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A369826 Number of the rightmost decimal digits of n^n that are the same as those of n^(n^n). 9
1, 0, 1, 0, 1, 1, 4, 3, 2, 1, 2, 10, 3, 1, 2, 1, 9, 3, 2, 0, 2, 20, 3, 0, 1, 2, 9, 6, 1, 1, 2, 30, 3, 2, 2, 1, 5, 3, 2, 0, 2, 40, 3, 0, 2, 1, 6, 3, 1, 1, 4, 50, 5, 1, 2, 1, 7, 3, 6, 0, 2, 60, 3, 0, 1, 1, 18, 3, 1, 3, 2, 70, 3, 1, 2, 2, 5, 6, 2, 0, 2, 80, 3, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET
-1,7
COMMENTS
The common digits might include leading 0's (such as at n = 43) and those 0's are included in the count.
Common digits do not extend beyond the most significant digit of n^n, so that for instance a(5) = 4 stops at all digits of n^n = 3125.
a(-1) = 1 since (-1)^((-1)^(-1)) = (-1)^(-1) = -1 (which is a one-digit number);
a(0) = 0 since 0^0 = 1 and consequently 0^(0^0) = 0 and they have no digits in common.
a(n) = 0 if and only if n=0 or n == 2 or 18 (mod 20), see Links for details.
For n = k*10^c with c >= 1 and k != 0 (mod 10), a(n) = c*n which is the rightmost 0's of n^n.
From Jianing Song, Feb 13 2024: (Start)
For n >= 3, n^(n^n) - n^n is divisible by 32: let vp(N) denote the p-adic valuation of N and write n^(n^n) - n^n = n^n * (n^(n*(n^(n-1)-1)) - 1). For even n >= 4, we have v2(n^(n^n) - n^n) = v2(n^n) = n*v2(n) >= 5. For odd n, we have v2(n^(n^n) - n^n) = v2(n^(n*(n^(n-1)-1)) - 1) is 3*v2(n-1) + 2*v2(n+1) - 2 >= 5, with the equality holding if and only if v2(n-1) = 1 and v2(n+1) = 2, i.e., n == 3 (mod 8). As a result, for most n, the number of common digits is completely determined by the 5-adic valuation of n^(n^n) - n^n. For example, for n > 1, a(n) = 1 if and only if v5(n^4-1) = 1 and n == 3, 4, 7, 8, 12, 14 (mod 20), namely n being congruent to 3, 4, 8, 12, 14, 23, 27, 28, 34, 44, 47, 48, 52, 54, 63, 64, 67, 72, 83, 84, 87, 88, 92, 94 modulo 100.
Every number appears infinitely many times in this sequence: for a given m >= 1, let n be a solution to n == 0 (mod 2^max{m,2}) and n == -1 (mod 5^m), then n^(n^n) - n^n is divisible by 2^m. On the other hand, n*(n^(n-1)-1) is divisible by 4 but not by 5, so v5(n^(n^n) - n^n) = v5(n^(n*(n^(n-1)-1)) - 1) = v5(n^4-1) + v5(n*(n^(n-1)-1)) = v5(n^4-1) = m. (End)
LINKS
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, Tetration.
FORMULA
For any n > 1 and n <> 5, a(n) is such that n^n == n^(n^n) (mod 10^(a(n))) and n^n !== n^(n^n) (mod 10^(a(n)+1)).
EXAMPLE
For n=5, a(n)=4 since 5^5 = 3125 and 5^(5^5) == 203125 (mod 10^6).
PROG
(PARI)
\\ As n^(n^n) - n^n = n^n * (n^(n*(n^(n-1)-1)) - 1), it suffices to calculate the 2-adic and the 5-adic valuations of n^(n*(n^(n-1)-1)) - 1.
a(n) = {if(n<=1, abs(n), if(n==5, 4, my(coeff = [0, 3, 0, 1, 1, 0, 3, 1, 1, 2, 0, 3, 1, 2, 1, 0, 3, 2, 0, 2], v2, v5);
v2 = if(n%2, 3*valuation(n-1, 2) + 2*valuation(n+1, 2) - 2, n*valuation(n, 2));
v5 = n*valuation(n, 5) + coeff[1+n%20]*valuation(n^4-1, 5);
min(v2, v5); ))
} \\ Jianing Song, Feb 13 2024
CROSSREFS
Cf. A002488, A317905, A349425, A369771 (n^^3 and n^^4).
Sequence in context: A284801 A090284 A369110 * A093580 A196535 A106655
KEYWORD
nonn,base,easy
AUTHOR
Marco Ripà, Feb 02 2024
STATUS
approved

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Last modified June 13 22:21 EDT 2024. Contains 373391 sequences. (Running on oeis4.)