. J. A. Sloane, Jul 02 2015

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Eric M. Schmidt, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179.

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

def a_list(n) : return crt(0, 1, 3^n, 4^n).digits(12) # Eric M. Schmidt, Jul 09 2015

More terms from Eric M. Schmidt, Jul 09 2015

A054869 Digits of an idempotent 6-adic number.

Original entry on oeis.org

3, 1, 2, 0, 5, 3, 1, 2, 2, 2, 5, 1, 5, 5, 1, 4, 1, 3, 1, 2, 5, 5, 5, 0, 5, 2, 5, 5, 5, 3, 1, 4, 3, 3, 0, 4, 2, 2, 4, 0, 1, 3, 3, 1, 4, 0, 2, 0, 1, 2, 5, 2, 4, 0, 2, 3, 3, 0, 3, 4, 5, 5, 2, 5, 5, 4, 3, 2, 3, 1, 5, 4, 5, 4, 0, 1, 1, 0, 4, 2, 0, 1, 3, 0, 1, 5, 0, 4, 3, 5, 0, 1, 0, 2, 4, 0, 3, 4, 2

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

( a(0) + a(1)*6 + a(2)*6^2 + ... )^k = a(0) + a(1)*6 + a(2)*6^2 + ... for each k. Apart from 0 and 1, in base 6 there are only 2 numbers with this property. For the other see A055620.

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Kenny Lau, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

n=10000;res=1-pow((3**n+1)//2,n,3**n)*2**n
for i in range(n):print(i,res%6);res//=6
# Kenny Lau, Jun 09 2018

a(n) == 3^(2^n) (mod 6^n). - Robert Dawson, Oct 28 2022

A216092 a(n) = 2^(2*5^(n-1)) mod 10^n.

Original entry on oeis.org

4, 24, 624, 624, 90624, 890624, 2890624, 12890624, 212890624, 8212890624, 18212890624, 918212890624, 9918212890624, 59918212890624, 259918212890624, 6259918212890624, 56259918212890624, 256259918212890624

Offset: 1

V. Raman, Sep 01 2012

nonn

a(n) is the unique positive integer less than 10^n such that a(n) is divisible by 2^n and a(n) + 1 is divisible by 5^n. - Eric M. Schmidt, Sep 01 2012

Robert Israel, Table of n, a(n) for n = 1..996

Cf. A007185, A016090, A216093, A091664, A018247.

f:= n -> 2&^(2*5^(n-1)) mod 10^n:
map(f, [$1..100]); # Robert Israel, Mar 13 2025

Table[PowerMod[5,2^n,10^n],{n,20}]-1 (* Harvey P. Dale, Dec 17 2017 *)

def A216092(n) : return crt(0, -1, 2^n, 5^n) # Eric M. Schmidt, Sep 01 2012

a(n) = (5^(2^n) mod 10^n) - 1.
a(n)^3 == a(n) (mod 10^n).
a(n-1) == a(n) (mod 10^(n-1)). - Robert Israel, Mar 13 2025

A376842 Asymptotic phase shift of the tetration base n written by juxtaposing its representative congruence classes modulo 10 (i.e., the 1, 2, or 4, distinct sfasamenti taken by starting from the minimum height such that the congruence speed of n is constant), and a(n) = -1 if n is a multiple of 10.

Original entry on oeis.org

8, 46, 6248, 5, 4268, 2684, 6842, 2, -1, 4, 46, 6248, 8, 5, 64, 4, 28, 4862, -1, 6248, 6248, 2486, 8, 5, 46, 2684, 4862, 6842, -1, 8426, 8426, 28, 28, 5, 4, 2684, 28, 82, -1, 64, 4268, 46, 4268, 5, 4862, 4, 2, 6842, -1, 5, 6, 8, 6248, 5, 8426, 2684, 4268, 2

Offset: 2

Marco Ripà, Oct 06 2024

By Definition 1.2 of “Number of stable digits of any integer tetration” (see Links), let bar_b be the smallest hyperexponent of the tetration m^^b such that the congruence speed of m is constant.
Then, assuming that n is not a multiple of 10, we define as asymptotic phase shift (original name "sfasamento asintotico” - see References, “La strana coda della serie n^n^...^n”, Chapter 7) the subsequence of the 4 (or 2 or even 1) congruence classes of the differences between the rightmost non-stable digits of n^^(bar_b) (i.e., the least significant digit of the tetration n^^(bar_b) that is not frozen as we move to n^^(b + 1)) and the corresponding digit of n^^(bar_b + 1), and ditto for (n^^(bar_b + 1) and n^^(bar_b + 2)), (n^^(bar_b + 2) and n^^(bar_b + 3)), (n^^(bar_b + 3) and n^^(bar_b + 4)).
Now, let us indicate this subsequence as [s_1, s_2, s_3, s_4] and then, if s_1 = s_3 and also s_2 = s_4, we transform [s_1, s_2, s_3, s_4] into [s_1, s_2], and lastly, if s_1 = s_2, we transform [s_1, s_2] into [s_1].
Finally, we get a(n) by juxtaposing all the 4 (or 2, or 1) digits inside [...] (e.g., if n = 3, then bar_b = 2 so that we get s_1 = 4 = s_3 and s_2 = 6 = s_4, and consequently a(3) = 46 instead of 4646 - see Definition 3.2 of "Graham's number stable digits: an exact solution" in Links).
Assuming that n is not a multiple of 10, in general, for any given pair of nonnegative integers c and k, the congruence class modulo 10 of the difference between the least significant digits of n^^(bar_b+c+4*k) and n^^(bar_b+c+1+4*k) does not depend on k. Moreover, if s_1 <> s_2, then s_1 + s_3 = s_2 + s_4 = 10.
In detail, if the last digit of n is 5, then a(n) = 5, while if n is coprime to 5, a(n) mandatorily belongs to the following set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 28, 37, 46, 64, 73, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.
The author conjectures that if n is not a multiple of 10, then a(n) is necessarily an element of the subset {2, 4, 5, 6, 8, 9, 19, 28, 46, 64, 82, 2486, 2684, 3971, 4268, 4862, 6248, 6842, 7931, 8426, 8624}.
As a mere consequence of a(3) = 46 and the fact that congruence speed of 3 is 0 at height 1 and 1 at any height above 1, it follows that 4 is equal to the difference between the slog_3(G)-th least significant digit of Graham's number, G, and the slog_3(G)-th least significant digit of any power tower of the form 3^3^3^... whose height is above slog_3(G) (where slog(...) indicates the super-logarithm of the argument).
By definition, a(n) consists of a circular permutation of the digits of A376446(n).

			a(3) = 46 since the congruence speed of 3^^b becomes constant starting from height 2 and its value is 0 for b = 1 and 1 for any b >= 2, then (3^3 - 3^(3^3))/10 == 4 (mod 10) while (3^(3^3) - 3^(3^(3^3)))/10^2 == 6 (mod 10).

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, Pages 245-260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.
Marco Ripà, Graham's number stable digits: an exact solution, arXiv:2411.00015 [math.GM], 2024.
Marco Ripà, Twelve Python Programs to Help Readers Test Peculiar Properties of Integer Tetration, ResearchGate, 2024. See p. 22.
Wikipedia, Graham's Number.
Wikipedia, Tetration.

Cf. A317905, A370211, A372490, A373387, A376446.

Cf. A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.

A376446 Modular phase shift of the tetration base n written by juxtaposing its representative congruence classes modulo 10 taken by starting from height 2 + v(n), and a(n) = -1 if n is a multiple of 10.

Original entry on oeis.org

8, 64, 2486, 5, 4268, 8426, 8426, 2, -1, 4, 64, 4862, 8, 5, 46, 4, 82, 6248, -1, 4862, 6248, 4862, 8, 5, 64, 6842, 8624, 4268, -1, 4268, 2684, 28, 82, 5, 4, 8426, 28, 82, -1, 46, 4268, 46, 2684, 5, 4862, 4, 2, 2684, -1, 5, 6, 8, 2486, 5, 4268, 2684, 4268, 2

Offset: 2

Marco Ripà, Sep 23 2024

Let b := 2 + v(n), where v(n) is equal to u_5(n - 1) iff n == 1 (mod 5), u_5(n^2 + 1) iff n == 2,3 (mod 5), u_5(n + 1) iff n == 4 (mod 5), u_2(n^2 - 1) - 1 iff n == 5 (mod 10), while u_5 and u_2 indicate the 5-adic and the 2-adic valuation of the argument (respectively).
Assuming that n is not a multiple of 10, we define as modular phase shift (original name "sfasamento modulare" - see References, “La strana coda della serie n^n^...^n”, Chapter 7) the subsequence of the 4 (or 2 or even 1) congruence classes of the differences between the rightmost non-stable digits of n^^b (i.e., the least significant digit of the tetration n^^b that is not frozen as we move to n^^(b + 1)) and the corresponding digit of n^^(b + 1), and ditto for (n^^(b + 1) and n^^(b + 2)), (n^^(b + 2) and n^^(b + 3)), (n^^(b + 3) and n^^(b + 4)).
Now, let us indicate this subsequence as [s_1, s_2, s_3, s_4] and then, if s_1 = s_3 and also s_2 = s_4, we transform [s_1, s_2, s_3, s_4] into [s_1, s_2], and lastly, if s_1 = s_2, we transform [s_1, s_2] into [s_1].
Finally, we get a(n) by juxtaposing all the 4 (or 2, or 1) digits inside [...] (e.g., if n = 3, then b = 2 + 1 so that we get s_1 = 6 = s_3 and s_2 = 4 = s_4, and consequently a(3) = 64 instead of 6464).
Assuming that n is not a multiple of 10, in general, for any given pair of nonnegative integers c and k, the congruence class modulo 10 of the difference between the least significant digits of n^^(b+c+4*k) and n^^(b+c+1+4*k) does not depend on k. Moreover, if s_1 <> s_2, then s_1 + s_3 = s_2 + s_4 = 10.
In detail, if the last digit of n is 5, then a(n) = 5, while if n is coprime to 5, a(n) mandatorily belongs to the following set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 28, 37, 46, 64, 73, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.
The author conjectures that if n is not a multiple of 10, then a(n) is necessarily an element of the subset {2, 4, 5, 6, 8, 9, 19, 28, 46, 64, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.
As a mere consequence of a(3) = 64 and the fact that congruence speed of 3 is 0 at height 1 and 1 at any height above 1, it follows that 4 is equal to the difference between the slog_3(G)-th least significant digit of Graham's number, G, and the slog_3(G)-th least significant digit of any power tower of the form 3^3^3^... whose height is above slog_3(G) (where slog(...) indicates the super-logarithm of the argument).

			a(3) = 64 since the congruence speed of 3 at height 3^^(2 + u_5(3^2 + 1)) is constant and its value is 1, so (3^(3^3) - 3^(3^(3^3)))/10^2 == 6 (mod 10) while (3^(3^(3^3)) - 3^(3^(3^(3^3))))/10^3 == 4 (mod 10).

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, Pages 245—260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43—61.
Marco Ripà, Twelve Python Programs to Help Readers Test Peculiar Properties of Integer Tetration, ResearchGate, 2024. See pp. 7, 9, 12, 14, 27.
Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441—457.
Wikipedia, Graham's Number.
Wikipedia, Tetration.

Cf. A317905, A370211, A372490, A373387.

Cf. A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.

# Function to calculate the p-adic valuation
def p_adic_valuation(n, p):
    count = 0
    while n % p == 0 and n != 0:
        n //= p
        count += 1
    return count
# Function to calculate tetration (tower of powers)
def tetration(base, height, last_digits=500):
    results = [base]
    for n in range(1, height):
        result = pow(base, results[-1], 10**last_digits)  # Only the last last_digits digits
        results.append(result)
    return results
# Function to find the first non-zero difference and compute modulo 10
def find_difference_mod_10(tetrations):
    differences = []
    for n in range(len(tetrations) - 1):
        string_n = str(tetrations[n]).zfill(500)  # Pad with zeros if needed
        string_n_plus_1 = str(tetrations[n+1]).zfill(500)
        # Find the first difference starting from the rightmost digit
        for i in range(499, -1, -1):  # From right to left
            if string_n[i] != string_n_plus_1[i]:
                difference = (int(string_n[i]) - int(string_n_plus_1[i])) % 10
                differences.append(difference)
                break
    return differences
# Function to determine the first hyperexponent based on modulo 5 congruences
def calculate_initial_exponent(a):
    mod_5 = a % 5
    if mod_5 == 1:
        valuation = p_adic_valuation(a - 1, 5)
        initial_exponent = valuation + 2
    elif mod_5 in [2, 3]:
        valuation = p_adic_valuation(a**2 + 1, 5)
        initial_exponent = valuation + 2
    elif mod_5 == 4:
        valuation = p_adic_valuation(a + 1, 5)
        initial_exponent = valuation + 2
    else:
        valuation = p_adic_valuation(a**2 - 1, 2)
        initial_exponent = valuation + 1
    return initial_exponent
# Main logic
try:
    a = int(input("Enter a positive integer greater than 1: "))
    # Check if the number ends with 0
    if a % 10 == 0:
        print(-1)
    elif a <= 1:
        raise ValueError("The number must be greater than 1.")
    else:
        # Calculate the initial exponent based on modulo 5 congruence
        initial_exponent = calculate_initial_exponent(a)
        # Generate tetrations for 30 iterations and the last 500 digits
        tetrations = tetration(a, 30, last_digits=500)
        # Find the modulo 10 differences for the 4 required iterations
        mod_10_differences = find_difference_mod_10(tetrations[initial_exponent-1:initial_exponent+4])
        # Optimization of the output
        if mod_10_differences[:2] == mod_10_differences[2:]:
            mod_10_differences = mod_10_differences[:2]
        if len(set(mod_10_differences)) == 1:
            mod_10_differences = [mod_10_differences[0]]
        # Convert the list of differences into a string without brackets or commas
        result_str = ''.join(map(str, mod_10_differences))
        # Print the optimized result
        print(f"a({a}) = {result_str}")
except Exception as e:
    print(f"ERROR!\n{e}")

A224477 (5^(2^n) + (10^n)/2) mod 10^n: a sequence of trimorphic numbers ending (for n > 1) in 5.

Original entry on oeis.org

0, 75, 125, 5625, 40625, 390625, 7890625, 62890625, 712890625, 3212890625, 68212890625, 418212890625, 4918212890625, 9918212890625, 759918212890625, 1259918212890625, 6259918212890625, 756259918212890625, 7256259918212890625, 42256259918212890625

Offset: 1

Eric M. Schmidt, Apr 07 2013

a(n) is the unique nonnegative integer less than 10^n such that a(n) + 2^(n-1) - 1 is divisible by 2^n and a(n) is divisible by 5^n.

Eric M. Schmidt, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Trimorphic Number
Index entries for sequences related to automorphic numbers

Cf. A033819. Converges to the 10-adic number A018247. The other trimorphic numbers ending in 5 are included in A007185, A216093, and A224478.

def A224477(n) : return crt(2^(n-1)+1, 0, 2^n, 5^n)

a(n) = (A007185(n) + 10^n/2) mod 10^n.

A067270 Numbers m such that m-th triangular number (A000217) ends in m.

Original entry on oeis.org

0, 1, 5, 25, 625, 9376, 90625, 890625, 7109376, 12890625, 212890625, 1787109376, 81787109376, 59918212890625, 259918212890625, 3740081787109376, 56259918212890625, 256259918212890625, 7743740081787109376

Offset: 1

Joseph L. Pe, Feb 21 2002

Thanks to David W. Wilson for the proof that this sequence is a proper subset of A003226.
Also, numbers m such that the m-th k-gonal number ends in m for k == 1, 3, 5, or 9 (mod 10). - Robert Dawson, Jul 09 2018
This sequence is the intersection of A093534 and A301912. - Robert Dawson, Aug 01 2018

			The 5th triangular = 15 ends in 5, hence 5 is a term of the sequence.

Chai Wah Wu, Table of n, a(n) for n = 1..888
Robert Dawson, On Some Sequences Related to Sums of Powers, J. Int. Seq., Vol. 21 (2018), Article 18.7.6.

Proper subset of A003226. Cf. A007185, A018247, A016090, A018248.

Intersection of A093534 and A301912.

(* a5=A018247 less the commas; a6=A018248 less the commas; *)
b5 = FromDigits[ Reverse[ IntegerDigits[a5]]]; b6 = FromDigits[ Reverse[ IntegerDigits[a6]]]; f[0] = 1; f[n_] := Block[{c5 = Mod[b5, 10^n], c6 = Mod[b6, 10^n]}, If[ Mod[c5(c5 + 1)/2, 10^n] == c5, c5, c6]]; Union[ Table[ f[n], {n, 0, 20}]]

from itertools import count, islice
from sympy.ntheory.modular import crt
def A067270_gen(): # generator of terms
    a = 0
    yield from (0,1)
    for n in count(0):
        if (b := int(min(crt(m:=(1<<(n+1),5**n),(0,1))[0], crt(m,(1,0))[0]))) > a:
            yield b
            a = b
A067270_list = list(islice(A067270_gen(),15)) # Chai Wah Wu, Jul 25 2022

Edited and extended by Robert G. Wilson v, Nov 20 2002

0 prepended by David A. Corneth, Aug 02 2018

cp's OEIS Frontend

A318140 The 10-adic integer f = ...6510474330 satisfying f^2 + 1 = a, a^2 + 1 = b, b^2 + 1 = c, c^2 + 1 = d, d^2 + 1 = e, and e^2 + 1 = f.

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A055620 Digits of an idempotent 6-adic number.

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PARI

Python

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A259468 Digits of an idempotent 12-adic number.

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Sage

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A259469 Digits of an idempotent 12-adic number.

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Sage

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A054869 Digits of an idempotent 6-adic number.

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Python

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A216092 a(n) = 2^(2*5^(n-1)) mod 10^n.

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Maple

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A376446 Modular phase shift of the tetration base n written by juxtaposing its representative congruence classes modulo 10 taken by starting from height 2 + v(n), and a(n) = -1 if n is a multiple of 10.

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